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shtirl [24]
2 years ago
14

A rock is being twirled in a circle on the end of a string. The string provides the centripetal force needed to keep the ball mo

ving in a circle. Does the force of tension exerted by the string on the rock do work on the ball in this situation? Explain.
Physics
1 answer:
KonstantinChe [14]2 years ago
4 0

Answer:

No

Explanation:

The force of tension exerted by the string on the rock acts as centripetal force, so its direction is always towards the centre of the circle.

However, the direction of motion of the rock is always tangential to the circle: this means that the force is always perpendicular to the direction of motion of the rock.

As we know, the work done by a force on an object is

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the force and the displacement

In this situation, F and d are perpendicular, so \theta=90^{\circ}, therefore cos \theta = 0 and the work done is zero:

W=0

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Irina finds an unlabeled box of fine needles, and wants to determine how thick they are. A standard ruler will not do the job, a
Vsevolod [243]

Answer:

d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m

Explanation:

The expression which represent the first diffraction minima by a circular aperture is given by d sin\Theta =1.22\lambda--------eqn 1

The angle through which the first minima is diffracted is given by tan\Theta =\frac{y_1}{D}---------eqn 2

As \Theta is very small so we can write sin\Theta =tan\Theta

So from eqn 1 and eqn 2 we can write

y_1=\frac{1.22\lambda D}{d}--------eqn 3

Here y_1 is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture

It is given that diameter of circular aperture is 14.7 cm so y_1=\frac{14.7}{2}=7.35 \ cm

Now putting all these value in eqn 3

d=\frac{1.22\lambda D}{y_1}

d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m

8 0
3 years ago
What is the average velocity of a car that travels 450 km north in 9.0 hours?
Hatshy [7]

The average velocity of a car that travels 450 km north in 9.0 h is 5.0 x 10 km/h

3 0
3 years ago
Aluminum oxide can be produced during rocket launches. Show that the sum of positive and negative charges in a unit of Al2O3 equ
Triss [41]

Answer:

The sum of positive and negative charges in a unit of Al2O3 equals zero.

Aluminium has a charge of +3 while Oxygen has a charge of -2 on each ion.

Al203 has 2 Al atoms and 3 O atoms.

Charge on Al2O3 = 2(charge on Al ion) + 3(charge on O ion)

= 2(3) + 3(-2)

= 6 - 6

= 0

Explanation:

Aluminium has 3 electrons in the outermost shell and has the tendency to lose those 3 electrons to form a positive ion and have a complete outermost shell.

Whereas, Oxygen has 6 electrons in the outermost and has the tendency to accept two more electrons to form a negative ion and have a complete outermost shell.

4 0
3 years ago
An elevator was a full mass of 2400 kg has a power rating on 16.5 horsepower how much time will it take the elevator to lift a f
ehidna [41]
Jajjqjakwkwkwkjsjshdhdbxjixozkaknsns
6 0
3 years ago
A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
Viefleur [7K]

Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If  the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J -  F x 11 =  756.9 J

F x 11 = 4851 J -   756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N  

4 0
3 years ago
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