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Tomtit [17]
3 years ago
12

A red supergiant is observed in a globular cluster in another galaxy (located 150,000 light years away). It is predicted, based

on its mass and age of the star that it will become supernova in about 10,000 years. When will this supernova be visible on Earth? a) 10,000 years from now. b) 10,000 years ago. c) 150,000 years ago. d) 150,000 years from now. e) 160,000 years from now.
Physics
1 answer:
Karolina [17]3 years ago
4 0

Answer:

e) 160,000 years from now.

Explanation:

The distance of the supergiant from us is

d = 150,000 light years

This means that the light emitted by the star takes 150,000 years to reach us.

We are told that the star will become a supergiant in

t = 10,000 years

from now. When this will happen, the light from the supernova will not reach us instantaneously, but it will take 150,000 years to travel the distance between us and the star.

Therefore, the supernova will be visible on Earth only after:

T = 10,000 y + 150,000 y = 160,000 years

So, option E) is correct.

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Suppose you want to use this human engine to lift a 2.35 kg box from the floor to a tabletop 1.30 m above the floor. How much mu
morpeh [17]

Answer:

The increase in the gravitational potential energy is 29.93 joules.

Explanation:

Given that,

Mass of the box, m = 2.35 kg

It is lifted from the floor to a tabletop 1.30 m above the floor, h = 1.3 m

We need to find the increase the gravitational potential energy. Initial it will placed at ground i.e. its initial gravitational potential is equal to 0. The increase in the gravitational potential energy is given by :

U=mgh

U=2.35\ kg\times 9.8\ m/s^2\times 1.3\ m

U = 29.93 Joules

So, the increase in the gravitational potential energy is 29.93 joules. Hence, this is the required solution.

6 0
4 years ago
2. What is stroboscopic motion? -​
Oduvanchick [21]

Answer: The illusion of motion that occurs when a stationary object is first seen briefly in one location and, following a short interval, is seen in another location.

Explanation:

6 0
4 years ago
HELP
Free_Kalibri [48]

Answer:

Explanation:

Although there is absolutely NO regard for significant digits, I can help you with this, nonetheless.

The equation for Potential Energy is PE = mgh. We have everything but the height of the ball. We have to solve for that using a one-dimensional motion equation:

v² = v₀² + 2aΔx, where Δx is our displacement (the height we need for PE). Filling in and keeping in mind that at the max height of parabolic travel, the final velocity of the object is 0:

0 = (21.5)² + 2(-9.8)Δx and

0 = 462.25 - 19.6Δx and

-462.25 = -19.6Δx so

Δx = 23.58 m. Using this as the h in our PE equation:

PE = .19(9.8)(23.58) so

PE = 43.9 J, choice C.

8 0
3 years ago
Which is an inertial reference frame (or at least a very good approximation of one)? Which is an inertial reference frame (or at
ioda

Answer:

A jet plane flying straight and at level at constant speed

Explanation:

     The<em> inertial frame </em>of reference is a frame of reference in which all <em>Newton law  is valid</em> ie Newton second law of motion and therefore newton first law of motion holds good. <em>The frame of reference does not accelerate.</em>

All the object that is in the frame of reference are at rest or moving with constant rectilinear motion with constant velocity unless acted upon by any force.

4 0
3 years ago
Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
4 years ago
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