All categories

3 years ago

If a racquet ball hits the wall with 20 N of force, how much force will it bounce back with?

Physics

1 answer:

loris [4]3 years ago

8 0

Answer:

-20N

Explanation:

The racquet ball will bounce back with the same force.

This is in compliance with newton's third law of motion:

"action and reaction are equal but opposite"

If the ball hits the wall with an action force of 20N, the reaction force will be -20N.

The negative indicates an oppositely directed force.

You might be interested in

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

$\Delta P.E=mg(h_{2}-h_{1})$

$\Delta P.E=10000\times9.8\times(10000-0)$

$\Delta P.E=10000\times9.8\times10000$

$\Delta P.E=980000000\ J$

$\Delta P.E=980\ MJ$

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)$

$\Delta K.E=148298642\ J$

$\Delta K.E=148.3\ MJ$

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

7 0

3 years ago

Help please !! I need help with this!

belka [17]

Hello Again! I think the Answer might be 220 m! ( 1/2) ( 21 m/s + 0 m/s) (21 s) = 220 m

6 0

3 years ago

If the phase of the vibrating sources was changed so that they were vibrating completely out of phase, what effect would this ha

Over [174]

Answer:

There would be complete destructive interference.

Explanation:

This is because since the waves are completely out of phase, the phase difference is half wavelength, that is the phase angle is 180°. The vibrating sources are 180° out of phase with each other.

Since this is the case, the crest of the one source meets the trough of the other, this causes the resultant vibrational wave to cancel out, thus producing a destructive interference pattern.

Since the vibrating sources are completely out of phase, every point they meet is completely out of phase, so the resultant interference pattern would produce a complete destructive interference pattern of no wave.

4 0

3 years ago

Two identical conducting spheres are placed 80.0 cm apart. One is given a charge of 5.8 C and the other is given a charge of 6.4

Zepler [3.9K]

Answer: $5.214(10)^{11} N$

Explanation:

According to Coulomb's Law:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them".

Mathematically this law is written as:

$F_{E}=K\frac{q_{1}.q_{2}}{d^{2}}$

Where:

$F_{E}$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}=5.8 C$ and $q_{2}=6.4 C$ are the electric charges

$d=80 cm \frac{1 m}{100 cm}=0.8 m$ is the separation distance between the charges

Solving:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{(5.8 C)(6.4 C)}{(0.8 m)^{2}}$

$F_{E}=5.214(10)^{11} N$

7 0

3 years ago

Two objects of the same mass, but made of different materials, are initially at the same temperature. Equal amounts of heat are

adelina 88 [10]

Answer:

The final temperature of the two objects is the same.

Explanation:

The expression for the heat energy in terms of mass, specific heat and the change in the temperature is as follows:

$Q=mc(T_{f} -T_{i})$

Here, Q is the heat energy, m is the mass of the object, c is the specific heat and $T_{f},T_{i}$ are the final temperature and initial temperature.

According to the given question, Two objects of the same mass, but made of different materials, are initially at the same temperature. Equal amounts of heat are added to each object.

$Q=mc(T_{f}-T_{i})$ ............(1)

$Q=mc(T'_{f}-T_{i})$ .............(2)

From (1) and (2),

$mc(T_{f}-T_{i})=mc(T'_{f}-T_{i})$

$T_{f}-T_{i}=T'_{f}-T_{i}$

$T_{f}=T'_{f}$

Therefore, the final temperature of the two objects is the same.

6 0

3 years ago