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Katyanochek1 [597]
3 years ago
8

Suppose an empty grocery cart rolls downhill in a parking lot. The cart has a maximum speed of 1.3 m/s when it hits the side of

the store and comes to rest 0.30 s later. If an unbalanced force of —65 N stops the cart, what is the mass of the grocery cart?
Physics
1 answer:
Ket [755]3 years ago
4 0

The cart comes to rest from 1.3 m/s in a matter of 0.30 s, so it undergoes an acceleration <em>a</em> of

<em>a</em> = (0 - 1.3 m/s) / (0.30 s)

<em>a</em> ≈ -4.33 m/s²

This acceleration is applied by a force of -65 N, i.e. a force of 65 N that opposes the cart's motion downhill. So the cart has a mass <em>m</em> such that

-65 N = <em>m</em> (-4.33 m/s²)

<em>m</em> = 15 kg

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Answer:

a). 87.5 mA or 87.5 x10^{-3}A

b). 1.78 \frac{m}{s}

Explanation:

d=2.6 mm \\Q=420C\\t=80min\\n=5.8x10^{28} \\q=1.6x10^{-19}

n the number of free electrons is 28 in text reference and if they don't give q is take as the charge of electron.

a).

I=\frac{Q}{t}\\ I= \frac{420 C}{80 min}*\frac{1min}{60 s} =\frac{420 C}{4800s}\\  I=87.5 x10^{-3}A

b).

I=n*abs (q)*V_{d}*A

A= \pi * (\frac{d}{2})^{2} \\A=\pi (*\frac{2.6x10^{-3} m}{2})^{2}  \\A=5.309x10^{-6}

V_{d} =\frac{I}{n*abs(q)*A} \\V_{d}=\frac{87.5 x10^{-2} }{5.8x^10{28} *1.6x^{-19} *5.3x^{6} }\\V_{d}=1.78 \frac{m}{s}

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If the PLATE SEPARATION of an isolated charged parallel-plate capacitor is doubled: A. the electric field is doubled
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Explanation:

The electric field of an isolated charged parallel-plate capacitor is given by :

E=\dfrac{q}{A\epsilon_o}........(1)

Where

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It is clear from equation (1) that the electric field of a parallel plate capacitor is directly proportional to the charge on the plate and inversely proportional to the area of cross section of a plate.

So, the correct option is (E) i.e. "none of the above".

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Answer:

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Explanation:

Acceleration is given by

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u = 10 km/h \cdot \frac{1000 m/km}{3600 s/h}=2.78 m/s is the initial velocity

v = 8 km/h \cdot \frac{1000 m/km}{3600 s/h}= 2.22 m/s is the final velocity

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