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2 years ago

Which describes electricity?

Physics

1 answer:

SVETLANKA909090 [29]2 years ago

7 0

Answer:

all of the above

Explanation:

- a build up of electric charge

- the force and motion of electrically charged particles

- an electric current

are three different ways to describe electricity.

So the answer is all of the above.

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When the mass of an object increases, the forcé of gravity<br>

diamong [38]

Answer:

increace

Explanation:

they are both going up

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2 years ago

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A mass m = 75 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.2 m and finally

dalvyx [7]

Answer:

The velocity is 19.39 m/s

Solution:

As per the question:

Mass, m = 75 kg

Radius, R = 19.2 m

Now,

When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.

$F_{C} = F_{G}$

$\frac{mv^{2}}{R} = mg$

where

v = velocity

g = acceleration due to gravity

$v = \sqrt{2gR} = \sqrt{2\times 9.8\times 19.2} = 19.39\ m/s$

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2 years ago

The de Broglie wavelength of an electron traveling at 7.0 x 107m/s is 1.0 x 10-11m. (Remember that me = 9.1 x 10-31kg and h = 6.

ivolga24 [154]

The De Broglie's wavelength of a particle is given by:

$\lambda=\frac{h}{p}$

where

$h=6.6 \cdot 10^{-34} Js$ is the Planck constant

p is the momentum of the particle

In this problem, the momentum of the electron is equal to the product between its mass and its speed:

$p=m_e v=(9.1 \cdot 10^{-31} kg)(7.0 \cdot 10^7 m/s)=6.4 \cdot 10^{-23} kg m/s$

and if we substitute this into the previous equation, we find the De Broglie wavelength of the electron:

$\lambda=\frac{h}{p}=\frac{6.6 \cdot 10^{-34} Js}{6.4 \cdot 10^{-23} kg m/s}=1.0 \cdot 10^{-11} m$

So, the answer is True.

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2 years ago

A bowling ball has a mass of 5 kg. A student rolls the bowling ball one time at 1 m/s and a second time and 2 m/s. Compare the m

sasho [114]

The answer is 5kg m/s with the second momentum being 10 kg m/s. i took the test and it was right. hope this helps yalls

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2 years ago

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$