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3 years ago

A 1500 kg car, initially traveling at 22.0 m/s, hits its brakes and skids to a stop. Determine the work done by friction.

Physics

1 answer:

Sedbober [7]3 years ago

8 0

Answer:

The work done by the car is 363 kJ

Explanation:

Work : Work is said to be done when a Force moves an object through a certain distance. Work and Energy are interchangeable because they have the same unit. The unit of work is Joules (J).

Mathematically work done can be expressed as,

E = W = 1/2mv²

W = 1/2mv²................................ Equation 1

Where E = Energy, W = work done, m = mass of the car, v = velocity of the car

Given: m=1500 kg, v=22 m/s

Substituting these values into equation 1

W = 1/2(1500)(22)²

W = 750 × 484

W = 363000 J

W = 363 kJ

Thus the work done by the car is 363 kJ

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Two students, standing on skateboards, are initially at rest, when they give each other a shove! After the shove, one student (7

Elden [556K]

Answer:

The other student (59kg) moves right at 7.44 m/s

Explanation:

Given;

mass of the first student, m₁ = 77kg

mass of the second student, m₂ = 59kg

initial velocity of the first student, u₁ = 0

initial velocity of the second student, u₂ = 0

final velocity of the first student, v₁ = 5.7 m/s left

final velocity of the second student, v₂ = ? right

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(77 x 0) + (59 x 0) = (-77 x 5.7) + (59 x v₂)

0 = - 438.9 + 59v₂

59v₂ = 438.9

v₂ = 438.9 / 59

v₂ = 7.44 m/s to the right

Therefore, the other student (59kg) moves right at 7.44 m/s

6 0

3 years ago

The initial and final velocities of two blocks experiencing constant acceleration are respectively −7.45 m/s and 14.9 m/s. (a) T

ikadub [295]

Answer:

a) $a_{1}=3.7 m/s^{2}$

b) $a_{2}=3.68 m/s^{2}$

Explanation:

a) The displacement of the first object is 22.5 m, so we can use the next equation:

$v_{f}^{2}=v_{i}^{2}+2a\Delta x$

$a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}$

$a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.5}$

$a_{1}=3.7 m/s^{2}$

positive acceleration.

b) Using the same equation we can find the second value of the acceleration:

$a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}$

$a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.6}$

$a_{2}=3.68 m/s^{2}$

positive acceleration.

I hope it helps you!

8 0

3 years ago

"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two firs

Fantom [35]

Answer:

A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if

the dispersion is great

6 0

3 years ago

6. A girl pushes her little brother on his sled with a force of 300 N for 750 m.

Oduvanchick [21]

w=225000 J

E=225000 J

P=9000 W

P=22500 W

Ew=Fd=E

P=E/t

5 0

2 years ago

The SI unit that is used to measure time is the

Sunny_sXe [5.5K]

Answer:second

Explanation:

4 0

3 years ago

Read 2 more answers