Ask question

Ask question

All categories

3 years ago

8

A tennis ball with mass m travelling west,hits a wall with a velocity of v.It bounces back with a velocity 1 ,63 v an easterly d

irection.The change in momentum of the tennis ball will be?

1 answer:

Pepsi [2]3 years ago

6 0

Answer:

163 v an easterly direction

You might be interested in

Search about non-ohmic devices in 3 pages

Tomtit [17]

Answer:

electronic diode,

Explanation:

Non-ohmic conductors are materials that do not obey ohm's law and they are electronic diode, transistors, tungsten, thermistors and vacuum tube etc.

7 0

3 years ago

Necesito ayudaaaaaa por favor

natima [27]

MAnswer:

Explanation:

4 0

3 years ago

Read 2 more answers

PLZZZZZ ANSWER Which example describes a nonrenewable resource?

Zepler [3.9K]

Answer:

I believe that the answer is D. There are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity.

Explanation:

Solar panels use the sun, and that is renewable.

The power plant uses tides and waves, they are renewable.

Windmills use wind, that is renewable.

So, the answer is D.

7 0

3 years ago

Read 2 more answers

Equal masses are suspended from two separate wires made of the same material. The wires have identical lengths. The first wire h

choli [55]

D. I think is the correct answer

6 0

3 years ago

Answer True or Flase1-Electric potential due to a uniform E field doesn’t change with location.2-The equipotential surfaces asso

TEA [102]

Answer:

1. False

2. True

3. True

Explanation:

1- False —> The relation between electric potential and electric field is given such that

$-\int\limits^a_b \vec{E}d\vec{l} = V_{ab}$

Therefore, for a uniform E field, electric potential is linearly proportional to the distance.

2- True —> The electric field lines always cross the equipotential lines perpendicularly.

3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$

There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.

3 0

3 years ago