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3 years ago

A 750 kg car moving at 25 m/s has Kinetic Energy. To stop the car requires work. On snowy roads, the force of friction that can

Physics

1 answer:

liraira [26]3 years ago

4 0

a. By the work-energy theorem, the total work required to stop the car is equal to the change in its kinetic energy,

W = 0 - 1/2 (750 kg) (25 m/s)² ≈ -230 kJ

b. The car covers a distance x as it stops such that

W = (-725 N) x ==> x ≈ 320 m

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What is a scientific model?Question 2 options:A group of objects that interact with each other.A complex system made of many sma

HACTEHA [7]

Answer:

A scientific model is a way of representing a system to better understand it's behavior

Explanation:

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3 years ago

A 71-kg swimmer dives horizontally off a 500-kg raft. If the diver's speed immediately after leaving the raft is 6m/s, what is t

Sloan [31]

Answer:

The answer is below

Explanation:

Momentum is used to measure the quantity of motion in an object. Momentum is the product of mass and velocity.

Momentum = mass * velocity

The principle of conservation of momentum states that momentum cannot be created or destroyed but can be transferred. Therefore the momentum before and after an action is equal.

Initial momentum = Final momentum

Let m be the mass of the diver, M be the mass of the raft, u be the initial velocity of the diver, U be the initial velocity of the raft, v be the final velocity of the diver and V be the final velocity of the raft.

m = 71 kg, M = 500 kg, v = 6 m/s

Initial both the raft and diver are at rest, hence u and U is zero, hence:

mu + MU = mv + MV

71(0) + 500(0) = 71(6) + 500(V)

0 = 426 + 500(V)

500(V) = -426

V = -426/500

V = -0.852 m/s

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3 years ago

When a falling meteoroid is at a distance above the earth's surface of 2.60 times the earth's radius, what is its acceleration d

Mice21 [21]

The gravitational acceleration at any distance r is given by

$g= \frac{GM}{r^2}$

where G is the gravitational constant, M the Earth's mass and r is the distance measured from the center of the Earth.

The Earth's radius is $r_e=6.37 \cdot 10^6 m$ , so the meteoroid is located at a distance of:

$r=r_e+2.60 r_e =3.60 r_e = 2.29 \cdot 10^7 m$

And by substituting this value into the previous formula, we can find the value of g at that altitude:

$g= \frac{GM}{r^2} = \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(2.29 \cdot 10^7 m)^2} =0.75 m/s^2$

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3 years ago

Read 2 more answers

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$