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MakcuM [25]
4 years ago
14

Running longer would be an example of how you might improve your running performance

Physics
1 answer:
AURORKA [14]4 years ago
8 0
This is true, however if it wee better worded it would be obvious. This is simply stating that you have a higher endurance which would be a visual improvement in your running performance. Hope this helped <3
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Which statement best describes beaches? Beaches change over time as waves move further up the shore. Beaches are always changing
NemiM [27]

Answer:

I think the 1st statement is right.

Explanation:

Wind patterns doesn't stay the same.

Waves don't follow the same patterns.

Waves move further up the shore.

I didn't hear about "waves adding" before..so i guess 1st statement is right.

3 0
3 years ago
What is mass?Explain
Alexxx [7]

Answer:

The term mass is used to refer to the amount of matter in any given object For instance, a person or object may be weightless on the moon because of the lack of gravity but that same person or object maintains the same mass regardless of location

Explanation:

8 0
3 years ago
Jessica does not like how her BestFriend has been treating her around others. When confronted, the friend says Jessica is being
Eva8 [605]
She is sad and feels left out because they are treating her badly. Hope this helps :)
7 0
3 years ago
A vertical spring stretches 3.4 cm when a 12-g object is hung from it. The object is replaced with a block of mass 26 g that osc
Mariulka [41]

Answer:

Period of motion is approximately 0.5447  seconds

Explanation:

We start by calculating the constant "k" of the spring which can be derived from the fact that an object of mass 12 g produced a stretch of 3.4 cm: (we write everything in SI units)

F = k * x

0.012 kg * 9.8  m/s^2 = k 0.034 m

k = 0.012 kg * 9.8  m/s^2 / (0.034 m)

k = 3.46 N/m

now we use the formula for the period (T) of a spring of constant k with a hanging mass 'm':

T=2\pi\,\sqrt{\frac{m}{k} }

which in our case becomes:

T=2\pi\,\sqrt{\frac{0.026}{3.46} } \approx 0.5447\,\,sec

4 0
3 years ago
Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W
jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

T^{2}=\frac{4\pi^{2}}{GM}r^{3}    (1)

Where;

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24}kg is the mass of the Earth

r  is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period T_{X} is:

T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}    (2)

Where r_{X}=1.2(10)^{6}m

T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}    (3)

T_{X}=413.712 s    (4)

For satellite Y, the orbital period T_{Y} is:

T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}    (5)

Where r_{Y}=1.9(10)^{5}m

T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}    (6)

T_{Y}=26.064 s    (7)

This means T_{X}>T_{Y}

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

V_{X}=\sqrt{\frac{GM}{r_{X}}} (8)

V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}

V_{X}=18224.783 m/s (9)

<u>For Satellite Y:</u>

V_{Y}=\sqrt{\frac{GM}{r_{Y}}} (10)

V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}

V_{Y}= 45801.13 m/s (11)

This means V_{Y}>V_{X}

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0
3 years ago
Read 2 more answers
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