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quester [9]
3 years ago
15

A 2,100 kg car drives east toward a 55 kg shopping cart that has a velocity of 0.50 m/s west. The two objects collide, giving th

e car a final velocity of 6.31 m/s east and the shopping cart a velocity of 13.5 m/s east. What was the
initial velocity of the car?
O A. 6.68 m/s east
OB. 7.19 m/s east
OC. 6.68 m/s west
OD. 7.19 m/s west
Physics
2 answers:
Marta_Voda [28]3 years ago
6 0

Answer:

C. 6.68m/s east

Explanation:

A p e x :)

Vanyuwa [196]3 years ago
6 0

Answer: C 6.68m/s west

Explanation:the west still dont change because its going the same way as the shopping cart is going

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Simple yet quick mechanical vibrations of various elastic substances produce sound. When they are moved or struck to vibrate, the auditory nerve of the ear receives the same sort of vibrations, which are subsequently perceived by the mind.

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X=16.2 cos(4.68t+0.420). What is the velocity of the oscillator at t=7.97 s?
pochemuha

Answer:

If using radians: 16.2

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Which best describes the definition for the atomic mass of an element
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Describe the difference between balanced forces and action/reaction forces
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A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and
Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

q_{out}=q_{2}-q_{1}

q_{out}=-6.43-(-4.00)

q_{out}=-2.43\ \mu C

The charge on the inner surface is q.

q+(-2.43)=-6.43

q=-6.43+2.43= 4.00\ \mu C

We need to calculate the electric field outside the shell

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}

E=33927618.73\ N/C

E=3.39\times10^{7}\ N/C

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

5 0
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