All categories

3 years ago

A force of 20.0 N is applied to a 3.00 kg object for 4.00 seconds. Calculate the impulse experienced by the object.

Physics

1 answer:

GenaCL600 [577]3 years ago

7 0

Answer:

Impulse = 80Ns

Explanation:

Given the following data;

Mass = 3kg

Force = 20N

Time = 4 seconds

To find the impulse experienced by the object;

Impulse = force * time

Impulse = 20*4

Impulse = 80Ns

Therefore, the impulse experienced by the object is 80 Newton-seconds.

You might be interested in

If you fall and skid to a stop on a carpeted floor, you can get a rug burn. Much of this discomfort comes from abrasion,but ther

just olya [345]

Explanation:

When you run, your body has a kinetic energy and when you fall while running, the friction between the carpet and your foot, transforms the kinetic energy into thermal energy or heat energy. This can even cause, real burn if the skin were too hot.

3 0

4 years ago

The fastest man alive can run the 100 meter dash in 9.58 seconds, calculate average speed in meters per second

densk [106]

Answer:

100 ÷ 9.58 = 10.44 (approximate answer)

3 0

4 years ago

HELP QUICK!!!WILL GIVE BRAINLIEST!! According to the Sternberg's triarchic theory of intelligence, information synthesis is a st

maxonik [38]

Formulated in the 1980s by Robert Sternberg, the triarchic theory distinguishes three aspects of intelligence: analytic skills, such as the ability to think abstractly and evaluate information; creativity, the ability to invent novel solutions or ideas; and practical skills, which enable one to cope with concrete ...

4 0

3 years ago

Read 2 more answers

Study the interaction of the light rays with the mediums. Classify the medium in each image as being transparent or opaque.

Tresset [83]

Answer:

a) transparent

b) transparent

c) opaque

Explanation:

In the first one, the light rays go completely through, so it is transparent.

The second one I'm not too sure about. It is refraction so it's going through a different material, but the fact that it went through makes me say transparent.

Last one, the light rays are reflecting off the surface so it's opaque.

Please feel free to correct me if I'm wrong. This is just my understanding

6 0

3 years ago

Read 2 more answers

Two objects carry initial charges that are q1 and q2, respectively, where |q2| > |q1|. They are located 0.160 m apart and beh

mart [117]

Answer:

$\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges $\rm q_1$ and $\rm q_1$ , separated by a distance $\rm r$ , is given by

$\rm F = \dfrac{kq_1q_2}{r^2}.$

where k is the Coulomb's constant.

Initially,

$\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.$

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

$\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).$

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, $\rm F_f = +1.30\ N.$

The new charges on the two objects are

$\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.$

The new force is given by

$\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\$

Using (1),

$\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0$

$\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0$

$\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.$

Using (1),

When $\rm q_1 = \pm 8.00\times 10^{-7}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 8.00\times 10^{-7}\times 9\times 10^9}=\mp4.6\times 10^{-6}\ C.$

When $\rm q_1=\pm 4.6\times 10^{-6}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 4.64\times 10^{-6}\times 9\times 10^9}=\mp7.97\times 10^{-7}\ C\approx 8.0\times 10^{-7}\ C.$

Since, $\rm |q_2|>|q_1|$

Therefore, $\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

7 0

3 years ago

A force of 20.0 N is applied to a 3.00 kg object for 4.00 seconds. Calculate the impulse experienced by the object.​

A force of 20.0 N is applied to a 3.00 kg object for 4.00 seconds. Calculate the impulse experienced by the object.