<h3><u>Answer;</u></h3>
C) He could injure or pull a muscle.
<h3><u>Explanation</u>;</h3>
- <em><u>When muscles are stretched the muscle fibers are temporary lengthened. Muscles have a unique ability to undergo contraction and lengthening since they are elastic.</u></em>
- <em><u>When the muscles are warm they are more elastic, therefore, warming up the muscles is very important for the purpose of avoiding injuries.</u></em> Ir is important to engage in moderate cardiovascular warm-up prior to stretching, which increases the blood flow to the active and thus avoiding injuries to the muscles.
- Stretching muscles when they are cold may bring injuries or cause muscle pulls, which are painful.
Answer:
<em>The mass of the object is 745000 units of the sun</em>
Explanation:
We know that the centripetal force with which the stars orbit the object is represented as
= 
and this centripetal force is also proportional to
= 
where
m is the mass of the stars
M is the mass of the object
v is the velocity of the stars = 10^6 m/s
r is the distance between the stars and the object = 10^14 m
k is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2
We can equate the two centripetal force equations to give
= 
which reduces to
= 
and then finally
M = 
substituting values, we have
M =
= 1.49 x 10^36 kg
If the mass of the sun is 2 x 10^30 kg
then, the mass of the the object in units of the mass of the sun is
==> (1.49 x 10^36)/(2 x 10^30) = <em>745000 units of sun</em>
Out of the given options, weight is influenced by mass and gravity
Answer: Option A
<u>Explanation:
</u>
The object's mass is defined as the quantity of a matter with which the object is formed. It can change its state of matter but the quantity will remain the same. However, the weight is defined as how much force gravity exerts on the object's mass to pull it.
The mass is always same irrespective the location but the weight may vary from one place to the other while talking for the bigger picture. For example, the object's weight may be 60 kg on Earth but when it is measured on the moon, it will be lesser.
The weight of an object generally has nothing doing with the volume and it doesn't depend solely on the gravitational pull. The mass plays a crucial role.

Answer:
t = 4.08 s
R = 40.8 m
Explanation:
The question is asking us to solve for the time of flight and the range of the rock.
Let's start by finding the total time it takes for the rock to land on the ground. We can use this constant acceleration kinematic equation to solve for the displacement in the y-direction:
We have these known variables:
- (v_0)_y = 0 m/s
- a_y = -9.8 m/s²
- Δx_y = -20 m
And we are trying to solve for t (time). Therefore, we can plug these values into the equation and solve for t.
- -20 = 0t + 1/2(-9.8)t²
- -20 = 1/2(-9.8)t²
- -20 = -4.9t²
- t = 4.08 sec
The time it takes for the rock to reach the ground is 4.08 seconds.
Now we can use this time in order to solve for the displacement in the x-direction. We will be using the same equation, but this time it will be in terms of the x-direction.
List out known variables:
- v_0 = 10 m/s
- t = 4.08 s
- a_x = 0 m/s
We are trying to solve for:
By using the same equation, we can plug these known values into it and solve for Δx.
- Δx = 10 * 4.08 + 1/2(0)(4.08)²
- Δx = 10 * 4.08
- Δx = 40.8 m
The rock lands 40.8 m from the base of the cliff.
The phases of the moon are the changing appearances of the moon, as seen from Earth. Which phase happens immediately after a third quarter moon are the following
Explanation:
- After the full moon (maximum illumination), the light continually decreases. So the waning gibbous phase occurs next. Following the third quarter is the waning crescent, which wanes until the light is completely gone -- a new moon.
waning gibbous phase
- The waning gibbous phase occurs between the full moon and third quarter phases. The last quarter moon (or a half moon) is when half of the lit portion of the Moon is visible after the waning gibbous phase.
Time takes by the moon to go through all the phases
about 29.5 days
- It takes 27 days, 7 hours, and 43 minutes for our Moon to complete one full orbit around Earth. This is called the sidereal month, and is measured by our Moon's position relative to distant “fixed” stars. However, it takes our Moon about 29.5 days to complete one cycle of phases (from new Moon to new Moon).
- At 3rd quarter, the moon rises at midnight and sets at noon. Then we see only a crescent. At new, the moon rises at sunrise and sets at sunset, and we don't see any of the illuminated side!