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3 years ago

A book that weighs 16N is on top of a shelf that is 3m high. What is the work required to

Physics

1 answer:

DENIUS [597]3 years ago

8 0

48J

Explanation:

Given parameters:

Weight of the book = 16N

Height of the shelf = 3m

Unknown:

Work done to raise the book = ?

Solution:

Work done is defined as the product of force and distance. The force here is the weight of the book which signifies that it was carried in the vicinity of gravitational force.

Work done = Force x distance = weight x distance

Weight is a type of force;

Input the parameters;

Work done = 16 x 3 = 48J

The unit of workdone is in Joules

learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

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We should continue to send robotic spacecrafts into space
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3 years ago

Denise is conducting a physics experiment to measure the acceleration of a falling object when it slows down and comes to a stop

77julia77 [94]

Answer: The answer is D. (9.8 m/s2)

Explanation:

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3 years ago

A ___ force is not path-dependent and does not change the overall mechanical energy of an object.

finlep [7]

Answer:

conservative

Explanation:

Nonconservative force is the force that depends on a path, however conservative does not depend on a path and it is not associated with the potential energy. When the work is done by an unconservative force, mechanical energy is added or removed. Friction is the best example for a non-conservative force. When these non-conservative forces are acting, the mechanical energy changes but these are not preserved.

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3 years ago

Read 2 more answers

If an atomic nucleus were the size of a dime how far away might one of its electrons be?

mihalych1998 [28]

The radius of a nucleus of hydrogen is approximately $r_{n1}=1\cdot 10^{-15}m$ , while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom: $r_{e1}=5.3 \cdot 10^{-11}m$

The radius of a dime is approximately $r_{n2} = 9\cdot 10^{-3}m$ : if we assume that the radius of the nucleus is exactly this value, then we can find how far is the electron by using the proportion
$r_{n1}:r_{e1}=r_{n2}:r_{e2}$
from which we find
$r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m$

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.

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3 years ago

A neutral solid metal sphere of radius 0.1 m is at the origin, polarized by a point charge of 2 × 10−8 C at location m. At locat

liraira [26]

Answer: E = $1.8 *10 ^{4} N$

Explanation: The formulae for intensity of an electric field of a solid metal sphere relative to a point is given below

$E =\frac{Kq}{r^{2} }$ r

where $k=9* 10^{9}N/m^{2}$ , $q=2 *10 ^{-8} c$ , r = 0.1m r = is the position vector of the charge.

it has been stated in the question that the charge is placed at the center thus it has no position vector.

$E=\frac{9 * 10^{9}* 2* 10^{-8} }{0.1^{2} }\\ =\frac{18* 10^{1} }{0.01} \\=\frac{18* 10^{1} }{1 *10^{-2} } = 1.8*10^{4} N$

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3 years ago