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3 years ago

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A cart falls from a track with an acceleration of 10 m/s2 (which is always the acceleration due to gravity). If it falls for 9 s

econds before hitting the ground, what is it’s velocity at impact? (show work plz)

1 answer:

schepotkina [342]3 years ago

8 0

Answer:

90m/s

Explanation:

Given parameters:

Acceleration = 10m/s²

Time of fall = 9s

Unknown:

Final velocity = ?

Solution:

We can assume that the cart falls from rest.

Initial velocity = 0m/s

Using

v = u + gt

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

t is the time

v = 0 + 10 x 9 = 90m/s

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Answer:

Explanation:

Let the magnitude of magnetic field be B .

flux passing through the coil's = area of coil x field x no of turns

Φ = 3.13 x 10⁻⁴ x B x 135 = 422.55 x 10⁻⁴ B .

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current i = dΦ /dt x 1/R

charge through the coil = ∫ i dt

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= 1 / R ∫ dΦ

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Total resistance R = 61.1 + 44.4 = 105.5 ohm .

3.44 x 10⁻⁵ = 422.55 x 10⁻⁴ B / 105.5

B = 3.44 x 10⁻⁵ x 105.5 / 422.55 x 10⁻⁴

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3 years ago

A baseball pitcher throws a baseball horizontally at a linear speed of 49.4 m/s. Before being caught, the baseball travels a hor

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Answer:

$v = 3.61 m/s$

Explanation:

As we know that ball travels horizontal distance of 24.7 m with uniform speed 49.4 m/s

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4 years ago

A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni

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Answer:

Induced emf, $\epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}$

Explanation:

The varying magnetic field with time t is given by according to equation as :

$B=B_{max}e^{-t/\tau}$

Where

$B_{max}\ and\ t$ are constant

Let $\epsilon$ is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=-A\dfrac{d(B)}{dt}$

$\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}$

$\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}$

So, the induced emf in the loop as a function of time is $A\dfrac{B_{max}e^{-t/\tau}}{\tau}$ . Hence, this is the required solution.

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3 years ago

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal

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Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

$v=\sqrt{v_x^2+v_y^2}$ (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

$v_y^2=v_{oy}^2+2gh$ (2)

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g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

$v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}$

vx is calculated by using the information about the horizontal range of the ball:

$R=v_o\sqrt{\frac{2h}{g}}$ (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

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The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

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Answer:

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b )

When power is switched off , it will decelerate because of frictional torque .

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3 years ago