Energy is the capacity to do what ? motion or work

The creation of electricity in a coil of a wire through the movement of a magnet is called what?

emmasim [6.3K]

Answer:

D. Electromagnetic Induction.

Explanation:

6 0

3 years ago

Moving by short leaps on one foot at a time.

Evgen [1.6K]

The answer is:

Hopping.

Explanation:

Hopping is done by taking off on one foot and landing back on that same foot. Hopping is done in shorter intervals, meaning you usually don't travel large distances through a single hop. Hopping is categorized as short leaps for its small distance covered singularly, and each hop is done only on one foot.

(Think about if one of your feet/legs is injured or asleep. If you wanted to go from say the living room to the kitchen, but didn't wish to move that leg or foot, you would likely hop on one foot to get to the destination.)

6 0

3 years ago

A sprinter accelerates at 7.5 m/s from rest in 2.0 s, what distance did she go? (15 m)

a_sh-v [17]

Answer:

The sprinter traveled a distance of 7.5 m

Explanation:

Motion With Constant Acceleration

It's a type of motion in which the rate of change of the velocity of an object is constant.

The equation that rules the change of velocities is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Using the equation [1] we can solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

The sprinter travels from rest (vo=0) to vf=7.5 m/s in t=2 s. Computing the acceleration:

$\displaystyle a=\frac{7.5-0}{2}$

$a=3.75\ m/s^2$

Now calculate the distance:

$\displaystyle x=0*2+\frac{3.75*2^2}{2}$

$\displaystyle x=7.5\ m$

The sprinter traveled a distance of 7.5 m

8 0

3 years ago

A car moving at 10 m/s crashes into a large bush and stops in 1.3 m. Using the Work-Energy theorem, calculate the average force

ad-work [718]

Answer:

The magnitude of the average force the seat belt exerts on the passenger is 2692.3 N.

It takes 0.26 s to bring the passenger in the car to a halt.

Explanation:

Hi there!

The negative work (W) needed to bring a moving object to stop is equal to its kinetic energy (KE):

W = KE

F · s = 1/2 · m · v²

Where:

F = applied force on the passenger.

s = displacement of the passenger.

m = mass of the passenger.

v = velocity of the passenger.

Solving the equation for F:

F = 1/2 · m · v² / s

Replacing with the data:

F = 1/2 · 70 kg · (10 m/s)² / 1.3 m

F = 2692.3 N

The magnitude of the average force the seat belt exerts on the passenger is 2692.3 N.

According to the second law of Newton:

F = m · a

Where "a" is the acceleration of the passenger.

We also know from kinematics that the velocity of an object can ve calculated as follows:

v = v0 + a · t

Where:

v = velocity of the passenger at time t.

v0 = initial velocity.

t = time.

a = acceleration.

When the passenger stops, its velocity is zero. So replacing a = F/m, let´s solve the equation for the time it takes the passenger to stop:

v = v0 + a · t

0 = 10 m/s + (-2692.3 N/ 70 kg) · t

-10 m/s / (-2692.3 N/ 70 kg) = t

t =0.26 s

It takes 0.26 s to bring the passenger in the car to a halt.

5 0

3 years ago

A truck tire rotates at an initial angular speed of 21.5 rad/s. The driver steadily accelerates, and after 3.50 s the tire's ang

erma4kov [3.2K]

Given:

initial angular speed, $\omega _{i}$ = 21.5 rad/s

final angular speed, $\omega _{f}$ = 28.0 rad/s

time, t = 3.50 s

Solution:

Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

$\alpha = \frac{\omega_{f} - \omega _{i}}{t}$

Now, putting the given values in the above formula:

$\alpha = \frac{28.0 - 21.5}{3.50}$

$\alpha = 1.86 m/s^{2}$

Therefore, angular acceleration is:

$\alpha = 1.86 m/s^{2}$

5 0

3 years ago