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OverLord2011 [107]
2 years ago
15

If Weight=mg and Force=ma, and the only force acting on an object is gravity, then the masses cancel each other out and you are

left with g=a. Therefore, that must mean everything on earth must fall at the same rate. Is this true Why or why not?
Physics
2 answers:
anygoal [31]2 years ago
8 0

The correct answer to question:

Explanation:

gregori [183]2 years ago
5 0

When g=a, that means everything on earth fall at the same rate.

<h3>Why does everything fall to the earth at the same rate?</h3>

As such, all objects free fall at the same rate regardless of their mass. Because the 9.8 N/kg gravitational field at Earth's surface causes a 9.8 m/s/s acceleration of any object placed there, we often call this ratio the acceleration of gravity.

<h3>Why is gravity equal to acceleration?</h3>

When objects fall to the ground, gravity causes them to accelerate. Acceleration is a change in velocity, and velocity, in turn, is a measure of the speed and direction of motion. Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls.

Know more about gravity here

brainly.com/question/4014727

#SPJ2

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A machine, modeled as a simple spring-mass system, oscillates in simple harmonic motion. Its acceleration is measured to have an
ANTONII [103]

a=5000\dfrac{mm}{s}=5\dfrac{m}{s}

f=10{Hz}\Longrightarrow t=\dfrac{1}{10}s

a_{max}=\dfrac{50\frac{m}{s}}{\frac{s}{10}}\cdot\dfrac{1}{\sqrt{2}}=\dfrac{50\dfrac{m}{s^2}}{\sqrt{2}}\approx\boxed{35.4\dfrac{m}{s^2}}

Hope this helps.

r3t40

5 0
2 years ago
From Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to:
bazaltina [42]

Answer:

The correct option is (B).

Explanation:

The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,

T^2\propto a^3\\\\T^2=ka^3

It is mentioned that, an asteroid with an orbital period of 8 years. So,

(8)^2=ka^3\\\\64=ka^3\\\\a=(64)^{\dfrac{1}{3}}\\\\a=4\ AU

So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.

7 0
2 years ago
Your car is traveling at 70/mi/h (1000ft/s) your sneeze and your eyes close for 0.33 seconds how far did you travel during your
Novay_Z [31]
If you're moving at 70 mi/hr, then you cover 33.88 feet in 0.33 sec. If you're moving at 1,000 ft/sec, then you cover 33 feet in 0.33 sec. 70 mph and 1,000 fps are not equivalent. 1,000 fps is about 682 mph, whereas 70 mph is about 103 fps.
6 0
3 years ago
Four discs are spinning at constant angular velocity. Four ideantical balls are dropped on the discs at different distances from
Arlecino [84]

Answer: They will have equal final angular velocity.

Explanation:

Since the discs starts from rest and rotates about a fixed axis, they are subject to a constant net torque. The work done by the torque during the second revolution is as the work done during the first revolution.

The four disc will turn through equal angles in equal times.

8 0
2 years ago
The quantity of charge through a conductor is modeled as Q = (3.00 mC/s4)t4 − (2.00 mC/s)t + 9.00 mC. What is the current (in A)
Mumz [18]

Answer:

The current at time t = 4.00 s is 0.766 A.

Explanation:

Given that,

The quantity of charge through a conductor is modeled as :

Q=(3t^4-2t+9)\ mC

We need to find the current (in A) at time t = 4.00 s. We know that the rate of change of electric charge is called electric current. It is given by :

I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(3t^4-2t+9)}{dt}\\\\I=12t^3-2

At t = 4 s

I=12(4)^3-2=766\ mC/s\\\\I=0.766\ C/s=0.766\ A

So, the current at time t = 4.00 s is 0.766 A. Hence, this is the required solution.

7 0
2 years ago
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