If electric field lines are close together versus farther apart, the strength of the electric field is greater.
The force on a 0.0050 C charge in an electric field of 300 V/m is 1.5 N.
The strength of an electric field is 1.1 × 10⁴ N/C.
What is field line?A graphic tool for viewing vector fields is a field line. It is made up of a fictitious integral curve that is perpendicular to the field vector over its entire length.
The force on the charge in the electric field is = 300×0.0050 N = 1.5 N.
The strength of an electric field is = (9 ×10⁹×3.0×10⁻⁹)/(0.05)² N/C = 1.1 × 10⁴ N/C.
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Answer:
1. The strength of the electric field is greater.
2. 1.5 N
3. 1.1 × 104 N/C
(for honors ppl)
1. They represent non-contact electrical forces.
2. The strength of the electric field is greater.
3. 1.5 N
4. 1.1 × 104 N/C
5. approximately 1.45 V/m directed away from the positive charge
6. It increases.
What is the maximum speed with which a 1200- kg car can round a turn of radius 90.0 m on a flat road if the coefficient of static friction between tires and road is 0.55?
The maximum speed with which a 1200- kg car can make a turn is 22.02 m/s.
Maximum speed of the carv(max) = √μrg
where;
μ is coefficient of frictionr is radius of the curved roadg is acceleration due to gravityv(max) = √(0.55 x 90 x 9.8)
v(max) = 22.02 m/s
Thus, the maximum speed with which a 1200- kg car can make a turn is 22.02 m/s.
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What affects fuel consumption in automobiles?
A. Drag
B. Nothing
C. Air resistance
D. Time of day
Answer:
A and C
Explanation:
drag (the area of lower air pressure behind the car when moving) and mostly air resistance (the work to push the air in front of us away to move through - the faster we go, the stronger the air resists to move aside).
A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?
The vertical component of the force being applied to the beam by the high-tension cable is 36.37 N.
Calculation of the cable's tension-
Utilizing the moment principle, determine the cable's tension:
Torque clockwise = TL sin∅
Counterclockwise torque = 1/2WL
TL sin∅ = 1/2WL
⇒T sin∅ = 1/2W
⇒T = W/2sin∅
⇒T = (29* 9.8)/ (2*sin57)
⇒T = 169.43 N
Calculation of the vertical component of the force-
Forces that operate perpendicular to the surface vertical plane are called the vertical force. Gravity always pulls objects straight down to the earth's core.
T+F = W
⇒F = W-T
⇒F = (21*9.8)-169.43
⇒F = 36.37 N
So, the force on the beam has a vertical component of 36.37 N.
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A golf ball is driven from a point A with a speed of 40 m/s at an angle of elevation of 30°. On its downward flight, the ball hits an advertising hoarding at a height 15.1 m above the level of A, as shown in the diagram below. Find A. The time taken by the ball to reach its greatest height above A, B. The time taken by the ball to travel from A to B, C. The speed with which the ball hits the hoarding. 20
(a) The time taken by the ball to reach its greatest height is 2 seconds.
(b) The time taken by the ball to travel from A to B, C is 4.08 m/s.
(c) The speed with which the ball hits the hoarding is 69.3 m/s.
Maximum height of the projectile
H = u²sin²θ/2g
H = (40² x (sin 30)²) / (2 x 9.8)
H = 20.4 m
Time for the ball to travel 20.4 mh = ut - ¹/₂gt²
20.4 = (40 x sin 30)t - (0.5)(9.8)t²
20.4 = 20t - 4.9t²
4.9t² - 20t + 20.4 = 0
solve the quadratic equation using formula method;
t = 2 seconds
Time taken for the ball to travel A, B , CThis is the time of motion of the ball;
T = 2usinθ/g
T = (2 x 40 sin 30)/9.8
T = 4.08 s
Speed of the ball when it hits the hoardingv = u + gt
vy = 40 sin30 + (9.8 x 4.08)
vy = 59.98 ms
vx = 40 x cos30
vx = 34.64 m/s
vf = √(vy² + vx²)
vf = √(59.98² + 34.64²)
vf = 69.3 m/s
Thus, the time taken by the ball to reach its greatest height is 2 seconds.
The time taken by the ball to travel from A to B, C is 4.08 m/s.
The speed with which the ball hits the hoarding is 69.3 m/s.
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A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .
a) What is the magnitude of the magnetic field at the center of the coil?
b) What is the magnitude of the magnetic field at a point on the axis of the coil a distance of 9.50 cm from its center?
Hi there!
a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}[/tex]
dB = Differential Magnetic field element
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
R = radius of loop (2.15 cm = 0.0215 m)
i = Current in loop (0.460 A)
For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}[/tex]
Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
[tex]B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}[/tex]
Taking out constants from the integral:
[tex]B =\frac{\mu_0 i}{4\pi R^2} \int ds[/tex]
Since we are integrating around an entire circle, we are integrating from 0 to 2π.
[tex]B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds[/tex]
Evaluate:
[tex]B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}[/tex]
Plugging in our givens to solve for the magnetic field strength of one loop:
[tex]B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T[/tex]
Multiply by the number of loops to find the total magnetic field:
[tex]B_T = N B = 0.00631 = \boxed{6.318 mT}[/tex]
b)
Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}[/tex]
In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}[/tex]
Using the diagram, if 'z' is the point's height from the center:
[tex]r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}[/tex]
Substituting this into our expression:
[tex]dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }[/tex]
Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
[tex]B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds[/tex]
Evaluate:
[tex]B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}[/tex]
Multiplying by the number of loops:
[tex]B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}[/tex]
Plug in the given values:
[tex]B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ = 0.00006795 = \boxed{67.952 \mu T}[/tex]
Two separate but nearby coils are mounted along the same axis. A power supply controls the flow of current in the first coil, and thus the magnetic field it produces. The second coil is connected only to an ammeter. The ammeter will indicate that a current is flowing in the second coil:.
The ammeter will indicate that a current is flowing in the second coil only when current changes in the first coil.
Induced emf
An emf is induced in a coil placed in a magnetic field when a current carrying conductor moves in the field.
emf = NdФ/dt
where;
dФ is change in flux of the fieldN is number of turnsdt is change in timeThus, the ammeter will indicate that a current is flowing in the second coil only when current changes in the first coil.
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Determine the total electric potential energy for the charge distribution with three chargers in a straight line
The total electric potential energy is [tex]\frac{kq_{1} q_{3} }{r_{13} } + \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1} q_{2} }{r_{12} }[/tex].
Electric Potential Energy of a System of Charges :
The system's electric potential energy is equal to the amount of work necessary to create a system of charges by guiding them toward their designated locations from infinity against the electrostatic force without accelerating them. The symbol for it is U.U=W=qV. Electrostatic fields are conservative, therefore the work is independent of the path.
Assume three charges q₁ , q₂ and q₃ bring from infinity to point P.
To bring q₁ no work is done,
[tex]V_{p} = \frac{kq_{1} }{r_{1} }[/tex]
where, V = electric potential energy.
q = point charge.
r = distance between any point around the charge to the point charge.
k = Coulomb constant; k = 9.0 × 109 N.
Now bring q₂,
[tex]V_{2} = \frac{kq_{2} }{r_{2} }[/tex]
Work done by q₁ ;
[tex]W_{1} = V_{p} q_{2} = \frac{kq_{1}q_{2} }{r_{12} }[/tex]
Now bring q₃,
[tex]V_{3} = \frac{kq_{3} }{r_{3} }[/tex]
Work done on q₃ by q₁ and q₂
[tex]W= q_{3} [ V_{1} + V_{2} ][/tex]
[tex]=\frac{kq_{1} q_{3} }{r_{13} }[/tex][tex]+ \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1}q_{3} }{r_{12} }[/tex]
This work done is stored in the form of potential energy.
∴U=W= potential energy of three systems.
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The total electric potential energy is [tex]\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}[/tex]
Electric Potential Energy of a System of Charges
The labor required to establish a system of charges by guiding them toward their intended positions from infinity against the electrostatic force without accelerating them is equal to the electric potential energy of the system. Its identifier is U.U=W=qV. Due to the conservatism of electrostatic fields, the work is independent of the path.
Consider having three charges. Q1, Q2, and Q3 bring point P from infinity.
No work has been done to bring q1,
[tex]V_{1} = \frac{kq1}{r1}[/tex]
where, V = electric potential energy.
q = point charge.
r = distance between any point around the charge to the point charge.
k = Coulomb constant; k = 9.0 × 109 N.
Now bring q₂,
[tex]V_{2} = \frac{kq2}{r2}[/tex]
Work done by q₁ ;
W1 = [tex]V_{p} q2[/tex] = [tex]\frac{kq1q2}{r12}[/tex]
Now bring q₃,
[tex]V_{3} = \frac{kq3}{r3}[/tex]
Work done on q₃ by q₁ and q₂
W= q3{[tex]V_{1} + V_{2}[/tex]}
W = [tex]\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}[/tex]
This work done is stored in the form of potential energy.
∴U=W= potential energy of three systems.
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6. A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. (3pt) a) How long are her feet in the air? b) What is her highest point above the board? c) What is her velocity when her feet hit the water?
The results of the calculation are;
a) The feet spends 0.41 s in air
b) The highest point above board is 2.62 m
c) The velocity when her feet hit the water is 7.2 m/s
What is the time spent in air?From the data presented;
v = u + at
But v = 0 m/s at the maximum height thus;
0 = 4 - (9.8 * t)
4 = 9.8 * t
t = 4/9.8
t = 0.41 s
b) from;
h = ut - 1/2gt^2
h = (4 * 0.41) - (0.5 * 9.8 * (0.41)^2)
h = 1.64 - 0.82
h = 0.82 m
The total height above board = 0.82 m + 1.8 m = 2.62 m
c) The total time in air is obtained from;
h = ut + 1/2gt^2
u = 0m/s because she dropped off the board
h = 1/2gt^2
2.62 = 0.5 * 9.8 * t^2
t = √2.62/0.5 * 9.8
t = 0.73 seconds
Hence, the velocity when her feet hit the water is obtained from;
v = u + gt
when u = 0 m/s
v = gt
v = 9.8 * 0.73 s
v = 7.2 m/s
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The Venn diagram compares protons with electrons. Which shared property belongs in the region marked "B”?
The shared property of electron and proton that belongs in the region marked "B” is the presence of charge on both of them.
What charge is present on proton and electron?We know that proton is positively charge particle while on the other hand, electron is negatively charge particle. Due to opposite charges, these particles attract each other.
So we can conclude that the shared property of electron and proton that belongs in the region marked "B” is the presence of charge on both of them.
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Answer:
electrically charged
Explanation:
got it right
A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?
26.10 N is the vertical component of the force.
Rx represents the Horizontal component of force
Ry represents The Vertical component of force
According to the given diagram
Rx - Tcosθ = 0
Rx = Tcosθ
And,
Ry + Tsinθ = mg
Ry = mg - Tsinθ
The horizontal component of force =The Vertical component of force
Rx = Ry
Tcosθ = mg - Tsinθ
T(cosθ + sinθ) = 29 × 9.8 = 284.2 N
T√2 cosθ = 284.2 N
T × √2 ×0.544 = 284.2 N
T × 0.769 = 284.2 N
T = 370 N (app)
So,
Ry = 284.2 - 370 (sin 57°)
= 284.2 - 310.3 = -26.10 N
Hence, 26.10 N is the vertical component of the force exerted.
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A sled is initially given a shove up a frictionless 37.0 ∘ incline. It reaches a maximum vertical height 1.20 m higher than where it started at the bottom.
What was its initial speed?
Express your answer to three significant figures and include the appropriate units.
The initial speed of the sled at the given height is 4.85 m/s.
Initial speed of the sledApply the principle of conservation of energy;
K.E = P.E
¹/₂mv² = mgh
v² = 2gh
v = √2gh
where;
h is the vertical height reachedg is acceleration due to gravityv = √(2 x 9.8 x 1.2)
v = 4.85 m/s
Thus, the initial speed of the sled at the given height is 4.85 m/s.
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A crate with a mass of M = 62.5 kg is suspended by a rope from the endpoint of a uniform boom. The boom has a mass of m = 116 kg and a length of l = 7.65 m. The midpoint of the boom is supported by another rope which is horizontal and is attached to the wall as shown in the figure.
1. The boom makes an angle of θ = 57.7° with the vertical wall. Calculate the tension in the vertical rope.
2. What is the tension in the horizontal rope?
Answer:
Tension in the vertical rope: approximately [tex]613 \; {\rm N}[/tex].
Tension in the horizontal rope: approximately [tex]3.74 \times 10^{3}\; {\rm N}[/tex].
Assumption: [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].
Explanation:
Since the system is not moving, the tension in the vertical rope would be equal to the weight of the crate:
[tex]\begin{aligned}\text{weight of crate} &= m\, g \\ &= 62.5\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \\ &= 613.25 \; {\rm N}\end{aligned}[/tex].
Note that [tex]\theta = 57.7^{\circ}[/tex] is the angle between the beam (the lever) and the vertical rope. The torque that this vertical rope exert on the beam would be:
[tex]\begin{aligned} \tau &= r\, F\, \sin(\theta) \\ &=(7.65\; {\rm m}) \, (613.25\; {\rm N})\, (\sin(57.7^{\circ})) \\ &\approx 3.965 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].
This torque is in the clockwise direction.
The weight of the beam ([tex]m = 116\; {\rm kg}[/tex]) would be:
[tex]\begin{aligned}\text{weight of beam} &= m\, g \\ &= 116 \; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \\ &= 1.138 \times 10^{3}\; {\rm N}\end{aligned}[/tex].
Note that [tex]\theta = 57.7^{\circ}[/tex] is also the angle between the beam and the direction of the (downward) gravitational pull on this. Since this beam is uniform, it would appear as if the weight of this beam is applied at the center of this beam (with a distance of [tex](7.65\; {\rm m}) / 2[/tex] from the pivot.) Thus, the torque gravitational pull exerts on this beam would be:
[tex]\begin{aligned} \tau &= r\, F\, \sin(\theta) \\ &= \genfrac{(}{)}{}{}{7.65\; {\rm m}}{2} \, (1.138 \times 10^{3}\; {\rm N})\, (\sin(57.7^{\circ})) \\ &\approx 3.679 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].
This torque is also in the clockwise direction.
The tension in the horizontal rope would need to supply a torque in the counterclockwise direction. The magnitude of that torque would be approximately:
[tex]\begin{aligned} & 3.965\times 10^{3}\; {\rm N \cdot m} + 3.679\times 10^{3}\; {\rm N \cdot m} \\ \approx \; & 7.645 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].
Note the angle between the direction of this tension and the beam is [tex](90^{\circ} - \theta) = 32.3^{\circ}[/tex]. This force is applied [tex](7.65\; {\rm m}) / 2[/tex] from the pivot. Hence, achieving that torque of [tex]7.645 \times 10^{3}\; {\rm N \cdot m}[/tex] would require:
[tex]\begin{aligned} F &= \frac{\tau}{r\, \sin(90^{\circ} - \theta)} \\ &\approx \frac{7.645\times 10^{3}\; {\rm N \cdot m}}{((7.65\; {\rm m}) / 2) \times \sin(32.3^{\circ})} \\ &\approx 3.74 \times 10^{3}\; {\rm N} \end{aligned}[/tex].
During long jump athlete runs before taking the jump by doing so he
A:provides himself a larger inertia
B:decreases his inertia
C:decreases his momentum
D:decreases his K.E
A. During long jump athlete runs before taking the jump by doing so he provides himself a larger inertia.
What is inertia?
Inertia is the reluctance of an object to stop moving once in motion or start moving when it is at rest.
When an athlete runs before taking the jump, he is trying to increase his inertia, that is his reluctance to stop, thereby increasing his forward motion or jump.
Thus, during long jump athlete runs before taking the jump by doing so he provides himself a larger inertia.
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2. Find the value of A in the unit vector 0.4î+ 0.8ĵ+ λk.
There is no A in the vector, so I assume you mean λ.
The magnitude of any unit vector is 1, so
[tex]\|0.4\,\vec\imath + 0.8\,\vec\jmath + \lambda \,\vec k\| = \sqrt{0.4^2 + 0.8^2 + \lambda^2} = 1[/tex]
Square both sides and solve.
[tex]0.4^2 + 0.8^2 + \lambda^2 = 1^2 \implies \lambda = \boxed{\pm \sqrt{0.2}}[/tex]
K donates, or transfers, one electron to bromine, which has 7 electrons. Both K and Br are now stable with 8 electrons. K becomes a positive ion and Br becomes a negative ion. The positive K ion and the negative Br ion attract each other to form an ionic bond.
K is cation by losing of electron whereas Br is anion due to accepting of electrons.
Is charge appears when an atom lose or accept electron?Yes, the positive ion appears on K and become cation whereas the negative ion bears on Br which make it anion because of losing and gaining of electron by these atoms. This transferring of electrons leads to formation of ionic bonds between them.
So we can conclude that K is cation by losing of electron whereas Br is anion due to accepting of electrons.
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Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a spitting cobra rears up to a height of 0.420 m above the ground and launches venom at 2.50 m/s, directed 35.0° above the horizon. Neglecting air resistance, find the horizontal distance (in m) traveled by the venom before it hits the ground.
The horizontal distance traveled by the venom before it hits the ground is 0.6 m
What is Projectile ?A stone or ball or anything projected is known as a projectile.
Given that a spitting cobra rears up to a height of 0.420 m above the ground and launches venom at 2.50 m/s, directed 35.0° above the horizon.
Assuming air resistance is neglected, the parameters to be considered are;
Height h = 0.42 mVelocity v = 2.5 m/sAngle Ф = 35°The ball we reach the ground at the same time when it is dropped vertically or horizontally.
In a vertical direction,
h = 1/2g[tex]t^{2}[/tex]
Substitute all the parameters into the formula
0.42 = 1/2 x 9.8 x [tex]t^{2}[/tex]
0.42 = 4.9[tex]t^{2}[/tex]
[tex]t^{2}[/tex] = 0.42/4.9
[tex]t^{2}[/tex] = 0.0857
t = [tex]\sqrt{0.0857}[/tex]
t = 0.29 s
The horizontal distance traveled by the venom before it hits the ground will be
Distance = ucosФ x t
Distance = 2.5cos35 x 0.29
Distance = 0.599 m
Distance = 0.6 m
Therefore, the horizontal distance (in m) traveled by the venom before it hits the ground is 0.6 m
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Find the orbital speed of an ice cube in the rings of Saturn. The mass of Saturn is 5.68 x 1026 kg, and use an orbital radius of 3.00 x 105 km. (G = 6.67 × 10-11 N ∙ m2/kg2)
The orbital speed of an ice cube in the rings of Saturn is determined as 11,237.7 m/s.
What is orbital speed?The orbital speed of an astronomical body or object is the speed at which it orbits around the center of mass of the most massive body.
Orbital speed of ice cube in the rings of SaturnThe orbital speed of ice cube in the rings of Saturn is calculated as follows;
v = √GM/r
where;
G is universal gravitation constantM is mass of Saturnr is the distance of the ice cube = 3 x 10⁸ mv = √(6.67 x 10⁻¹¹ x 5.68 x 10²⁶)/(3 x 10⁸)
v = 11,237.7 m/s
Thus, the orbital speed of an ice cube in the rings of Saturn is determined as 11,237.7 m/s.
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Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.
1. What is the First Cosmic Speed i.e. the speed of a satellite on a low lying circular orbit around this planet? (Planet-X doesn't have any atmosphere.)
2. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
3. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?
(a) The speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.
(b) The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.
(c) The radius of the synchronous orbit of a satellite is 69,801 km .
Speed of the satellitev = √GM/r
where;
M is mass of the planetr is radius of the planetv = √[(6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 7,338.93 m/s
Escape velocity of the satellitev = √2GM/r
v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 10,378.82 m/s
Speed of the satellite at the given periodv = 2πr/T
r = vT/2π
r = (7,338.93 x 16.6 x 3600 s) / (2π)
r = 69,801 km
Thus, the speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.
The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.
The radius of the synchronous orbit of a satellite is 69,801 km .
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A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal.
The horizontal component of the force exerted by the hi.nge on the beam is 8.662x10^1 N.
What is the magnitude of the force that the beam exerts on the hi.nge?
288.51 N is the magnitude of the force that the beam exerts on the hi.nge.
Given
Mass 0f beam = 40 Kg
The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N
Angle between the beam and cable is = 90°
Angle between beam and the horizontal component = 31°
As the system of the beam, hi_nge and cable are in equilibrium.
The magnitude of the force that the beam exerts on the hi_nge can be calculated by -
F =The horizontal component of force + the vertical component of force
F = 86.62 N + 40 × 9.8 × sin 31°
F =86.62 N + 201.89 N
F = 288.51 N
Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51 N.
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10. When the force applied on an object is doubled, how does the pressure exert on the object change?
Answer:
It doubles as well.
Explanation:
Assuming that the area on which the force acts remains constant, and knowing that by definition:
P = F/A
We see that if we have a double force:
p = 2 F/A
Then the pressure is doubled.
Answer:
pressure doubles if area is constant
Explanation:
we know,
p=F/A
when force is double by keeping area constant
F=2F
Then,
Change in pressure (p') will be
P'=2F/A
or,p'=2×F/A
or,p'=2p(since, P=F/A)
therefore, when force is double by keeping area constant the pressure will increase by 2 times
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) Define magnetic flux density
Your friend's 10.8 g graduation tassel hangs on a string from his rearview mirror. When he accelerates from a stoplight, the tassel deflects backward toward the rear of the car at an angle of 5.23 ∘ relative to the vertical.
A) Find the tension in the string holding the tassel.
B) At what angle to the vertical will the tension in the string be twice the weight of the tassel?
The tension in the string holding the tassel and the vertical will the tension in the string
T = 0.1953 NФ = 34.4 °What is the tension in the string holding the tassel. ?Generally, the equation for Tension is mathematically given as
[tex]TCos\theta = mg[/tex]
Therefore
[tex]TCos6.58^{o} = 19.8*10^{-3}*9.8[/tex]
T = 0.1953 N
b).
Where
[tex]T* sin \theta = ma[/tex]
[tex]0.1953*Sin6.58 \textdegree = 19.8*10^{-3}*a[/tex]
a = 1.13 m/s^2
In conclusion
T* sinФ = ma
2msinФ = ma
2sinФ = a
[tex]sin\theta = \frac{a}{2}[/tex]
[tex]\theta = sin^{-1}\frac{a}{2} \\\\\theta= sin^{-1}\frac{1.13}{2}[/tex]
Ф = 34.4 °
In conclusion, The tension in the string holding the tassel and the vertical will the tension in the string
T = 0.1953 N
Ф = 34.4 °
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how we will measure centimeters
Answer:
.Explanation:
——»To measure centimeters, we can use ruler.
Use a ruler with the side marked either cm or mm. Align the edge of the object with the first centimeter line on the ruler, then find the length in whole centimeters, or the larger numbers on the ruler.Four canisters contain helium gas.
If all the canisters had the same amount of particles, which canister would have the fastest moving particles?
W
X
Y
Z
D. The canister with the fastest moving particle is Z due to concentration of particle.
Canister with the fastest moving particle
The speed of the particles depend on the mean distance of the particles.
The canister with the largest concentration per particle will contain particles with the greatest speed.
If the particle concentration is increasing from W to Z, then Z will have fastest moving particle.
Thus, the canister with the fastest moving particle is Z due to concentration of particle.
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Answer:
d. Z
Explanation:
If all the canisters had the same amount of particles, which canister would have the fastest moving particles?
Z
Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b).
By what angle has planet Y rotated through during this time?
Planet Y has rotated by 135.5° through during this time.
To find the answer, we need to know about the relation between angle and radius of orbit.
What's the expression of angle in terms of radius?Angle= arc/radiusAs arc = orbital velocity × time,angle= (orbital velocity × time)/radius
Orbital velocity= √(GM/radius), G= gravitational constant and M = mass of sunSo, angle = (√(GM)× time)/radius^3/2What's is the angle rotated by planet Y after 5 years, if ratio of the radius of orbit of planet X and Y is 4:3 and planet X is rotated by 88°?Let Ф₁= angle rotated by planet Y, Ф₂= angle rotated by planet XAs time = 5 years ( a constant)Ф₁/Ф₂= (radius of planet X / radius of planet Y)^(3/2)Ф₁= (radius of planet X / radius of planet Y)^(3/2) × Ф₂= (4/3)^(3/2) × 88°
= 135.5°
Thus, we can conclude that Planet Y has rotated by 135.5° through during this time.
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The small amount of missing mass that occurs when _____ atoms fuse to form helium
atoms is converted into energy in the form of heat and light.
nitrogen
hydrogen
barium
Ochromium
hydrogen
Explanation:Atomic fusion occurs when 2 or more atoms combine to create a new element.
Helium
Before figuring out what elements create helium, we need to understand its properties. Helium is the second element on the periodic table. It has 2 protons and usually 2 electrons in the base state.
There are technically ions but base helium will have 2 electrons, especially because it is a noble gas.
Fusion
In most cases, including this one, when 2 atoms fuse together you can add their protons together to find the atomic number of the new element. Since helium only has 2 protons, the elements fused to form helium must both have 1 proton.
The element with 1 proton is hydrogen. So, 2 hydrogen atoms fuse to form helium.
Real World Applications
Fusion can be seen across the universe. One of the most common examples of hydrogen fusing to become helium is in stars. Due to the heat within stars, 2 hydrogen atoms can fuse together to create helium. This process also creates energy in the form of heat and light that the star is made of.
An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.
1. Determine the density of the object.
2. When the object is immersed in oil, the force scale reads 35.6 N. Calculate the density of the oil.
Answer:
The density of this object is approximately [tex]1.36\; {\rm kg \cdot L^{-1}}[/tex].
The density of the oil in this question is approximately [tex]0.600\; {\rm kg \cdot L^{-1}}[/tex].
(Assumption: the gravitational field strength is [tex]g =9.806\; {\rm N \cdot kg^{-1}}[/tex])
Explanation:
When the gravitational field strength is [tex]g[/tex], the weight [tex](\text{weight})[/tex] of an object of mass [tex]m[/tex] would be [tex]m\, g[/tex].
Conversely, if the weight of an object is [tex](\text{weight})[/tex] in a gravitational field of strength [tex]g[/tex], the mass [tex]m[/tex] of that object would be [tex]m = (\text{weight}) / g[/tex].
Assuming that [tex]g =9.806\; {\rm N \cdot kg^{-1}}[/tex]. The mass of this [tex]63.8\; {\rm N}[/tex]-object would be:
[tex]\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}[/tex].
When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.
When this object was suspended in water, the buoyancy force on this object was [tex](63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}[/tex]. Hence, the weight of water that this object displaced would be [tex]47.0 \; {\rm N}[/tex].
The mass of water displaced would be:
[tex]\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}[/tex].
The volume of that much water (which this object had displaced) would be:
[tex]\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}[/tex].
Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately [tex]4.793\; {\rm L}[/tex].
The mass of this object is [tex]6.50\; {\rm kg}[/tex]. Hence, the density of this object would be:
[tex]\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}[/tex].
(Rounded to [tex]\text{$3$ sig. fig.}[/tex])
Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately [tex]4.793\; {\rm L}[/tex].
The weight of oil displaced would be equal to the magnitude of the buoyancy force: [tex]63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}[/tex].
The mass of that much oil would be:
[tex]\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}[/tex].
Hence, the density of the oil in this question would be:
[tex]\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}[/tex].
(Rounded to [tex]\text{$3$ sig. fig.}[/tex])
1. Is the image projected on a movie screen real or virtual? What about the image of yourself seen in a bathroom mirror?
2. Hold a shiny spoon in front of you. What differences do you notice about the image of your face seen in the convex and concave sides?
3. Where are the images formed by each side of the spoon? In front or behind the spoon? (Try the parallax method. Look at the image of an overhead light. Hold the tip of a pencil where you think the image is. Move your head from side to side. If the image and pencil tip appear to move relative to each other, adjust the position of the pencil back and forth until they appear as one)
Answer: 1. The movie one is virtual and the bathroom mirror is real
2. The image is distorted in a way
3. Behind the spoon
Explanation:
The movie one is virtual and the bathroom mirror is real and The image is distorted in a way and Behind the spoon.
What is virtual screen?In virtual desktops, the desktop environment is segregated from the physical device being used to access it. They are preloaded images of operating systems and apps. Over a network, users can remotely view their virtual desktops.
A virtual desktop can be accessed from any endpoint device, including a laptop, smartphone, or tablet. The user interacts with the client software that was installed on the endpoint device by the virtual desktop provider.
A virtual desktop mimics the appearance and feel of a real workstation. Because robust resources, like storage and back-end databases, are easily accessible, the user experience is frequently even better than that of a physical workstation.
Therefore, The movie one is virtual and the bathroom mirror is real and The image is distorted in a way and Behind the spoon.
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A 1400-kg sports car (including the driver) crosses the rounded top of a hill (radius = 87.0 m ) at 18.0 m/s .
a) Determine the normal force exerted by the road on the car.
b)Determine the normal force exerted by the car on the 65.0- kg driver.
c)Determine the car speed at which the normal force on the driver equals zero.
(a) The normal force exerted by the road on the car is 8,506.2 N.
(b) The normal force exerted by the car on the driver is 394.9 N.
(c) The speed of the car at which the normal force on driver is zero is 29.2 m/s.
Normal force exerted by the road on the carThe normal force exerted by the road on the car is calculated as follows;
Normal force = weight of the car - centripetal force of car
Weight of the car = (1400 x 9.8) = 13,720 N
Centripetal force of the car = (1400 x 18²)/87 = 5,213.8 N
Normal force = 13,720 N - 5,213.8 N
Normal force = 8,506.2 N
Normal force exerted on the driverNormal force = weight of driver - centripetal force of driver
Weight of driver = (65 x 9.8) = 637 N
Centripetal force of driver = (65² x 18²)/(87) = 242.07 N
Normal force = 637 N - 242.07 N = 394.9 N
Speed at which normal force on driver is zeroN = mg - mv²/r
0 = mg - mv²/r
mv²/r = mg
v²/r = g
v² = rg
v = √rg
v = √(87 x 9.8)
v = 29.2 m/s
Thus, the normal force exerted by the road on the car is 8,506.2 N.
The normal force exerted by the car on the driver is 394.9 N.
The speed of the car at which the normal force on driver is zero is 29.2 m/s.
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In an experiment replicating Millikan’s oil drop experiment, a pair of parallel plates are placed 0.0200 m apart and the top plate is positive. When the potential difference across the plates is 240.0 V, an oil drop of mass 2.0 × 10-11 kg gets suspended between the plates. (e = 1.6 × 10-19 C)
a) Draw a free-body diagram for the charge.
b) What is the charge on the oil drop?
c) Is there an excess or deficit of electrons on the oil drop? How many electrons are in excess or deficit?
Answer: See below
Explanation:
Given:
The potential between plates, V = 240 V
Distance between plates, d = 0.02 m
The mass of drop, m = 2x10^-11
Charge on electron, e = 1.6x10^-19
Part (a)
The free-body diagram is attached below
Part (b)
The electric field is given by,
[tex]E=\frac{V}{d}[/tex]
On applying force balance, the force on oil drop is equal to the weight of the oil,
[tex]$$\begin{aligned}F_{E} &=m g \\q E &=m g \\q \frac{V}{d} &=m g \\q &=\frac{m g d}{V}\end{aligned}$$[/tex]
Substituting the given values in the above equation,
[tex]\begin{aligned}&q=\frac{2 \times 10^{-11} \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2} \times \frac{1 \mathrm{~N}}{1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}} \times 0.02 \mathrm{~m}}{240 \mathrm{~V} \times \frac{1 \mathrm{~N} \cdot \mathrm{m} / \mathrm{C}}{1 \mathrm{~V}}} \\&q=1.63 \times 10^{-14} \mathrm{C}\end{aligned}[/tex]
Therefore, the charge on the oil drop is 1.63x10^-14 C
Part (c)
There will be an excess of electrons on the oil drop.
The number of electrons on oil drop can be calculated as,
[tex]\begin{aligned}q &=n e \\1.63 \times 10^{-14} \mathrm{C} &=n \times 1.6 \times 10^{-19} \mathrm{C} \\n &=1.01 \times 10^{5}\end{aligned}[/tex]
Therefore, the number of excess electrons is 1.01x10^5