All categories

3 years ago

When you are running around a track, what kind of energy are you using?

Physics

1 answer:

Sergeeva-Olga [200]3 years ago

7 0

Answer:

kinetic energy

Explanation:

we are using chemical energy in our bodies to produce movement, whitch in turn convents to warmth.

You might be interested in

Which units are used to measure both velocity and speed? Check all that apply. m/s d/t km/h mph lph

GaryK [48]

The units used to measure speed and velocity are as follows: m/s, Km/h, and mph.

5 0

4 years ago

Read 2 more answers

During the operation of a laser, as photons interacts with atoms inside the laser the photons _________.

Rashid [163]

are ejected as an avalnche from excited states

7 0

4 years ago

If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can st

liberstina [14]

Answer:

The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Explanation:

Given;

coefficient of kinetic friction, μ = 0.84

speed of the automobile, u = 29.0 m/s

To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration

x is the shortest distance

First we determine a;

From Newton's second law of motion

∑F = ma

F is the kinetic friction that opposes the motion of the car

-Fk = ma

but, -Fk = -μN

-μN = ma

-μmg = ma

-μg = a

- 0.8 x 9.8 = a

-7.84 m/s² = a

Now, substitute in the value of a in the equation above

v² = u² + 2ax

when the automobile stops, the final velocity, v = 0

0 = 29² + 2(-7.84)x

0 = 841 - 15.68x

15.68x = 841

x = 841 / 15.68

x = 53.64 m

Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

4 0

3 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

Ira Lisetskai [31]

The elastic potential energy (Ep) is given by $Ep = \frac{1}{2}*k*x^2$

Data:
Ep = 5184 J
k = 16200 N/m
x (displacement) = ?

Solving:
$Ep = \frac{1}{2}*k*x^2$
$5184 = \frac{1}{2}*16200*x^2$
$5184*2 = 16200x^2$
$10368 = 16200x^2$
$16200x^2 = 10368$
$x^2 = \frac{10638}{16200}$
$x^2 = 0.64$
$x = \sqrt{0.64}$
$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\quad\checkmark$

8 0

4 years ago

Read 2 more answers

A 4000-kg car bumps into a stationary 6000kg truck. The Velocity of the car before the collision was +4m/s and -1m/s after the c

Goryan [66]

Answer:

<em>The velocity of the truck is 3.33 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is

P=mv.

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2+...+m_nv_n$

If some collision occurs, the velocities change to v' and the final momentum is:

$P'=m_1v'_1+m_2v'_2+...+m_nv'_n$

In a system of two masses:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

There are two objects: The m1=4000 Kg car and the m2=6000 Kg truck. The car was moving initially at v1=4 m/s and the truck was at rest v2=0. After the collision, the car moves at v1'=-1 m/s. We need to find the velocity of the truck v2'. Solving for v2':

$\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2}$

Substituting:

$\displaystyle v'_2=\frac{4000*4+6000*0-4000(-1)}{6000}$

$\displaystyle v'_2=\frac{16000+4000}{6000}$

$\displaystyle v'_2=3.33$

The velocity of the truck is 3.33 m/s

7 0

3 years ago