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2 years ago

Question 1

Physics

2 answers:

Murljashka [212]2 years ago

7 0

The answer is A.) Gravity

mixer [17]2 years ago

5 0

Answer:

A) Gravity.

Explanation:

Earth is stuck in orbit around the sun, because of the Sun's gravitational pull. In this scenario, the string represents gravity, keeping the Earth attracted towards the sun.

Hope this helps! Feel free to give me Brainliest if you feel this helped. Have a good day and good luck with your assignment :)

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Consider an isolated system, by which we mean a system free of external forces. If the sum of the particle energies in the syste

Airida [17]

Answer:

The system's potential energy is -147 J.

Explanation:

Given that,

Energy = 147 J

We know that,

System is isolated and it is free from external forces.

So, the work done by the external forces on the system should be equal to zero.

$W=0$

We need to calculate the system's potential energy

Using thermodynamics first equation

$\Delta U=W-\Delta E$

Put the value into the formula

$\Delta U=0-147$

$\Delta U=-147\ J$

Hence, The system's potential energy is -147 J.

5 0

2 years ago

What does the area under the velocity-time graph represent.

V125BC [204]

Answer:

The distance traveled!

Explanation:

This is a velocity time graph of an object moving in a straight line due North.

8 0

2 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

3 0

2 years ago

Why do we balance chemical equations

antiseptic1488 [7]

So it could follow the correct mass for the atom

5 0

2 years ago

PLEASE HELP WITH GIVE BRAINLIEST!!! ❤️

sasho [114]

The force that the book exerts on the table is a normal force, not a weight force. (The book's weight doesn't act on the table, it acts on the book.) It's equal in magnitude to the weight of the book, again, because of the first law.

6 0

2 years ago