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2 years ago

Worth 20 points Please answer showing the conversions

Physics

1 answer:

vladimir1956 [14]2 years ago

4 0

Answer:

b) 18 m/s is the correct answer.

Explanation:

initial velocity(u) = 30m/s

acceleration(a) = -4m/s²

time(t) = 3 seconds

final velocity(v) = ?

using the formula,

a = (v-u)/t

or, -4 = (v-30)/3

or, -12 = v-30

or, v = -12+30

so, v = 18 m/s

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Will water pressure be greater at the bottom of a container that you put a hole in or at the top of a container?

ddd [48]

Pressure=hrg
pressure is great at the bottom because the weight lf water exerts pressure below the container due to the gravitational pull of the earth.

4 0

4 years ago

A train leaves the train station at noon and travels at a constant speed of vt = 50 mi/hr on a straight track. 2 hr later, a car

asambeis [7]

Answer:

350 miles

Explanation:

When the car starts 2 hours later, the train would have a head start of

50 * 2 = 100 miles

The speed of the car relative to the train is

70 - 50 = 20 mi/hr

For the car to catch up with the train, it must cover the 100 miles difference at the rate of 20mi/hr. So the time it would need to cover this difference is

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4 years ago

A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is<u> positive.</u>

<u></u>

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is <u>positive.</u>

<u></u>

<u></u>

<u>Given:</u>

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$