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gizmo_the_mogwai [7]
2 years ago
8

What is the energy equivalent of an object with a mass of 1. 05 g? 3. 15 Ă— 105 J 3. 15 Ă— 108 J 9. 45 Ă— 1013 J 9. 45 Ă— 1016 J

.
Physics
1 answer:
Molodets [167]2 years ago
5 0

Considering the equivalence between mass and energy given by the expression of Einstein's theory of relativity, the correct answer is the last option: the energy equivalent of an object with a mass of 1.05 kg is 9.45×10¹⁶ J.

The equivalence between mass and energy is given by the expression of Einstein's theory of relativity, where the energy of a body at rest (E) is equal to its mass (m) multiplied by the speed of light (c) squared:

E=m×c²

This indicates that an increase or decrease in energy in a system correspondingly increases or decreases its mass, and an increase or decrease in mass corresponds to an increase or decrease in energy.  

In other words, a change in the amount of energy E, of an object is directly proportional to a change in its mass m.

In this case, you know:

  • m=1.05 kg
  • c= 3×10⁸ m/s

Replacing:

E= 1.05 kg× (3×10⁸ m/s)²

Solving:

<u><em>E= 9.45×10¹⁶ J</em></u>

Finally, the correct answer is the last option: the energy equivalent of an object with a mass of 1.05 kg is 9.45×10¹⁶ J.

Learn more:

  • brainly.com/question/9477556
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ANSWER FAST PLEASE HELP
KengaRu [80]

Answer:

B. 175 N

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

Fnet = Fapp + Fg

Where;

Fnet is the net force

Fapp is the applied force

Fg is the force due to gravitation

In this scenario, we observed that both forces are acting in the same direction.

Therefore:

Net force = 100 N + 75 N

Net force = 175 Newton

6 0
3 years ago
In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp
MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

V_{Ax}=V_{A}cos\theta

V_{Ax}=(21m/s)cos(-14^{o})

V_{Ax}=20.38 m/s

V_{Ay}=V_{A}sin\theta

V_{Ay}=(21m/s)sin(-14^{o})

V_{Ay}=-5.08 m/s

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

t= -0.9159s    or   t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

x_{B}=V_{Bx}t

x_{B}=(20.88m/s)(0.468s)

x_{B}=9.77m

So once we got the two distances we can now find the difference between them:

x_{B}-x_{A}=9.77m-6.46m=3.31m

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0
3 years ago
What H &amp; N can friction produce?
Angelina_Jolie [31]
Friction produces heat hope this helps

5 0
3 years ago
If a photon of Light leaves the sun than reachers the earth 494 second later. What is the distance from the earth to sun
denis-greek [22]

The distance from the Earth to the Sun is 92.96 million mi.

4 0
3 years ago
Read 2 more answers
Joe drove at the speed of 45 miles per hour for a certain distance. He then drove at the speed of 55 miles per hour for the same
Snowcat [4.5K]

Answer:

v_{avg} = 49.5 mph

Explanation:

Let the distance moved by Joe is "d"

so the time taken by him to drove it by speed 45 mph is given as

t_1 = \frac{d}{v_1}

t_1 = \frac{d}{45}

now the same distance is traveled by him with speed 55 mph

so the time taken by him

t_2 = \frac{d}{55}

so total time taken by him for complete distance 2d

t = t_1 + t_2

t = \frac{d}{45} + \frac{d}{55}

t = 0.0404 d

now the average speed is given as

v_{avg} = \frac{2d}{t}

v_{avg} = \frac{2d}{0.0404d}

v_{avg} = 49.5 mph

5 0
3 years ago
Read 2 more answers
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