What is the symbol that is used to indicate the fulcrum in a lever?
A triangle is the symbol used to indicate a fulcrum in a lever, a triangle has three sides and three vertices.
What is a fulcrum?A fulcrum is a pivot that has a base and a point that serves as a balance for a lever.
What is a lever?Basically a lever is a simple machine that is a beam with two sides pivoted to fulcrum.
A lever has Load, Effort and the last part is fulcrum
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it's raining in 60 degree and man is walking in an inclined plain of base angle 30 degree find angle with vertical at which the man must hold his umbrella to be safe
Which of the following experiments demonstrates Newton's Third Law? Select two answers
a rocket goes forward by burning fuel behind it
A student jumps 2 feet by using his muscles to exert a force on the ground
Matter cannot be created or destroyed
A person claps because they enjoy Physical Science
A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge density λ = 3.5 nC/m. The point P is located on the postive y-axis at a distance y0 = 15 cm from the origin. The z-axis points out of the screen. 25% Part (a) By symmetry the electric field at point P has no component in the _____________. 25% Part (b) Choose the correct expression for the y-component of the electric field at P due to a thin slice of the rod of thickness dx located at point x. 25% Part (c) Integrate your correct choice in part (b) over the length of the rod and choose the correct expression for the y-component of the electric field at point P. 25% Part (d) Calculate the magnitude of the electric field at point P, in newtons per coulomb.
Answer:
a) The electric field at point P has no component in the x and z directions.
b) dEy = kλdxy₀ / (√( x² + y₀²))^3/2
c)Ey = = 2kλd / y₀( d² + 4y₀²))^1/2
d) Ey = 411.84 N/C
Explanation:
a)
from the uploaded image;
the electric field will only be in y-direction.
Therefore the electric field at P have no component in the x and z directions.
b)
dEy = dEsin θ
dE = kdq / r²
from figure
sinθ = y₀ / √( x² + y₀²)
r = √( x² + y₀²)
dq = λdx
dEy = kdq sinθ / r²
dEy = kλdxy₀ / (√( x² + y₀²))^3/2
c)
the net electric field at p is ,
Ey = ∫^d/2_-d/2 kλdxy₀ / (√( x² + y₀²))^3/2
= kλy₀ ∫^d/2_-d/2 dx / (√( x² + y₀²))^3/2
= 2kλd / y₀( d² + 4y₀²))^1/2
d)
let y₀ = 15 × 10⁻²m, λ = 3.5 nC/m , d = 1.5m
Ey = 2kλd / y₀( d² + 4y₀²))^1/2
Ey = 2(9×10⁹ N.m²/C²)(3.5×10⁻⁹m)(1.5m) / 0.15×(1.5)² + 4(0.15)²))^1/2
Ey = 411.84 N/C
A 100V battery is connected to two oppositely charged plates that are 10cm apart.
i. What is the magnitude of the electric field?
ii. Calculate the Electric Force exerted on a +200 C point charge.
iii. What is the electric potential energy of the charge when it’s 8cm and 2cm from the negatively
charged plate?
iv. How much work was required to move the charge from 8cm to 2cm?
v. If the 10gram point charge was at rest at point A, what is the final speed at point B?
Answer:
A 100V battery is connected to two oppositely charged plates that are 10cm apart.
i. what is the magnitude of the electric field?
ii. Calculate the electric force exerted on a +200uC point charge.
Explanation:
The small piston of a hydraulic lift has a crosssectional area of 2.87 cm2 and the large piston 314 cm2 . What force must be applied to the small piston for the lift to raise a load of 2.4 kN
Answer:
P1 = P2 pressure is uniform
F1 / A1 = F2 / A2
F1 = F2 (A1 / A2) = 2,400 N * (2.87 / 314) = 21.9 N
A 5-kilogram bowling ball is traveling down a lane with a velocity of 40ms. What is the kinetic energy and momentum of the bowling ball
The Kinetic energy of the bowling ball is 4000 J while the momentum of the ball is 200 kg m/s
what is kinetic energy?Kinetic energy, KE is the energy exerted by a body in motion. This is represented mathematically as half the product of mass, m and velocity, v squared.
Kinetic energy
KE = 1 / 2 m v^2
m = 5 kg
v = 40 m/s
KE = 0.5 * 5 *40 ^2
KE = 4000 J
Momentum, M
M = mass * velocity
M = 5 * 40
M = 200 kg m/s
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A raging bull of mass 700 kg runs at 36 km/h .How much kinetic energy does it have ?
Answer:
Use equation for kinetic energy: Ek=mV²/2
m=700 kg
V=10m/s
Ek=700kg*100m²7s²/2
Ek=35000 J=35kJ
Explanation:
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A small object with mass m, charge q, and initial speed v0 = 5.00 * 103 m>s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm (Fig. P21.78). The electric field between the plates is directed downward and has magnitude E = 800 N>C. Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distance d = 1.25 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance. Calculate the object’s charge-to-mass ratio, q>m
Answer:
q/m = 2177.4 C/kg
Explanation:
We are given;
Initial speed v_o = 5 × 10³ m/s = 5000 m/s
Now, time of travel in electric field is given by;
t_1 = D_1/v_o
Also, deflection down is given by;
d_1 = ½at²
Now,we know that in electric field;
F = ma = qE
Thus, a = qE/m
So;
d_1 = ½ × (qE/m) × (D_1/v_o)²
Velocity gained is;
V_y = (a × t_1) = (qE/m) × (D_1/v_o)
Now, time of flight out of field is given by;
t_2 = D_2/v_o
The deflection due to this is;
d_2 = V_y × t_2
Thus, d_2 = (qE/m) × (D_1/v_o) × (D_2/v_o)
d_2 = (qE/m) × (D_1•D_2/(v_o)²)
Total deflection down is;
d = d_1 + d_2
d = [½ × (qE/m) × (D_1/v_o)²] + [(qE/m) × (D_1•D_2/(v_o)²)]
d = (qE/m•v_o²)[½(D_1)² + D_1•D_2]
Making q/m the subject, we have;
q/m = (d•v_o²)/[E((D_1²/2) + (D_1•D_2))]
We have;
E = 800 N/C
d = 1.25 cm = 0.0125 m
D_1 = 26.0 cm = 0.26 m
D_2 = 56 cm = 0.56 m
Thus;
q/m = (0.0125 × 5000²)/[800((0.26²/2) + (0.26 × 0.56))]
q/m = 312500/143.52
q/m = 2177.4 C/kg
The ratio of the charge to mass of the object will be equal to:
[tex]\dfrac{q}{m}=2177.4\ \frac{C}{kg}[/tex]
what is electric field?
The electric field is defined as the whenever an atom carries a charge whether positive or negative the their influence of force is spread around the charge at a particular distance this effect of force is called as the electric field.
We are given;
Initial speed [tex]V_o[/tex] = 5 × 10³ m/s = 5000 m/s
Now, time of travel in electric field is given by;
[tex]t_1=\dfrac{D_1}{v_o}[/tex]
Also, deflection down is given by;
[tex]d_1=\dfrac{1}{2}at^2[/tex]
Now,we know that in electric field;
F = ma = qE
Thus, [tex]a=\dfrac{qE}{m}[/tex]
So;
[tex]d_1=\dfrac{1}{2}\times(\dfrac{qE}{m})(\dfrac{D_1}{v_o})^2[/tex]
Velocity gained is;
[tex]V_y=(a\timest_1)=(\dfrac{qE}{m})\times (\dfrac{D_1}{v_o})[/tex]
Now, time of flight out of field is given by;
[tex]t_2=\dfrac{D_2}{v_o}[/tex]
The deflection due to this is;
[tex]d_2=V_y\times t_2[/tex]
Thus, [tex]d_2=(\dfrac{qE}{m}\times (\dfrac{D_1}{v_o})\times(\dfrac{D_2}{v_o})[/tex]
[tex]d_2=(\dfrac{qE}{m})\times (\dfrac{D_1\times D_2}{(v_o)^2}[/tex]
Total deflection down is;
[tex]d=d_1+d_2[/tex]
[tex]d=[\dfrac{1}{2}\times(\dfrac {qE}{m}) \times (\dfrac{D_1}{v_o^2})]+[(\dfrac{qE}{m})\times (\dfrac{D_1\times D_2}{v_o^2}})][/tex]
Making q/m the subject, we have;
[tex]\dfrac{q}{m}=\dfrac{(d\times v_o^2)}{[E(\dfrac{D_1^2}{2})+(D_1\times D_2))]}[/tex]
We have;
E = 800 N/C
d = 1.25 cm = 0.0125 m
D_1 = 26.0 cm = 0.26 m
D_2 = 56 cm = 0.56 m
Thus;
[tex]\dfrac{q}{m}=\dfrac{(0.0125\times 5000^2)}{[800((\dfrac{0.26^2}{2})+(0.26\times 0.56))]}[/tex]
[tex]\dfrac{q}{m}= 2177.4[/tex]
Hence the ratio of the charge to mass of the object will be equal to:
[tex]\dfrac{q}{m}=2177.4\ \frac{C}{kg}[/tex]
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A 57 kg person in a rollercoaster moving through the bottom of a curved track of radius 42.7 m feels a normal force of 995 N. How fast is the car moving?
Answer:
Use Fc centripetal force as positive and W the weight as negative
N = m v^2 / R + m g
v^2 = (N - m g) R / m
v^2 = (995 - 57 * 9.8) 42.7 / 57 = 327 m^2/s^2
v = 18.1 m/s
Note: N - m g is the net force producing the centripetal force
A plane travels 1743 KM in 2 hours 30 minutes. How fast was the plane traveling?
Answer:
[tex]v=697.2km/h[/tex]
Explanation:
Hello.
In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:
[tex]V=\frac{d}{t}[/tex]
The distance is clearly 1743 km and the time is:
[tex]t=2h+30min*\frac{1h}{60min} =2.5h[/tex]
Thus, the velocity turns out:
[tex]v=\frac{1743km}{2.5h}\\ \\v=697.2km/h[/tex]
Which is a typical velocity for a plane to allow it be stable when flying.
Best regards.
A crate (100 kg) is in an elevator traveling upward and slowing down at 6 m/s2. Find the normal force exerted on the crate by the elevator. Assume g
Answer:
380 N
Explanation:
Forces: gravity mg (downward), Normal force N (upward)
Acceleration: Note velocity is up, but slowing down, so acceleration is opposite to velocity, or downward
Newton's 2nd law:
mg - N = ma
N = m(g - a) = 100(9.8 - 6) = 380 N
For the following distance vs time graph
Answer:
3
5
4
Explanation:
x = (8=(9+9)
(9+9) = 4, 5, 3
A teenager is standing at the rim of a large horizontal uniform wooden disk that can rotate freely about a vertical axis at its center. The mass of the disk (in kg) is M and its radius (in m) is R. The mass of the teenager (in kg) is m. The disk and teenager are initially at rest. The teenager then throws a large rock that has a mass (in kg) of m_rock. As it leaves the thrower's hands, the rock is traveling horizontally with speed v (in m/s) relative to the earth in a direction tangent to the rim of the disk. The teenager remains at rest relative to the disk and so rotates with it after throwing the rock. In terms of M, R, m, m_rock and v, what is the angular speed of the disk
Answer:
[tex]\omega = \frac{m_{rock} \cdot v}{\left(\frac{M}{2}+m\right) R}[/tex]
Explanation:
[tex](inertia.of.the.disk) Id =\frac{M R^{2}}{2}\\(inertia.of.the.solid) Is = m R^{2}\\\\Itotal = Id + Is = \left(\frac{M}{2}+m\right) R^{2}\\\\\\L=\vec{\gamma} \times \vec{p}={mvr} \quad \& \quad L=I \omega\\\\Mrock \timesv \times R=\left(\frac{M}{2}+m\right) R^{2} \times \omega\\\\\omega = \frac{m_{rock} \cdot v}{\left(\frac{M}{2}+m\right) R}\\[/tex]
The required angular speed of disk is [tex]\dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex].
Given data:
The mass of disk is, M.
The radius of disk is, R.
The mass of teenager is, m.
The mass of rock is, [tex]m_{rock}[/tex].
The speed of rock is, v.
Here we need to apply the concept of angular momentum to obtain the value of angular speed of disk. The expression for the angular momentum is given as,
[tex]L = I_{total} \times \omega[/tex] .......................................(1)
Here, [tex]\omega[/tex] is the angular speed and [tex]I_{total}[/tex] is the total moment of inertia of system.
And its value is,
[tex]I_{total} = I_{disk}+I_{solid}\\\\I_{total} = \dfrac{MR^{2}}{2}+mR^{2}[/tex] ..........................................(2)
And the angular momentum is also expressed as,
[tex]L = p \times R\\L =( m_{rock}v) \times R[/tex] ....................................................(3)
Then, using the equation (1), (2) and (3) we have,
[tex]( m_{rock} \times v) \times R = (\dfrac{MR^{2}}{2}+mR^{2}) \times \omega\\\\m_{rock} \times v = (\dfrac{MR}{2}+mR) \times \omega\\\\\\\omega = \dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex]
Thus, we can conclude that the required angular speed of disk is [tex]\dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex].
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A skier of mass 58.0 kg starts from rest at the top of a ski slope of height 70.0 m . Part APart complete If frictional forces do −1.04×104 J of work on her as she descends, how fast is she going at the bottom of the slope? Take free fall acceleration to be g = 9.80 m/s2 . 31.8 m/s Previous Answers Correct Part BPart complete Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.250. If the patch is of width 68.0 m and the average force of air resistance on the skier is 150 N , how fast is she going after crossing the patch? 18.1 m/s Previous Answers Answer Requested Part C After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.30 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?
Answer:
A) v = 31.8 m/s
B) v_f = 18.1 m/s
C) F_avg = 4130.7 N
Explanation:
A) From law of conservation of energy, we know that;
mgh = W_f + ½mv²
v is the speed at which she is going at the bottom of the slope.
Thus, making v the subject, we have;
v = √[2gh - (2W_f)/m)]
We are given;
h = 70 m
m = 58 kg
W_f = 1.04 × 10⁴ J = 10400 J
Thus;
v = √[(2 × 9.8 × 70) - (2 × 10400)/58)]
v = 31.8 m/s
B) Total force will be given by the formula;
F_t = F_k + F_r
Where;
F_k is force of kinetic friction = mg•μ_k
μ_k = 0.25
So, F_k = 58 × 9.8 × 0.25
F_k = 142.1 N
We are given force of air resistance(F_r) as 150 N
Thus;
F_t = 142.1 + 150
F_t = 292.1 N
Final velocity is gotten from the formula;
v_i² - v_f² = 2F_t•L/m
Thus;
v_f = √[v_i² - (2F_t•L/m)]
Where, v_i = 31.8 m/s, F_t = 292.1, m = 58 kg, L = 68 m
Thus;
v_f = √[31.8² - (2 × 292.1 × 68/58)]
v_f = 18.1 m/s
C) the average force exerted on her by the snowdrift as it stops her is given by the formula;
F_avg = m•v_f²/2l
F_avg = 58 × 18.1²/(2 × 2.3)
F_avg = 4130.7 N
What is the reactance of a 60.0- mH inductor when connected to an AC current source that has a frequency of 180 Hz
Answer:
67.9 Ω
Explanation:
L = 60 mH = 60 x 10⁻³ H = 0.060 H
f = 180 Hz, ω = 2πf = 1131 rad/s
reactance = ωL = 1131 x 0.060 = 67.9 Ω
A boat and a trailer are being pulled over an undulating road at a velocity v. The contour of the road is such that it can be approximated by a sine wave having a wavelength of l = 10 ft and an amplitude of Y = 0.5 in. The total static deflection of the springs and tires of the trailer due to the weight of the boat and trailer has been measured as 1.5 in. Assume that the damping inherent in the system is viscous and the damping ratio is 0.05. Determine: (a) The speed v at which the amplitude of the boat and the trailer will be a maximum; (b) The value of this maximum amplitude referred to in part (a); (c) The amplitude when the boat and trailer are traveling at the speed of 55 mph
Answer:
a) v = 25.54 ft/s
b) Xmax = 0.4180
c) Xmax = 0.0048 ft
Explanation:
a) determine the speed v at which the amplitude of the boat and the trailer will be a maximum;
to calculate the distance travelled
S = vt
given an expression of wavelength
l = vt
l = 2πv / ω
ω = 2πv / l
equation for counter of the road
y = Y sin ωt
= Y sin ( (2πv / l)t)
next we calculate the angular frequency
ω = √ ( k/m)
= √ ( g / Sst )
= √( 32.2 / ( 1.5/12))
= 16.05 rad/s
Now we calculate the speed v at which amplitude of the boat will be maximum
ω = 2πv / l
16.05 = (2 × π × v ) / 10
160.5 = 2πv
v = 160.5 / 2π
v = 25.54 ft/s
b)
we calculate the maximum amplitude using the following expression;
X/Y = [ √( 1 + ( 2Sr)²)] / [ √(( 1 - r²)² + ( 2Sr)²))]
Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×1)²)] / [ √(( 1 - 1²)² + ( 2×0.05×r)²))]
Xmax / 0.0416 = 1.004987 / 0.1
Xmax 0.1 = 0.0418
Xmax = 0.0418 / 0.1
Xmax = 0.4180
c)
we convert speed of the boat from mph velocity m/s
v = 55 mph × 1.46667ft/s / 1mph
v = 80.66 ft/s
next we calculate angular velocity
ω = 2πv / l
= 2π × 80.66 / 10
= 50.68 rad/s
next is the frequency ratio
r = 50.68 / 16.05
= 3.16
finally we find the amplitude of the boat
X/Y = [ √( 1 + ( 2Sr)²)] / [ √(( 1 - r²)² + ( 2Sr)²))]
Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×3.16)²)] / [ √(( 1 - 3.16²)² + ( 2×0.05×3.16)²))]
Xmax / 0.0416 = √1.0998 / √ ( 80.74 + 0.0998)
Xmax / 0.0416 = 1.0487 / 8.9910
Xmax 8.9910 = 0.0436
Xmax = 0.0436 / 8.9910
Xmax = 0.0048 ft
How is the speed of an object related to the amount of
kinetic energy the object has?
Answer:
directly proportional to the square of its speed.
Explanation:
;)
If a→=3i^-2j^-k^ and b→=i^+4j^+k^, find a unit vector n^ normal to the plane containing a→ and b→ such that a→, b→ and n^, in this order, form a right-handed system.
Since vector a = 3i - 2j - k and vector b = i + 4j + k, the unit vector, n, normal to the plane is (7i - j + 7k)/3√11 where a, b and n, in this order, form a right-handed system.
The normal vector to the planeThe normal vector to the plane is c = a × b
Since a = 3i - 2j - k and b = i + 4j + k,
c = a × b
c = (3i - 2j - k) × (i + 4j + k)
= 3i × i + (- 2j) × i + (-k) × i + 3i × 4j + (- 2j) × 4j + (-k) × 4j + 3i × k + (- 2j) × k + (-k) × k
= 3i × i - 2j × i - k × i + 12i × j - 8j × j - 4k × 4j + 3i × k - 2j × k - k × k
Given that
i × j = k, k × i = j, j × k = i, j × i = -k, i × k = -jk × j = -i, i × i = 0, j × j = 0 and k × k = 0. we havec = 3(0) - 2(-k) - (-j) + 12k - 8(0) - 16(-i) + 3(-j) - 2i - 0
c = 0 + 2k + j + 12k - 0 + 16i - 3j - 2i
c = 16i - 2i + j - 3j + 2k + 12k
c = 14i -2j + 14k
Unit vector normal to the plane nThe unit vector normal to the plane n = c/|c| where |c| is the magnitude of c = √(14² + (-2)² + 14²) = √(196 + 4 + 196)
= √396
= √(36 × 11)
= 6√11
So, n = c/|c|
= (14i -2j + 14k)/6√11
= 2(7i - j + 7k)/6√11
= (7i - j + 7k)/3√11
So, the unit vector normal to the plane is (7i - j + 7k)/3√11
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For the circuit shown in (Figure 1), find the potential difference between points a and b. Each resistor has
R = 160 N and each battery is 1.5 V
The potential difference between points a and b is zero.
Total emf of the series circuit
The total emf in the circuit is the sum of all the emf in the circuit.
emf(total) = 1.5 + 1.5 = 3.0 V
Potential differenceThe potential difference between two points, a and b is calculated as follows;
V(ab) = Va - Vb
V(ab) = 1.5 - 1.5
V(ab) = 0
Thus, the potential difference between points a and b is zero.
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An object whose weight is 10kg is placed on smooth plane inclined at 30° to the horizontal. find the acceleration of the object as it moves down the plane
please i need help
Answer:
4.9 m/s²
Explanation:
Draw a free body diagram. There are two forces on the object:
Weight force mg pulling straight down,
and normal force N pushing perpendicular to the plane.
Sum the forces in the parallel direction.
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 30°)
a = 4.9 m/s²
Two students are discussing how the speed of the car compares to the speed of the truck when both vehicles are in front of the house. Student 1 says, "The distance traveled by the car and the truck is the same, and the time is the same, so they must have the same speed." Student 2 says, "I don’t see how that can be. The car catches up to the truck, so the car has to be going faster."
a. Which aspects of Student 1’s reasoning, if any, are correct? Support your answer in terms of relevant features of your graphs in part (a).
b. Which aspects of Student 2’s reasoning, if any, are correct? Support your answer in terms of relevant features of your graphs in part (a).
c. Derive an expression for the acceleration of the car. Express your answer in terms of D and vt
d. Determine the time at which the speed of the car is equal to the speed vt of the truck. Express your answer in terms of tD. Justify your answer.
Answer:
a) Student 1 is right in the way he calculates the speed cid d of the truck, since going at constant speed
b) Student 2 is right that for the car to reach the truck it must have some relationship,
c) t = √d/a
Explanation:
In this exercise you are asked to analyze the movement in a mention using kinematics.
a) Student 1 is right in the way he calculates the speed cid d of the truck, since going at constant speed he can use the equation
v = d / t
b) Student 2 is right that for the car to reach the truck it must have some relationship, which is given by
v = v₀ + a t
x = v₀ t + ½ a t²
c) let's use the equations
v = d / t
v = v₀ + at
d / t = vo + at
If we can assume that the car starts from rest, so v₀ = 0
d / t = a t
t² = d / a
t = √d / a
if the speed of the car is not zero the result is a little more complicated
d = vo t + a t²
at² + vo t - d = 0
let's solve the quadratic equation
t = [-v₀ ± √ (v₀² + 4a d)] / 2a
Help as soon as possible
PLS
Answer:
B-A-C
Explanation:
The idea is to evaluate the slopes of the three graphs. Notice that the three graphs are a representation of position (x) as a function of time (t).
Then, the slope of those lines are giving information of the change in position over the change in time (rise over the run). For the graph with positive slope the velocity is positive, for those with negative slope (line going down) the velocity is negative. But we are asked to compare speeds, which are the magnitude of each velocity (slope) so it is important to determine the graph with the largest slope (graph C), and that with the smallest (graph B)
Then the order is: B-A-C (third option answer option)
How does the gravitational force that the Earth exerts on the Moon which has a much smaller mass compare to the gravitational force that the Moon exerts on the Earth?
a. The gravitational force the Moon exerts on the Earth is larger
b. The gravitational force the Earth exerts on the Moon is larger
C. Both the Earth and the Moon exerts equal gravitational force on each other.
I
Which scenario is an example of a scientific way of thinking ?
A example of a scientific way of thinking is making observations, forming questions, making hypotheses, doing an experiment, analyzing the data, and forming a conclusion.
This interaction is not possible in nature because it violates a conservation law. Which quantity is not conserved?
p+ e-+ --> π+n+ve
Answer:
its D i just did it on the app
A student conducts an investigation on electricity and magnetism. Which relationship will the student discover between the current and the magnetic field strength in a coiled wire?
Answer:
As current increases, the magnetic field strength increases.
So you're correct! :)
As the strength of the magnetic field increases, the current also increases.
From classical electromagnetism, we know that current is induced on conductor owing to relative motion between the conductor and the magnet. This is the principle upon which electromagnetic induction is based.
In the experiment, the student will find that the strength of the magnetic field is directly proportional to the induced current. Hence, as the strength of the magnetic field increases, the current also increases.
Learn more: https://brainly.com/question/9352088
a.
13. A planet orbiting the sun
Newton's 1st Law
b. Newton's 2nd Law
c. Newton's 3rd Law
Newtons first law of motion.
“Relative” is an important word. Block L of mass mL = 1.90 kg and block R of mass mR = 0.460 kg are held in place with a compressed spring between them. When the blocks are released, the spring sends them sliding across a frictionless floor. (The spring has negligible mass and falls to the floor after the blocks leave it.) (a) If the spring gives block L a release speed of 1.10 m/s relative to the floor, how far does block R travel in the next 0.740 s? (b) If, instead, the spring gives block L a release speed of 1.10 m/s relative to the velocity that the spring gives block R, how far does block R travel in the next 0.740 s?
Answer:m
R
ν
R
+m
L
ν
L
=0 ⇒ (0.500kg)ν
R
+(1.00kg)(−1.20m/s)=0
which yields ν
R
=2.40m/s . Thus , Δx=ν
R
t=(2.40m/s)(0.800s)=1.92m .
(b) Now we have m
R
ν
R
+m
L
(ν
R
−1.20m/s)=0 which yields
ν
R
=
m
L
+m
R
(1.2m/s)m
L
=
1.00kg+0.500kg
(1.20m/s)(1.00kg)
=0.800m/s .
Consequently , Δx=ν
R
t=0.640m .
Explanation:
One scientific investigation may result in
A.
further investigations.
B.
the generation of new ideas or phenomena to study.
C.
the development of new methods or procedures for investigation.
D.
all of these