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Komok [63]
3 years ago
11

Which two statements offer evidence that the clocks energy is conserved?

Physics
2 answers:
NeTakaya3 years ago
8 0

Answer: D, and C.

Explanation: I just took a quiz on this. Hope this helps!

Natali [406]3 years ago
3 0

Answer:

C is the best answer because we all know that clock is part of our daily lives but we don't know the about its background

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Can somebody tell me what time is it
kodGreya [7K]

Answer:

8:21am

Explanation:

I hate stupid school

5 0
2 years ago
Read 2 more answers
Experiment 1: A study is done to determine which of two fuel mixtures allows a rocket to travel farther over a period of time. R
leva [86]

Answer: D

Experiment 1 has a confounding variable related to the mass of the rockets. Any variation in mass may cause a discrepancy in the distance traveled.

This is the answer to the question because:

  • Both experiments do have a confounding variable.
  • Experiment 1 doesn't have to stay constant.
  • A double-blind experiment will not do anything to the placebo.
  • High blood pressure people will not make the results confusing.

The answer has to be the option D. Hope this helps you!

3 0
3 years ago
What happens to the gravitational force between two objects when their distance is changed?
kykrilka [37]
If they become closer, it is increased, and if the objects become farther away is decreased.
7 0
3 years ago
The maximum force that a grocery bag can withstand without ripping is 250 n. Suppose that the bag is filled with 20 kg of grocer
zhannawk [14.2K]

Answer:

Groceries stay in the bag.

Explanation:

Given:

Maximum force = 250 N

Bag filled with = 20 kg

Lifted acceleration = 5.0\ m/s^2

Solution:

We need to calculate the exerted force on the grocery bag by using Newton's second law.

F = ma

Where:

F = Exerted force on the object.

m = Mass of the object in kg

a = Acceleration of the object in m/s^2

Now, we substitute m = 20 kg and a = 5.0\ m/s^2 in Newton's second law,

F = 20\times 5.0

F = 100\ m/s^2

Since, the exerted force on the bag is less than 250 N, the groceries will stay in the bag.

3 0
3 years ago
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
3 years ago
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