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3 years ago

8

What is the acceleration that earth experiences due to your gravitational pull?

1 answer:

djverab [1.8K]3 years ago

4 0

Answer:

The acceleration that earth experiences due to gravitational pull is = 9.81 $\frac{m}{s^{2} }$ .

Explanation:

The acceleration that earth experiences due to gravitational pull is called the acceleration due to gravity. its value is 9.81 and its unit is $\frac{m}{s^{2} }$ .

When the object move upwards than in that case the earth gravitational force pulls down the body.

The formula of force due to gravity on the body is given as

F = mg

where g = acceleration due to gravity.

Due to this acceleration the body falls upon the surface of the earth.

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The tendency for an object to sink or float has to do with the object's density.<br> True<br> False

Stella [2.4K]

Answer:

TRUE

Explanation:

BECAUSE AN OBJECTS DENSITY SHOWS HOW MUCH WEIGHT IS PER CUBIC UNIT. LIKE SAD HAS MORE DENSITY THAN WATER SO IT SINKS WHILE WOOD HAS LESS DENSITY THAN WATER SO IT FLOATS

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3 years ago

If I travel 100 m N in 5 seconds, what is my average velocity?

Nina [5.8K]

Answer:

Explanation:

$Average- velocity = \frac{displacement }{time} \\A.V = \frac{100}{5} \\A.V = 20$

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3 years ago

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A lithium oxide compound is represented by which formula?

o-na [289]

Li2O is the molecular formula for lithium oxide.

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3 years ago

Read 2 more answers

The electric field between two parallel plates is uniform, with magnitude 646 N/C. A proton is held stationary at the positive p

Ahat [919]

Answer:

The distance from the positive plate at which the two pass each other is 0.0023 cm.

Explanation:

We need to find the acceleration of each particle first. Let's use the electric force equation.

$F=Eq$

$ma=Eq$

<u>For the proton</u>

$m_{p}a_{p}=Eq_{p}$

$a_{p}=\frac{Eq_{p}}{m_{p}}$

$a_{p}=\frac{646*1.6*10^{-19}}{1.67*10^{−27}}$

$a_{p}=6.19*10^{10}\: m/s^{2}$

<u>For the electron</u>

$m_{e}a_{e}=Eq_{e}$

$a_{e}=\frac{Eq_{e}}{m_{e}}$

$a_{e}=\frac{646*1.6*10^{-19}}{9.1*10^{−31}}$

$a_{e}=1.14*10^{14}\: m/s^{2}$

Now we know that the plate separation is 4.26 cm or 0.0426 m. The travel distance of the proton plus the travel distance of the electron is 0.0426 m.

$x_{p}+x_{e}=0.0426$

Both of them have an initial speed equal to zero. So we have:

$\frac{1}{2}a_{p}t^{2}+\frac{1}{2}a_{e}t^{2}=0.0426$

$t^{2}(a_{p}+a_{e})=2*0.0426$

$t^{2}=\frac{2*0.0426}{a_{p}+a_{e}}$

$t=\sqrt{\frac{2*0.0426}{6.19*10^{10}+1.14*10^{14}}}$

$t=2.73*10^{-8}\: s$

With this time we can find the distance from the positive plate (x(p)).

$x_{p}=\frac{1}{2}a_{p}t^{2}$

$x_{p}=\frac{1}{2}6.19*10^{10}*(2.73*10^{-8})^{2}$

$x_{p}=0.0023\: cm$

Therefore, the distance from the positive plate at which the two pass each other is 0.0023 cm.

I hope it helps you!

7 0

3 years ago

You move a 75-kg box 35 m. This requires a force of 90 N. how much work is done while moving the box?

Luda [366]

W = F*d.

= 90*35 = 3150J.

4 0

3 years ago