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lions [1.4K]
3 years ago
15

Is It true that if you drop a heavy object and a lighter object at the same time, they will both drop at the same time due to gr

avity?
Physics
1 answer:
valkas [14]3 years ago
4 0

Answer:

\huge\boxed{True.}

Explanation:

This is true because gravity is independent of mass. Whether it's a heavy or a light object, when dropped, they both will hit the ground at the same time. The reason is that gravity does not depend on mass.

Hope this helped!

<h2>~AnonymousHelper1807</h2>
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Pls help me prove the decay constant equation
belka [17]
Decay constant, proportionality between the size of a population of radioactive atoms and the rate at which the population decreases because of radioactive decay
8 0
4 years ago
PA HELP PO NEED KO ANSWER AS FAST AS POSSIBLE <br>I will mark you as brainliest ​
uysha [10]

Answer:

is there paragraph or not ?

Explanation:

https://glaquarium.org/wp-content/uploads/2015/06/Final-Becoming-a-Scientist-Observation-and-Questions-Lesson-Plan.pdf

6 0
3 years ago
Determine the mass of the load hanging on the spring with a stiffness of 200 N / m if the extension of the spring is 0.5 cm
dexar [7]

P. E=mgh

P. E=200

200=m x 10 X 0.005

200=m x 0.05

200/0.05=m

4000Kg=m

Hope it helps.......

7 0
4 years ago
A circular loop of radius r carries a current i. at what distance along the axis of the loop is the magnetic field one-half its
lana [24]

At r = 0.766 R the magnetic field intensity will be half of its value at the center of the current carrying loop.

We have a circular loop of radius ' r ' carrying current ' i '.

We have to find at what distance along the axis of the loop is the magnetic field one-half its value at the center of the loop.

<h3>What is the formula to calculate the Magnetic field intensity due to a current carrying circular loop at a point on its axis?</h3>

The formula to calculate the magnetic field intensity due to a current carrying ( i ) circular loop of radius ' R ' at a distance ' x ' on its axis is given by -

B(x) = \frac{\mu_{o} iR^{2} }{2(x^{2} +R^{2})^{\frac{3}{2} } }

Now, for magnetic field intensity at the center of the loop can calculated by putting x = 0 in the above equation. On solving, we get -

B(0) = \frac{\mu_{o} i}{2R}

Let us assume that the distance at which the magnetic field intensity is one-half its value at the center of the loop be ' r '. Then -

\frac{\mu_{o} iR^{2} }{2(r^{2} +R^{2})^{\frac{3}{2} } } = \frac{1}{2} \frac{\mu_{o}i }{2R}

2R^{3} = (r^{2} +R^{2} )^{\frac{3}{2} }

4R^{6} = (r^{2} +R^{2} )^{3}

r^{2} =0.587R^{2}

r = 0.766R

Hence, at r = 0.766 R - the magnetic field intensity will be half of its value at the center of the current carrying loop.

To solve more questions on magnetic field intensity, visit the link below-

brainly.com/question/15553675

#SPJ4

6 0
2 years ago
A boy is listening to the radio. The loudspeaker plays a musical note by vibrating 256 rimes esch second. How many times does th
Svet_ta [14]

Answer:

 f_tympanum = 256 Hz.

Explanation:

The eardrum is the vibrating membrane of the human ear, it works by resonance, that is, an external stimulus (force) makes it vibrate, as the eardrum is extremely light it can vibrate at the same frequency of the incident sound.

Consequently if the incident vibration is f = 256 hz, the eardrum resonates at the same frequency

            f_tympanum = 256 Hz.

As a reference the response of the eardrum goes from f = 20 Hz f = 20000 Hz

6 0
4 years ago
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