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Kazeer [188]
3 years ago
13

Distant galaxies appear to be much larger than those nearby. true or false

Physics
1 answer:
jonny [76]3 years ago
3 0

Answer:True

Explanation: Distant galaxies appear to be larger than those close by because of cosmic expansion. light from nearer galaxies travel farther and makes the other to be billions of years away. The farther it takes to get to another galaxy, the bigger it appears

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I desperately need help with my Physics Exam I am failing this class will mark Brainliest to whoever helps the most: PART 1 (I d
GREYUIT [131]
  1. Option 2nd is correct

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2 years ago
PLease Help!
scoundrel [369]

Answer:

Option A is the correct answer.

Explanation:

The instantaneous acceleration = Change in velocity in velocity/Time taken

The slope of the graph should give instantaneous acceleration.

 Slope of a graph = Change in value of Y -axis / Change in values of X -axis

 Comparing both the equations

   Change in value of Y -axis = Change in velocity in velocity

   Change in values of X -axis = Time taken

So velocity values should be on the Y axis and Time values should be on the X axis.

Option A is the correct answer.

8 0
4 years ago
BRAINLIEST FOR THE FIRST TO ANSWER
Sonja [21]

Answer:

Wrong its B Use a different amount of mass in the cart for five different trials, roll the cart down a ramp with the same slope for each trial, and measure how long it takes the cart to roll one meter each time.

Explanation:

8 0
3 years ago
Why doesn't the motor work?
exis [7]

Answer:

c

Explanation: its weird

8 0
3 years ago
Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig
HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

y=-\frac{1}{2} gt^2+v_0t+y_0

With the given values in the question the equation has one unknown v₀:

v_0=\frac{y-y_0}{t}+\frac{1}{2}gt

Solving for t=1:

1) v_0=y-y_0+\frac{g}{2}

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) E=\frac{1}{2}m{v_0}^2+mgy_0

If h is the height of the tower, the energy on top of the tower:

3) E=mgh

Combining equation 2 and 3 and solving for h:

4) h=\frac{{v_0}^2}{2g}+y_0

Combining equation 1 and 4:

h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0

4 0
3 years ago
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