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astra-53 [7]
3 years ago
11

Dark matter is missing stars that have fallen from the sky. true or false

Physics
1 answer:
spayn [35]3 years ago
7 0

Answer:

The answer to this question is False

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6/10
g100num [7]

Answer:pounds

Explanation:

7 0
3 years ago
A 5kg mass is pushed with a force of 10N for a distance of 2.5 meters. The work done is​
Aliun [14]

W = 25 J

Explanation:

Work done on an object is defined as

W = Fd = (10\:\text{N})(2.5\:\text{m}) = 25\:\text{J}

7 0
3 years ago
A 1.0 kilogram ball is thrown into the air with an initial velocity of 30m/s. How much kinetic energy does the ball have? (B) ho
Gala2k [10]

Answer:

450 joules ; 450 joules ; 45.9 m

Explanation:

Given that :

Initial Velocity, u = 30m/s

Mass, m = 1 kg

Kinetic Energy of ball (KE) = 0.5mu²

K. E = 0.5 * 1 * 30^2

K.E = 0.5 * 900

K.E = 450 Joules

B.) Potential Energy (P. E)

P. E = mgh

At the highest point, all kinetic energy has would have become potential energy, hence

K. E = P. E = 450 Joules

C) Height of the ball :

From ; P. E = mgh

Where ; g = acceleration due to gravity = 9.8m/s² ; h = height

450 = 1 * 9.8 * h

450 = 9.8h

h = 450 / 9.8

h = 45.918

h = 45.9 m

7 0
3 years ago
Why does it rain a lot?
Vaselesa [24]

Answer:

Clouds are made up of tiny water droplets. ... As more and more droplets join together they become too heavy and fall from the cloud as rain. Warm air can hold more moisture than cool air. When the warmer air is cooled and the moisture condenses, it often rains more heavily.

Explanation:

hope it helps

3 0
3 years ago
A capacitor with an initial potential difference of 185 V is discharged through a resistor when a switch between them is closed
GrogVix [38]

Answer:

  • a. \tau =  2.1161 s
  • b. V(18.8 \ s) = 0.0256 \ V

Explanation:

<h3>a.</h3>

The equation for the voltage V of  discharging capacitor in an RC circuit at time t is:

V(t) = V_0 e^{(- \frac{t}{\tau}) }

where V_0 is the initial voltage, and \tau is the time constant.

For our problem, we know

V_0 = 185 \ V

and

V(10 \ s) = V_0 e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

So

185 \ V \ e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

e^{(- \frac{10 \ s}{\tau}) } = \frac{1.64 \ V}{ 185 \ V }

ln (e^{(- \frac{10 \ s}{\tau}) } ) = ln (\frac{1.64 \ V}{ 185 \ V })

- \frac{10 \ s}{\tau}  = ln (\frac{1.64 \ V}{ 185 \ V })

\tau =  \frac{-10 \s}{ln (\frac{1.64 \ V}{ 185 \ V }) }

This gives us

\tau =  2.1161 s

and this is the time constant.

<h3>b.</h3>

At t = 18.8 s we got:

V(18.8 \ s) = 185 \ V  \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 0.0256 \ V

4 0
3 years ago
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