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2 years ago

If an object's velocity changes from 25 meters per second to

Physics

2 answers:

damaskus [11]2 years ago

3 0

Answer:

Explanation:

the way to get b you have divion each other

11Alexandr11 [23.1K]2 years ago

3 0

Answer:

Explanation:

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What homemade things could I use for wheels on a small balloon powered car?

Fantom [35]

You can use mostly anything as long as it is circular. Depending on how big it is, you could use sturdy paper plates and use a stick/rod and tape to hold it together, or you could use bottle caps if the car you are trying to make is really small.

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3 years ago

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Which statement describes how the photoelectric effect causes an electric

alina1380 [7]

Answer:

A. Light acts as particles, causing electrons on the surface it strikes to be destroyed.

Explanation:

The photoelectric effect is a phenomenon that occurs when light shined onto a metal surface causes the ejection of electrons from that metal. It was observed that only certain frequencies of light are able to cause the ejection of electrons.

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3 years ago

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The drum of a washing machine spins at 200 rpm the drum had a radius of 0.5m what will the centripetal force experienced by a to

miskamm [114]

The centripetal force experienced by the towel is 55 N.

The given parameters;

angular speed of the washing machine, ω = 200 rpm
radius of the machine' drum, r = 0.5 m
mass of the towel, m = 0.25 kg

The centripetal force experienced by the towel spinning along the walls of the drum is calculated as follows;

Fc = mrω²

where;

<em>Fc is the centripetal force</em>

<em>ω is angular speed in rad/s</em>

The angular speed in rad/s is calculate as;

$\omega = \frac{200 \ rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} \\\\\omega = 20.95 \ rad/s$

The centripetal force experienced by the towel is calculated as;

$F_c = mr\omega ^2\\\\F_c = 0.25 \times 0.5 \times (20.95)^2\\\\F_c = 54.9 \ N \ \approx 55 \ N$

Thus, the centripetal force experienced by the towel is 55 N.

Learn more here: brainly.com/question/20905151

5 0

3 years ago

The equivalence between gravitational and inertial mass is explained in : the set of all of the events that happened on Earth

xxTIMURxx [149]

Answer:

The correct answer is theory of general relativity.

Explanation:

According to the statement of equivalence the gravitational mass force on an object standing on the surface of earth is same as the pseudo force that acts on it if it accelerated at acceleration equal to acceleration due to gravity.

According to Einestine both the forces are indistinguishable as both the forces produce same effects. Thus both are equivalent and thus gravity is a phenomenon that can be analysed in a radically different way which gives some strange results such as bending of light, existence of black holes,e.t.c

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3 years ago

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),

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

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

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