All categories

3 years ago

How much thermal energy is created when a 3000 kg suv brakes to a stop from 20 m/s on a level road?

Physics

1 answer:

JulsSmile [24]3 years ago

5 0

Answer:

b. 600,000 J

Explanation:

Applying the law of conservation of energy,

The thermal energy created = Kinetic energy of the suv.

Q' = 1/2(mv²)............... Equation 1

Where Q' = Thermal energy, m = mass of the suv, v = velocity of the suv.

From the question,

Given: m = 3000 kg, v = 20 m/s

Substitute these values into equation 1

Q' = 1/2(3000×20²)

Q' = 600000 J

Hence the right option is b. 600,000 J

You might be interested in

You take your pulse and observe 80 heartbeats in a minute. What is the period of your heartbeat? What is the frequency of your h

Mama L [17]

Answer:

120 beats per minute.

Explanation:

6 0

3 years ago

Read 2 more answers

A vector has a magnitude of 82 m

pentagon [3]

The Y-component of the vector is 34.65 m.

<h3>What is a vector?</h3>

Vectors are quantities that have both magnitude and direction.

The Y-component of the vector can be calculated using the formula below.

Formula:

Y = rsin∅............. Equation 1

Where:

Y = Y-component of the vector
r = magnitude of the vector
∅ = angle of the vector to the horizontal.

From the question,

Given:

r = 82 m
∅ = 25°

Substitute these given values into equation 1

Y = 82(sin25°)
Y = 34.65 m

Hence, The Y-component of the vector is 34.65 m.
Learn more about vectors here: brainly.com/question/24855749

7 0

2 years ago

49W

Nikolay [14]

Answer:

Explanation:

The magnitude of the electrostatic force between two charged particles is given by

$F=\frac{kq_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the charges of the two particles

r is the separation between the particles

In this problem, the initial force between the particles is F.

Later, the distance between the two particles is increased by four, so

r' = 4r

So, the new force between the particles will be

$F'=\frac{kq_1 q_2}{(4r)^2}=\frac{kq_1 q_2}{16r^2}=\frac{1}{16}F$

So, the new force decreases by a factor of 16.

4 0

3 years ago

An 1120 kg car traveling at 17.2 m/s is brought to a stop while skidding 40m. Calculate the work done on the car by the friction

Nesterboy [21]

Answer:

Work = 165670.4 J = 165.67 KJ

Explanation:

First, we will find the deceleration of the car, using the 3rd equation of motion:

$2as = v_{f}^2 - v_{i}^2\\$

where,

a = deceleration = ?

s = skid distance = 40 m

vf = final speed = 0 m/s

vi = initial speed = 17.2 m/s

Therefore,

$2a(40\ m) = (0\ m/s)^2 - (17.2\ m/s)^2\\a = - 3.698\ m/s^2$

the negative sign indicates deceleration here.

Now, we will calculate the braking force applied by the brakes on the car:

$F = ma\\F = (1120\ kg)(-3.698\ m/s^2)\\F = - 4141.76\ N$

the negative sign indicates braking force.

Now, we will calculate the work done using the magnitude of this force:

$Work = |F|s\\Work = (4141.76\ N)(40\ m)\\$

8 0

3 years ago

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t

Jet001 [13]

<h2>The impact is occurred at a distance 348.55 m far from person.</h2>

Explanation:

Let s be the distance to the impact position and t be the time when he hears the sound though concrete after impact.

Time when he hears impact through air = t + 0.90

The distance traveled by sound wave in both concrete and air is s.

Speed of sound in air = 343 m/s

Speed of sound in concrete = 3000 m/s

Distance traveled in air = Distance traveled in concrete

343 x Time when he hears impact through air = 3000 x Time when he hears impact through concrete

343 x (t + 0.90) = 3000 x t

3000t - 343 t = 343 x 0.90

2657t = 308.7

t = 0.116 s

Distance traveled in concrete = 3000 x t = 3000 x 0.116 = 348.55m

So the impact is occurred at a distance 348.55 m far from person.

4 0

3 years ago