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nikklg [1K]
3 years ago
5

A cell phone is released from the top with the speed of 10ms what is the speed 3s after?

Physics
1 answer:
sergeinik [125]3 years ago
6 0

Answer:

30ms

Explanation:

you need to multiple the 10ms by 3s which gives you 30ms

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What force is required to accelerate a body with the mass of 15 kilograms at a rate of 8m/s2?
Tom [10]
Force = mass * acceleration
F = 15 *8 = 120 Newton
8 0
3 years ago
This is “Fusion Reactions”.<br> Please answer number 8. Thank you.
Angelina_Jolie [31]

Answer:

²₁H + ³₂He —> ⁴₂He + ¹₁H

Explanation:

From the question given above,

²₁H + ³₂He —> __ + ¹₁H

Let ⁿₐX be the unknown.

Thus the equation becomes:

²₁H + ³₂He —> ⁿₐX + ¹₁H

We shall determine, n, a and X. This can be obtained as follow:

For n:

2 + 3 = n + 1

5 = n + 1

Collect like terms

n = 5 – 1

n = 4

For a:

1 + 2 = a + 1

3 = a + 1

Collect like terms

a = 3 – 1

a = 2

For X:

n = 4

a = 2

X =?

ⁿₐX => ⁴₂X => ⁴₂He

Thus, the balanced equation is

²₁H + ³₂He —> ⁴₂He + ¹₁H

8 0
3 years ago
Photovoltaic cells: a) have become more economical to produce and use over the past 25 years. b) are the most efficient means of
Yanka [14]

Photovoltaic cells are the most efficient means of converting solar energy to electricity. Option b is correct.

<h3>What is a cell?</h3>

A cell is a voltage and current-producing device that consists of a single anode and cathode separated by an electrolyte.

One or more cells can make up a battery. One cell, for example, is one AA battery.

Light intensity on a solar cell is often measured in "suns," with one sun roughly equivalent to 1 kW/m².

Concentrated sunlight improves the ratio of current generated while the device is lighted vs when it is dark, hence enhancing output voltage and efficiency.

Photovoltaic cells are the most efficient means of converting solar energy to electricity.

Hence, option b is correct.

To learn more about the cell refer to:

brainly.com/question/3142913

#SPJ1

8 0
2 years ago
What is the velocity of the object?
dmitriy555 [2]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2>\huge\boxed{\text{V = 9.5 m/s}}</h2><h2>_____________________________________</h2>

<h2>DATA:</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

{\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

                           W\ =\ \frac{1}{2}\:6\:x\:30

                           W   =  90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2

\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}

\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

8 0
3 years ago
A drag racer starts her car from rest and accelerates at 5.5 m/s2 for the entire distance of 523 m. How long did it take the car
ANEK [815]

Answer:

t = 13.7 s or t = 14 s with proper significant figures

Explanation:

The initial speed is 0 m/s since the car starts from rest, acceleration is 5.5 m/s2 and distance is 523 m.

Since we have initial speed, acceleration and distance we can use the following formula to find the time. We can now use algebra to work out our answer.

d = vt + \frac{1}{2}at²

523 = (0)t + (\frac{1}{2})(5.5)t²

523 = 2.8t²

186.8 = t²

13.7 s = t

(t = 14 s with proper significant figures)

3 0
3 years ago
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