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2 years ago

A 1500 kg car

1 answer:

5 0

<h2>We determine the final velocity, </h2><h2>v</h2><h2>f</h2><h2>, of the car. We do this by applying the kinematic equation,</h2><h2 /><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>a</h2><h2>d</h2><h2 /><h2>where </h2><h2>v</h2><h2>i</h2><h2> is the initial velocity, a is the acceleration, and d is the distance covered across the movement. We can express the acceleration using the equation,</h2><h2 /><h2>a</h2><h2>=</h2><h2>F</h2><h2>m</h2><h2 /><h2>where F is the applied force and m is the mass of the car. We use the following values for the variables:</h2><h2 /><h2>v</h2><h2>i</h2><h2>=</h2><h2>5.25</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>F</h2><h2>=</h2><h2>1250</h2><h2> </h2><h2>N</h2><h2>m</h2><h2>=</h2><h2>1500</h2><h2> </h2><h2>k</h2><h2>g</h2><h2>d</h2><h2>=</h2><h2>42.8</h2><h2> </h2><h2>m</h2><h2>We proceed with the solution.</h2><h2 /><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>a</h2><h2>d</h2><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>F</h2><h2>d</h2><h2>m</h2><h2>Take the positive root.</h2><h2>v</h2><h2>f</h2><h2>=</h2><h2>√</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>F</h2><h2>d</h2><h2>m</h2><h2>=</h2><h2>√</h2><h2>(</h2><h2>5.25</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>)</h2><h2>2</h2><h2>+</h2><h2>2</h2><h2>(</h2><h2>1250</h2><h2> </h2><h2>N</h2><h2>)</h2><h2>(</h2><h2>42.8</h2><h2> </h2><h2>m</h2><h2>)</h2><h2>1500</h2><h2> </h2><h2>k</h2><h2>g</h2><h2>≈</h2><h2>9.9</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>Help improve Study.com. Report an Error</h2>

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The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a

bekas [8.4K]

Answer:

The work done on the canister by the 5.0 N force during this time is

54.06 Joules.

Explanation:

Let the initial kinetic energy of the canister be

KE₁ = $\frac{1}{2} mv_1^{2}$ = $\frac{1}{2} *3*3.6^{2}$ = 19.44 J in the x direction

Let the the final kinetic energy of the canister be

KE₂ = $\frac{1}{2} mv_2^{2}$ = $\frac{1}{2} *3*7.0^{2}$ = 73.5 J in the y direction

Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final

= 73.5 J - 19.44 J = 54.06 J

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3 years ago

Read 2 more answers

Is it possible to apply the same amount of force and do different amounts of work?

Tamiku [17]

Yes, with simple machines

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3 years ago

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur

sp2606 [1]

Answer:

a) q = -9.23 cm, b) h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

1 / f = 0.6 / 4 = 0.15

f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

1 / q = 1 / f - 1 / p

1 / q = 1 / 6.67 - 1/24

1 / q = 0.15 - 0.04167 = 0.10833

q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

M = h ’/ h = - q / p

h’= - q / p h

h’= - (-9.23) / 24.0 0.150

h’= 0.05759 cm

h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

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3 years ago

The diagram shows a tray of marbles being shaken from side to side. As this happens some of the marbles jump out of the tray.

irakobra [83]

The marbles that are 'more energetic' fall out of the tray, in the same way particles have enough energy to escape and turn into a gas.

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3 years ago

There are two possibilities for final stage of extremely massive stars. The first is a

makkiz [27]

Answer: Black hole.

Explanation:

As the massive star "compacts" under its own gravity, it triggers a massive supernova, after this point the remains of the star can become a neutron star, which is a very compact star made primarily, as the name says, of neutrons. The other possibility is a black hole, which is a finite region of space wherein it's interior there is a big concentration of mass, which creates a gravitational field strong enough that there is no particle that can escape it.

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3 years ago