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poizon [28]
3 years ago
15

A 4900 n gear box with a mass of 500kg falls to the ground. What is it’s acceleration due to gravity?

Physics
1 answer:
Lena [83]3 years ago
8 0

Answer:

9.8 m/s^2, towards the ground

Explanation:

For an object in free fall, the only force acting on it is the force of gravity. We can write it as

F=mg

where

F is called weight of the object

m is the mass of the object

g is the acceleration due to gravity

We can re-arrange the equation as

g=\frac{F}{m}

For the box in the problem, we know:

F = 4900 N

m = 500 kg

Therefore, we can find the value of g:

g=\frac{4900}{500}=9.8 m/s^2

And the direction is the same as the force of gravity (downward).

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An electron is moving the east with a speed of 5.0 × 106 m/s. There is an electric field of
Tanzania [10]

The velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.

<h3>What is Electric field?</h3>

Electric field is the physical field that surrounds a charge.

<h3>How to find final velocity of the electron when it moves some distance in a certain electric field?</h3>
  • From Newton's second law, the acceleration the electron will be

a=F/m=qE/m

  • where q= charge of electron

E= electric field

m= mass of electron

=(−1.60×10^−19C)(3×10³N/C)/(9.11×10^-31kg)

=10¹⁵×0.526m/s²

  • The kinematics equation v²=v0²+2a(Δx)
  • where v=final velocity of the electron

v0=initial velocity of the electron =5×10⁶m/s

a=acceleration of the electron =10¹⁵×0.526m/s²

Δx=distance moved by the electron in east direction =1cm=10^-2m

  • Now v^2=(5×10⁶)²+2×10¹⁵×0.526×10^-2

=25×10¹²+10.52×10¹²

=35.52×10¹²

  • Now velocity of electron=5.95×10⁶m/s.

Thus , we can conclude that the velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.

Learn more about electric field here:

brainly.com/question/26199225

#SPJ1

5 0
2 years ago
Terra tosses a 0.20 kg volleyball straight up at 10.0 m/s. how high does it go?
Sliva [168]

Answer:

5.1 meters

Explanation:

Terra tosses a 0.20kg volleyball up at at a speed of 10 m/s

The height can be calculated as follows

= v^2/2g

= 10^2/2×9.8

= 100/19.6

= 5.1 meters

Hence the height is 5.1 meters

5 0
2 years ago
A charge of 1. 5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How
ELEN [110]

It is a type capacitor is in which two metal plates arranged in such a way so that they are connected in parallel . The potential energy is stored in the capacitor will be 2.7×10⁻⁵.

<h3>What is parallel plate capacitor ?</h3>

It is an type capacitor is an in which two metal plates arranged in such a way so that they are connected in parallel and have some distance between them.

A dielectric medium is a must in between these plates. helps to stop the flow of electric current through it due to its non-conductive nature .

The given data in the problem is;

Q is the charge= 1.5 µC

V is the change in voltage across the plates is = 36 V.

U is the potential energy=?

The formula for the potential energy is given by;

\rm U= \frac{1}{2} \times Q \times V \\\\ \rm U= \frac{1}{2} \times 1.5\times 10^{-6} \times 36 \\\\  \rm U=2.7\times10^{-5}

Hence the potential energy is stored in the capacitor will be 2.7×10⁻⁵.

To learn more about the parallel plate capacitor refer to the link;

brainly.com/question/12883102

3 0
2 years ago
You have two contentious friends, Chris and Pat, and you’ve quickly discovered that they need you to resolve arguments they have
pshichka [43]

Answer:

v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points,

Explanation:

To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.

             v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points, in the only case that it is so is when there is no acceleration.

Therefore neither of them is right.

3 0
3 years ago
Is it possible to put a distance versus time graph to be universal go mine
xz_007 [3.2K]
Yes I'm pretty sure you can
3 0
3 years ago
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