Answer:
C= 50km/hr west
Explanation:
150/3= 50
Because it asks for velocity, make sure to include the direction as well.
Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.
1. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
2. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?
The minimum speed required for a satellite in order to break free permanently from the planet and the radius of the synchronous orbit of a satellite are 7.3 Km/s and 3.1 × 10⁴ km respectively.
To find the answer, we need to know about the escape velocity and time period of revolving satellite.
What's the expression of escape velocity of a satellite?Mathematically, escape velocity= √(2GM/R)G = gravitational constant, M = mass of planet, R= radius of the planetHere, M = 4.74×10²⁴kg, R = 5870 kmEscape velocity= √(6.67×10^(-11)×4.74×10²⁴/5.870×10⁶)= 7.3 Km/s
What's the expression of time period of a circularly orbiting satellite?T= {2π×r^(3/2)}/√(GM)r= (T/2π)⅔× (GM)^(1/3)r is the radius of the orbitWhat's the radius of the circular orbit, if the time period of the satellite is 16.6 hours?T = 16.6 hours = 16.6×3600 second = 59760sr = (59760/2π)^⅔× (6.67×10^(-11)×4.74×10²⁴)^(1/3)= 3.1 × 10⁴ km
Thus, we can conclude that the escape velocity and the radius of the synchronous orbit of a satellite are 7.3 Km/s and 3.1 × 10⁴ km respectively.
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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
an electron is moving with a speed of 0.85c in a direction opposite to that of photon. calculate the relative velocity of the electron and photon.
Answer:
1.85c
Explanation:
a photon moves at c, the electron is moving at 0.85c, and since they are moving in opposing directions, the relative speed would be 1.85c
The relative velocity of the electron and photon is equal to the speed of light in a vacuum, which is 3 × 10⁸ m/s The negative sign indicates that they are moving in opposite directions.
The formula for adding velocities in special relativity is given by:
V(relative) = (V₁ - V₂) ÷ (1 - (V₁ × V₂ ÷ c²))
where:
V₁ = velocity of the electron
V₂ = velocity of the photon
c = speed of light in a vacuum
The relative velocity is:
V(relative) = (0.85c - c) ÷ (1 - (0.85c × c ÷ c²))
V(relative) = (0.85c - c) ÷ (1 - 0.85)
V(relative) = (-0.15c) ÷ (0.15)
V(relative) = -c
Therefore, The relative velocity of the electron and photon is equal to the speed of light in a vacuum, which is 3 × 10⁸ m/s The negative sign indicates that they are moving in opposite directions.
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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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If the velocity of an object is 9 m/s and its momentum is 72 kgm/s, what is its mass
An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.
What is momentum?In Newtonian mechanics, linear momentum, or simply momentum, is the product of the mass and velocity of an object.
It is a vector quantity, possessing a magnitude and a direction.
The mathematical expression for momentum is:
p = m . v
where,
p is the linear momentum of the object.m is the mass of the object.v is the velocity of the object.An object has a velocity (v) of 9 m/s and its linear momentum (p) is 72 kg.m/s. We will use the definition of linear momentum to calculate the mass of the object.
p = m . v
m = p / v
m = (72 kg.m/s) / (9 m/s) = 8 kg
An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.
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A 69-kg base runner begins his slide into second base when he is moving at a speed of 3.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
(a) How much mechanical energy is lost due to friction acting on the runner?
J
(b) How far does he slide?
m
The mechanical energy lost is 353.28 Joules . The distance he slid off is 0.16m.
we know:-
Mass = 69 kg
Speed = 3.2 m/s
Coefficient of friction ( ratio of friction force and normal force ) = 0.70
Acceleration due to gravity, g = 9.8 m/s^2
(a) To determine the amount of mechanical energy that is lost because of friction acting on the runner, we would calculate the change in kinetic energy:
[tex]KE = \frac{1}{2} m (v-u)^{2}[/tex]
[tex]= \frac{1}{2} 69 ( 3.2-0 )^{2}[/tex]
[tex]= 353.28[/tex] Joules
Mechanical energy = 353.28 Joules
(b) To determine how far (distance) the runner slide:
acceleration = ug
[tex]= 3.2[/tex] × [tex]9.8[/tex]
[tex]= 31.36[/tex] [tex]m/s^{2}[/tex]
distance ,
[tex]V^{2} = U^{2} + 2aS[/tex]
[tex]( 3.2)^{2} = 0 + 2 ( 31.36) S[/tex]
[tex]10.24 = 62.72 S[/tex]
[tex]S = {\frac{10.24}{62.72} }[/tex]
Distance, S = 0.16 m
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Drag each label to the correct location on the image. Jessica is visiting a park with her mother. Jessica sits on a swing. Her mother pulls the swing to a height of 3 meters above the ground and lets it go. The image shows Jessica at three positions on the swing. Jessica‘s mass is 44 kilograms and the maximum velocity of the swing is 5 meters/second. What’s the energy she has at each position shown? Ignore friction and air resistance. Use g = 9.8 m/s2, PE = m × g × h, and KE = 0 joules KE = 550 joules PE = 862.4 joules
The energy she has at each position shown are:
Position at maximum height -- 1,294 JPosition at minimum height --0 JPosition at maximum velocity - 550 JPosition at minimum velocity - 0 JWhat does velocity implies?Velocity is known to be a term that connote the direction of any kind of a moving body or an object.
Note that the Speed is known to be a a scalar quantity and as such, Velocity is said to be a vector quantity.
Note also that from the question given, Jessica's height of the swing 3 meters above the ground, therefore:
Jessica Position at maximum height :
PE = mgh
PE = 44kg x 9.8m/s² x 3
PE = 1,294 J
Jessica Position at minimum height:
PE = 0 J
Jessica Position at maximum velocity:
KE = 1/2 x mv²
KE = 1/2 x 44kg x (5m/s²)²
KE = 550 J
Jessica Position at minimum velocity:
KE = 0 J
Therefore, The energy she has at each position shown are:
Position at maximum height -- 1,294 JPosition at minimum height --0 JPosition at maximum velocity - 550 JPosition at minimum velocity - 0 JLearn more about velocity from
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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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A first year student projected a farm business brochure to a farmer at 30 degrees to horizontal. calculate the maximum height attained by the projectile if it was launched at 400m/s
Answer:
Maximum height = 2040 m
Explanation:
We can solve the problem using kinematics.
Consider the vertical motion of the object and use the equation:
[tex]\boxed{v^2 = u^2 + 2as}[/tex]
where:
• v = final velocity (0 m/s, because when the object is at max. height, it has no vertical velocity)
• u = initial velocity (400sin30° m/s ⇒ vertical component of 400 m/s at 30° to horizontal)
• a = acceleration (-9.81 m/s²; considering upward acceleration to be negative)
• s = displacement (? m; this represents the max. height of the object),
Substitute the values into the equation and solve for s :
[tex]0^2 = (400 sin (30 \textdegree))^2 + 2(-9.81)(s)[/tex]
⇒ [tex]2(9.81)(s) = (400 sin (30 \textdegree))^2[/tex]
⇒ [tex]s = 2040 \space\ m[/tex] (3 s.f.)
What adaptation of a cactus protects it from predators? a round cactus with many spines Broad leaves Sharp spines Thick stems Yellow flowers
A round cactus with many spines is the adaptation of a cactus that protects it from predators.
A cactus, unlike other plants, has unique adaptations in its roots, leaves, and stems that allow it to flourish in hot and dry surroundings.
The one adaptation that protects the cactus from predators is spines.
A cactus does not have any parts that resemble leaves if you could look at one closely.
Instead, the leaves are transformed into spines, which protrude from the plant's tiny bumps known as areoles.
Herbivores that live in the desert may be enticed to eat the cactus. The spines prevent these predators by modifying leaves into spines.
Other than protection, Spines perform many functions like
1) Since evaporation is a problem in a desert since water is scarce, the spines prevent excessive evaporation.
2) The spines also impede airflow and prevent evaporation by trapping air.
3) Collecting dew from the early-morning fog is another crucial job that the spines do.
The gathered dew turned into liquid water and ran down to the earth below. The plant then absorbs this water.
Hence, the adaptation of a cactus that protects it from predators is round cactus with many spines.
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A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in Figure 4.29(b). The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
Figure 4.29
(a) Calculate the minimum force F he must exert to get the block moving.
40.873
(b) What is its acceleration once it starts to move, if that force is maintained?
m/s2
(a) The minimum force F he must exert to get the block moving is 38.9 N.
(b) The acceleration of the block is 0.79 m/s².
Minimum force to be applied
The minimum force F he must exert to get the block moving is calculated as follows;
Fcosθ = μ(s)Fₙ
Fcosθ = μ(s)mg
where;
μ(s) is coefficient of static frictionm is mass of the blockg is acceleration due to gravityF = [0.1(36)(9.8)] / [(cos(25)]
F = 38.9 N
Acceleration of the blockF(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32
a = F(net)/m
a = 28.32/36
a = 0.79 m/s²
Thus, the minimum force F he must exert to get the block moving is 38.9 N.
The acceleration of the block is 0.79 m/s².
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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 exerted by the hi.nge on the beam is 114.77 N.
Tension in the cable
Apply the principle of moment and calculate the tension in the cable;
Clockwise torque = TL sinθ
Anticlockwise torque = ¹/₂WL
TL sinθ = ¹/₂WL
T sinθ = ¹/₂W
T = (W)/(2 sinθ)
T = (29 x 9.8)/(2 x sin57)
T = 169.43 N
Vertical component of the forceT + F = W
F = W - T
F = (9.8 x 29) - 169.43
F = 114.77 N
Thus, the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
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An air-filled 39.1 μH solenoid has a length 4.0 cm and a cross-sectional area 0.60 cm2. How many turns are in this solenoid?
a.21000
b.144
c.12
d.120
e.1200
No. of turns in the solenoid is an option (b) 144.
The self-inductance of a long solenoid depends only on its physical properties (such as the number of turns of wire per unit length and the volume), and not on the magnetic field or the current.
Self-inductance of solenoid = 39.1 μH
= 39.1 × [tex]10^-^6[/tex] H
Length of the solenoid = 4.0 cm
Cross-sectional area = 0.60 cm²
Expression for the self-inductance of a coil ;
L = µ₀N²A / [tex]l[/tex]
where,
L = Self- Inductance
N = No. of turns.
A = Cross-sectional area
[tex]l=[/tex] Length of the solenoid
L =( 4π × [tex]10^-^7[/tex] × N² × 0.60 ) / 4.0
39.1× 4.0 / 4π × [tex]10^-^7[/tex] × 0.60 = N²
N² = 2.07 × [tex]10^6[/tex]
N = 144
Therefore, the no. of turns of the solenoid is 144.
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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).
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.
Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in the vein is 35.7 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?
The minimum necessary height of the IV bag above the position of the needle is 0.37 m.
we know that,
P = ρgh
where,
P = 35.7mmHg
= 4759.609 Pa
g = 9.8[tex]\frac{m}{s^{2} }[/tex]
ρ = 1.308 kg / m^3
now, substituting all the values, we get,
4759.609 = 1.308 × 9.8 × h
h = 0.37 m
The minimum necessary height of the IV bag above the position of the needle = 0.37 m.
what is an IV bag ?
A reagent, also called as an analytical reagent, is a substance or compound that is added to a system in chemistry to bring about a chemical reaction or examine to see if one happens. Even though the terms "reagent" and "reactant" are frequently used synonymously, "reactant" refers to a substance that is consumed during a chemical reaction.
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A 40.0-kg child running at 8.00 m/s suddenly jumps onto a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. The child is traveling tangential to the edge of the merry-go-round just before jumping on. The moment of inertia about its axis of rotation is 600 kg ∙ m2 and very little friction at its rotation axis. What is the angular speed of the merry-go-round just after the child has jumped onto it?
The angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.
What is principle of conservation of angular momentum?
The principle of conservation of angular momentum states that the sum of initial angular momentum is equal to final angular momentum.
Li = Lf
Li = Ii ωi + Ic ωc
Li = Iiωi + MR²(V/R)
Li = Iiωi + MRV ----- (1)
Angular speed of the merry go round after the child jumps on itLf = If ωf
ωf = Lf/If
If = Im + MR²
ωf = Lf / ( Im + MR²)
Recall, Lf = Li
ωf = (Iiωi + MRV) / ( Im + MR²)
where;
Iiωi is the initial angular momentum of the merry - go round = 0M is mass of the childR radius of rotationV is tangential speed of the childIm is the moment of inertia of the merry go roundωf = (Iiωi + MRV) / ( Im + MR²)
ωf = (0 + 40 x 1.5 x 8) / (600 + 40(1.5)²)
ωf = (480) / (690)
ωf = 0.696 rad/s
Thus, the angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.
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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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Avogadro‘s number was calculated by determining The number of atoms in
Answer:
12 grams of the isotope carbon-12.
Explanation:
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A 1300-kg car moving on a horizontal surface has speed v = 60 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.3 m .
What is the spring stiffness constant of the spring?
Express your answer to two significant figures and include the appropriate units.
The stiffness constant of the spring is 68,290.3 N/m
Stiffness constant of the spring
Apply the principle of conservation of energy;
U = K.E
¹/₂kx² = ¹/₂mv²
kx² = mv²
k = mv²/x²
where;
m is massv is speed = 60 km/h = 16.67 m/sx is the distancek = (1300 x 16.67²)/(2.3²)
k = 68,290.3 N/m
Thus, the stiffness constant of the spring is 68,290.3 N/m.
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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 treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg.
Tension in rope A:
7.42×102 N
Tension in rope B:
7.52×102 N
What is the tension in rope C?
Tension in the rope C is 1.24× 10² N.
To find the answer, we need to know about the horizontal component of tension in the rope B.
What's the angle made by the rope B by horizontal?From the figure of the answer, in the triangle PQR, tan(θ)= PQ/QR = 6/1= 6 θ= tan inverse of 6 = 80.5°What's the horizontal component of the tension in rope B?Horizontal component= tension in rope B × cos80.5°
= 7.52×10² N × cos80.5°
= 1.24×10² N
What's the tension in the rope C?From the figure, we have found that the tension in rope C = horizontal component of the tension in rope BSo, tension in rope C= 1.24×10² NThus, we can conclude that the tension in the rope C is 1.24× 10² N.
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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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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 = 8.662×101 N
What is the magnitude of the force that the beam exerts on the hi.nge?
The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.
To find the answer, we need to know about the tension.
How to find the magnitude of the force that the beam exerts on the hi.nge?Let's draw the free body diagram of the system using the given data.From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.[tex]N_x=86.62N[/tex]
We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.[tex]N_y=F_V=mg-Tsin59\\[/tex]
To find Ny, we need to find the tension T.For this, we can equate the net horizontal force.[tex]F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N[/tex]
Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,[tex]N_y= (40*9.8)-(169.8*sin59)=246.4N[/tex]
Thus, the magnitude of the force that the beam exerts on the hi.nge will be,[tex]N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N[/tex]
Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.
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The hi.nge will be subjected to a force of 261.12N from the beam.
We must understand the tension in order to choose the solution.
How can the amount of force the beam applies on the height be determined?Let's use the provided information to create the system's free body diagram.We need to calculate the force the beam is exerting on the height using the diagram.For this, it is assumed that the horizontal component of force is 86.62N, the same as the horizontal component of the normal reaction that the beam exerts on the height.We need to identify the vertical component of the normal reaction the beam exerts on the height. We must equalize the total force acting in the vertical direction to achieve this.[tex]N_y=F_v=mg-Tsin59[/tex]
Finding the tension T is necessary to determine Ny. Thus, we can use the net horizontal force to equate this.[tex]F_H=N_x=Tcos59\\T=\frac{F_H}{cos59} =169.84N[/tex]
As a result, the normal reaction that the beam has on the height becomes, with a vertical component,[tex]N_y=(40*9.8)-(169.84*sin59)=246.4N[/tex]
As a result, the force the beam applies on the height will be of the order of,[tex]N=\sqrt{N_x^2+N_y^2} =261.12N[/tex]
Thus, we can infer that the force the beam applies to the height is 261.12N in size.
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An Airbus A380-800 passanger airplane is cruising at constant altitude on a straight line with a constant speed. The total surface area of the two wings is 395 m2. The average speed of the air just below the wings is 259 m/s, and it is 288 m/s just above the surface of the wings.
What is the mass of the airplane? The average density of the air around the airplane is ρ^air = 1.21 kg/m^3.
The mass of the airplane of area of two wings 395m², and the average speed of lower and upper surface of the wings are, 259 and 288m/s is 386.8×10^3 kg.
To find the answer, we need to know about the Bernoulli's principle.
How to find the mass of the airplane?The Bernoulli's principle states that, the sum of pressure energy, kinetic energy and potential energy of an incompressible, non-viscous, fluid in streamlined flow is a constant.It can be expressed as,P+ [tex]\frac{1}{2}[/tex] ρv²+ρgh=a constant.
Instead of ρ we take d as density.
We have given that,[tex]A= 395 m^2\\v_l=295 m/s\\v_u=288m/s\\density=1.21kg/m^3[/tex]
We equate the principle for lower and upper surfaces of the wing like,[tex]P_1+\frac{1}{2}v_1^2d+dgh_1=P_2+ \frac{1}{2}v_2^2d+dgh_2\\here\\h_1=h_2\\thus\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\P_1-P_2=\frac{1}{2}*1.21(288^2-259^2)=9597.12 atm\\[/tex]
Thus, the mass of the airplane from the above equation will be,[tex]\frac{F}{A}=9597.12 atm\\\\ F=9597.12*395m^2=37.9*10^5 N\\\\mg=37.9*10^5 N\\\\m=\frac{37.9*10^5 N}{9.8}\\\\ m=386.8*10^3kg[/tex]
Thus, we can conclude that, the mass of the airplane is 386.8×10^3 kg.
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The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.
We need to be aware of the Bernoulli principle in order to determine the solution.
How can I determine an airplane's mass?According to the Bernoulli's principle, the total amount of pressure energy, kinetic energy, and potential energy in a streamlined flow of an incompressible, non-viscous fluid is constant.It can be stated as follows:P+ [tex]\frac{1}{2}[/tex]dv²+dgh = constant.
We substitute d for to represent density.
We've done that,[tex]v_1=259m/s\\v_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]
We compare the governing idea for the wing's bottom and upper surfaces to:[tex]P_1+\frac{1}{2}dv_1^2+dgh= P_2+\frac{1}{2}dv_2^2+dgh\\\\P_1+\frac{1}{2}dv_1^2=P_2+\frac{1}{2}dv_2^2\\\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\\\P_1-P_2=9597 atm[/tex]
Consequently, using the aforementioned equation, the airplane's mass will be,[tex]F/A= 9597 atm\\mg=9597*395 =38*10^5N\\m=\frac{38*10^5}{9.8} = 387*10^3kg[/tex]
Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.
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Why should you use an iron hammer instead of a wooden
hammer?
Answer:
more durability and easier to pound a hammer with the heavy weight.
Explanation:
One should use an iron hammer instead of the wooden hammer because the force developed with the help of the iron hammer is far greater than the wooden hammer
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
The mathematical expression for Newton's second law is as follows
F = m*a
where F represents the force applied
m is the mass of the object
a is the acceleration of the object
As we know that force is a product of mass and acceleration, let us assume that both iron and wooden hammers have the same dimension , considering the acceleration applied is the same while hammering ,
As for the same dimension the mass of iron hammer would be greater as result force applied through the iron hammer would be grater
Since the iron hammer's force is so much stronger than the wooden hammer's force, one should use an iron hammer instead of a wooden hammer.
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A moonshiner makes the error of filling a glass jar to the brim and capping it tightly. The moonshine expands more than the glass when it warms up, in such a way that the volume increases by 0.6% (that is, ΔV/V0 = 6 ✕ 10-3) relative to the space available. Calculate the force exerted by the moonshine per square centimeter if the bulk modulus is 2.1 ✕ 109 N/m2, assuming the jar does not break.
The force exerted per square centimeter is 126 N/cm².
What is pressure?Pressure is the force acting per unit area.
Pressure = force/areaBased on the data given:
volume increase, ΔV/V0 = 6 * 10⁻³
Bulk Modulus, B = 2.1 * 10⁹ N/m²
Bulk modulus B of a material is ratio of change in pressure and change in volume as given below:
B = ΔP/ [(ΔV/V)]Solving for ΔP;
ΔP = B * [(ΔV/V)]
ΔP = (2.1 * 10⁹ N/m²) * (6 * 10⁻³)
ΔP = 1.26 * 10⁶ N/m²
Converting to per square centimeter
ΔP = (1.26 * 10⁶ N/m²)/10⁴
ΔP = 126 N/cm²
In conclusion, the force exerted per square centimeter is a measure of the pressure.
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Technetium -99 (half-life = 6.01) is used in the medical imaging. How many half-lives would go by in 44.0h?
The number of half-lives that would go by in 44 h is 7 half-lives
What is half life?This is the time taken for half a substance to decay
How to determine the number of half lifeThe number of half-lives that will elaspe after 44 h can be obtained as illustrated below:
Half-life (t½) = 6.0 hTime (t) = 44 hNumber of half-lives (n) =?n = t / t½
n = 44 / 6.01
n = 7
Thus, 7 half-lives will elaspe after 44 h
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Calculate the magnitude of the electric field at one corner of a square 1.10 m on a side if the other three corners are occupied by 4.05×10−6 C charges. What is the direction of the electric field?
The direction of magnetic field is south-east and the magnitude is
[tex]23.66[/tex] × [tex]10^{3} N/C[/tex].
Here, the magnitude of CD and BC will be cancelled, as they both are in the opposite direction and equal to each other.
the magnitude, towards the diagonal AC will result in CP, which is the direction of the electric field.
magnitude of electric field can be defined as :- The force per charge on the test charge is a straight forward way to define the size of the electric field.
To find the magnitude of the electric field use the formula
[tex]E = kq/ r^{2}[/tex]
inserting the values,
[tex]E = 9. 10^{9}[/tex] × [tex]4.05[/tex] × [tex]10^{-6} / 1.1 \sqrt{2}[/tex]
[tex]E= 36.45[/tex] × [tex]10^{3}[/tex] / [tex]1.54[/tex]
[tex]E = 23.66[/tex] × [tex]10^{3}[/tex] N/C
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