Sound is an electromagnetic wave.

A ball strikes a wall. It exerts a force of 10 N to the left against the wall and bounces off. What force does the wall exert on

Galina-37 [17]

The wall exerts a force of equal magnitude but in the opposite direction. So the force by the wall is 10 N to the right.

6 0

3 years ago

Read 2 more answers

The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh

Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

$v^2=v^2_o-2ax$

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

$\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}$

Next, consider that the formula for the force is:

$F=ma$

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

$F=(820kg)(0.66\frac{m}{s^2})=542.20N$

Hence, the required force to stop the car is 542.20N

4 0

1 year ago

According to Newton's 3rd Law, an object's momentum depends on it's velocity and mass. 4 dog-sled teams competed to see who coul

podryga [215]

Answer:

C.) Sled Team C 28 kg moving at 12m/s

I'm pretty sure.

7 0

3 years ago

A robot arm moves so that P travels in a circle about point B, which is not moving. Knowing that P starts from rest, and its spe

algol13

Answer

Acceleration=>10.20mm/s

Explanation

Below in image

4 0

3 years ago

How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest? a.3.41 x 10^5 J b.2.73 x 10^5 J

algol13

Answer:

Work done by the frictional force is $3.41\times 10^5\ J$

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Initial velocity of car, u = 26.1 m/s

Finally, it comes to rest, v = 0

We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.

$W=k_f-k_i$

$W=\dfrac{1}{2}m(v^2-u^2)$

$W=\dfrac{1}{2}\times 1000\ kg(0^2-(26.1\ m/s)^2)$

W = −340605 J

or

$W=3.41\times 10^5\ J$

Hence, the correct option is (a).

6 0

3 years ago