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3 years ago

8

An electron is released 7.2 cm from a very long nonconducting rod with a uniform 6.3 μc/m. what is the magnitude of the electron

's initial acceleration?

1 answer:

nignag [31]3 years ago

6 0

Acceleration = charge*Electric field / mass

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The tangential acceleration of a cart moving at a constant speed in a horizontal circle is:

qaws [65]

Answer:

$a=0$ ∵ $\alpha=0\ rad.s^{-1}$

Explanation:

The tangential acceleration of a cart moving at a constant speed in a circle is:

The angular velocity is constant when the circular speed is constant.

We know that the (instantaneous) tangential velocity of such object is given by:

$v=r.\omega$

Now for angular acceleration we have a constant angular speed:

$\alpha=0\ rad.s^{-1}$

And angular acceleration is related to tangential acceleration as:

$a=r.\alpha$

$\Rightarrow a=0$

7 0

3 years ago

shtirl [24]

5.6 g/ml. That is the density.

5 0

3 years ago

Read 2 more answers

The acceleration of automobiles is often given in terms of the time it takes to go from 0 mi/h to 60 mi/h. One of the fastest st

Rudiy27

Answer:

9.934 m/s²

Explanation:

Given:

Initial speed of the Bugatti Veyron Super Sport = 0 mi/h

Final speed of the Bugatti Veyron Super Sport = 60 mi/h

Now,

1 mi/h = 0.44704 m / s

thus,

60 mi/h = 0.44704 × 60 = 26.8224 m/s

Time = 2.70 m/s

Now,

The acceleration (a) is given as:

$a=\frac{\textup{Change in speed}}{\textup{Time}}$

thus,

$a=\frac{26.8224 - 0 }{2.70}$

or

a = 9.934 m/s²

6 0

2 years ago

how much gravitational potential energy do you give a 70 kg person when you lift him up 3 m in the air?

SCORPION-xisa [38]

Given gravitational potential energy when he's lifted is 2058 J.

Kinetic energy is transferred to the person.

Amount of kinetic energy the person has is -2058 J

velocity of person = 7.67 m/s².

<h3>Explanation:</h3>

Given:

Weight of person = 70 kg

Lifted height = 3 m

1. Gravitational potential energy of a lifted person is equal to the work done.

$PE_g=W=m\times g\times h\\Acceleration due to gravity = g = 9.8 \ m/s^2 \\PE_g= m = m\times g\times h= 70\times 9.8 \times 3 = 2058\ kg.m/s^2 = 2058\ J$

Gravitational potential energy is equal to 2058 Joules.

2. The Gravitational potential energy is converted into kinetic energy. Kinetic energy is being transferred to the person.

3. Kinetic energy gained = Potential energy lost = $-PE_g = -2058\ kg.m/s^2$

Kinetic energy gained by the person = (-2058 kg.m/s²)

4. Velocity = ?

Kinetic energy magnitude= $\frac{1}{2} m\times v^2 = m\times g \times h$

Solving for v, we get

$v=\sqrt{2gh} =\sqrt{2\times 9.8 \times 3} = \sqrt{58.8} = 7.67 m/s^2$

The person will be going at a speed of 7.67 m/s².

4 0

3 years ago

Red: Distance = 3 x Time

sukhopar [10]

2 because that’s correct I I I I I I I I I I I I I I I I I I I

5 0

3 years ago