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3 years ago

5

a car of mass 1,000 kilograms is moving initially at the speed of 22 meters/second. when the brakes are applied, it takes the ca

r 3.0 seconds to stop. what is the force required to stop the car?

1 answer:

Evgen [1.6K]3 years ago

5 0

It i believe it would be 7.3 × 10∧3.

correct me if im wrong

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Frank has a paperclip. It has a mass of 12g and a volume of 3cm3. What is its density?

Mkey [24]

Answer:

D = 4 g/cm³

Explanation:

Density = Mass / Volume

Step 1: Define

D = x

M = 12 g

V = 3 cm³

Step 2: Substitute

D = 12g/3 cm³

Step 3: Simplify

D = 4g / cm³

4 0

3 years ago

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

uranmaximum [27]

Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

7 0

3 years ago

Consider a box sitting in the back of a pickup. The pickup accelerates to the right, and because the bed of the pickup is sticky

andreyandreev [35.5K]

The force would be coming from the right causing the box the lean/ slide to left, if it wasnt sticky.

8 0

3 years ago

What is happening to the water molecules as it changes from ice to liquid water? *

g100num [7]

Answer:

They are gaining energy, can move and slide past each other.

Explanation:

Because as ice melts, the molocules gain energy/heat which warm them up and are able to slide/move apart from each other.

7 0

3 years ago

The power source with the lowest environmental impact is

shutvik [7]

Nuclear Fission s a power source with a very low environmental impact.

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3 years ago