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3 years ago

13

A book that weighs 5 newtons is sitting on a table 1.5 meters above the floor. How much gravitational potential energy does the

book have?6.5 joules
4.5 joules
7.5 joules
3.33 joules

1 answer:

valina [46]3 years ago

5 0

Answer:7.5J

Explanation:7.5

J

Explanation:

P

E

=

m

g

h

Since

F

=

m

g

in this case

P

E

=

F

h

P

E

=

5

N

⋅

1.5

m

=

7.5

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What electrical force dies a Uranium nucleus exert on one of its inner electrons, located at a distance of 175 picometers (= 1.7

AlekseyPX

Answer:

correct option is d) 7.0 x 10^-7 N

Explanation:

given data

distance = 175 picometers = 1.75 × $10^{-10}$ m

to find out

electrical force

solution

we know atomic no of uranium is 92

and charge on electron is = 1.6 × $10^{-19}$ C

and electrical force is express as

electrical force = $\frac{1}{4 \pi \epsilon _o} \frac{q1q2}{r^2}$ .............1

put here value we get

electrical force = $9*10^9 \frac{92*(1.6*10^{-19})^2}{(1.75*10^{-10})^2}$

electrical force = 6.921 × $10^{-7}$ N

so correct option is d) 7.0 x 10^-7 N

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How do streams change as they flow from mountains down to plains? Plz help URGENT I need it today

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They start as one stream, then get curvier, then empy into a delta. A delta is a ton of streams of water. It is one, then empties out into a ton of streams.
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The initial volume reading in a graduated cylinder is 30 mL. The mass of an irregular shape of an unknown metal piece is 55.3 g.

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Answer:

7.9 $\frac{gr}{cm^3}$

Explanation:

When we put the metal piece in the liquid (which is in the graduated cylinder), how much it goes up is equal to the volume of the piece we inserted.

So now we know that the volume of that piece of unknown metal is 7mL (which is the same as 7 $cm^3$ ).

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a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

$\tau_{net} = I \alpha$ (1)

where

$\tau_{net}$ is the net torque acting on the object in rotation

I is the moment of inertia of the object

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The angular acceleration is the rate of change of the angular velocity, so it can be written as

$\alpha = \frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time interval

So we can rewrite eq.(1) as

$\tau_{net}=I\frac{\Delta \omega}{\Delta t}$

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Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

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Therefore, this means that

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