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devlian [24]
3 years ago
9

Look at figure D . Explain what will happen if a larger mass it put on the force meter

Physics
1 answer:
aev [14]3 years ago
5 0

it will experience great force

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I’m so confused. please help. i don’t know what i’m suppose to do
aniked [119]

Answer:

Maybe put them in order ????

Explanation:

4 0
3 years ago
An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro
Mashutka [201]

1) The distance travelled by the electron is 1\cdot 10^{17} m

2) The time taken is 4.0\cdot 10^{10}s

Explanation:

1)

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

u=5\cdot 10^6 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

a=-1.25\cdot 10^{-4} m/s^2 is the acceleration

Solving for s, we find the distance travelled:

s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m

2)

The total time taken for the electron in its motion can also be found by using another suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

u=5\cdot 10^6 m/s

v = 0

a=-1.25\cdot 10^{-4} m/s^2

And solving for t, we find the time taken:

t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0
3 years ago
A rock is thrown downward from the top of a cliff with an initial speed of 20 m/s. If the rock hits the ground after 5.0 s, what
Vinil7 [7]

Answer

I hope it's helps you

7 0
2 years ago
The speed of light in vacuum is defined to be c = 299,792,458 m/s = 1 μ0ε0 . The permeability constant of vacuum is defined to b
Radda [10]

Explanation:

It is given that,

The speed of light in vacuum is, c = 299,792,458 m/s

The permeability constant of vacuum is, \mu_o=4\pi\times 10^{-7}\ N.s^2/C^2

Let \epsilon_o is the permittivity of free space. The relation between \mu_o,\epsilon_o\ and\ c is given by :

c=\dfrac{1}{\sqrt{\mu_o\epsilon_o}}

\epsilon_o=\dfrac{1}{c^2u_o}

\epsilon_o=\dfrac{1}{(299792458\ m/s)^2\times 4\pi\times 10^{-7}\ N.s^2/C^2}

\epsilon_o=8.85\times 10^{-12}\ C^2/N.m^2

Hence, this is the required solution.

3 0
3 years ago
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Why do people eat bo oty
IRINA_888 [86]

Answer: I don't know my dude

Explanation:

7 0
3 years ago
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