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2 years ago

6

The electric potential at a point in space is -890 V. If a 0.0285 C charge is placed there, what will its potential energy U be?

Include the sign, + or - (Unit = J)

1 answer:

BabaBlast [244]2 years ago

5 0

Answer: -25.4

Explanation:

Acellus don’t forget the negative sign

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At what speed do a bicycle and its rider, with a combined mass of 100 kg, have the same momentum as a 1400 kg car traveling at 2

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A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc

max2010maxim [7]

The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

$f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }$

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

$f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }$

$f = \frac{1}{2\pi } ( 13.522)$

$f = 2.1535 Hz$

Hence,

The frequency of oscillation is 2.153 Hz

Learn more about frequency here:

<u>brainly.com/question/13978015</u>

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1 year ago

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Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs

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The motion of the ball on the vertical axis is an accelerated motion, with acceleration
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where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
$v_f =0$
So we can rewrite the equation as
$2(-9.81 m/s^2) h=-v_i^2$
from which we can isolate h
$h= \frac{v_i^2}{19.62}$ (1)

Now let's assume that $v_i$ is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: $2v_i$ . So the maximum height of the second ball is
$h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62}$ (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.

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