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3 years ago

Which of the following is a field (action-at-a-distance) force?

Physics

1 answer:

Kobotan [32]3 years ago

3 0

Answer:

$F_{grav}$ is a field force, as gravity does not physically apply force and does not require proximity.

Explanation:

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Ashley swam 20 miles in .5 hours and 40 miles in an hour is it a proportional?

DIA [1.3K]

Yes it proportional.

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3 years ago

Which was a testable idea that J.J. Thomson developed through his experiments

Natalka [10]

Cathode ray tube, plumb pudding model, atoms were not hard spheres, but had smaller parts.

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3 years ago

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A ray diagram is shown.

Svetlanka [38]

Answer:

It is virtual and behind the mirror

Explanation:

When an image is produced by a mirror, we have the following situations:

- The image is said to be in front of the mirror if it is on the same side of the object, with respect to the mirror

- The image is said to be behind of the mirror if it is on the opposite side of the object, with respect to the mirror

Moreover:

- The image is said to be real if it is in front of the mirror

- The image is said to be virtual if it is behind the mirror

In this case, the image (located at point I) is on the opposite side of the object (located at point O), so it is virtual and behind the mirror.

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3 years ago

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if a 5-kg bowling ball is projected upward witha velocity of 2.0 m/s, then what is the recoil velocity of the Earth(mass=6.0×10^

vitfil [10]

by the concept of momentum conservation we can say

if net force on a system of mass is ZERO then its momentum will remain conserved

Here a ball is projected upwards so if we take Ball + Earth as a system then total momentum of the system will remain conserved

Initially when ball is on the surface of earth the system has zero momentum and hence we can say after throwing the ball momentum of earth + ball must be zero

now using same equation we can say

$P_{ball} + P_{earth} = 0$

$m_1v_1 + m_2v_2= 0$

given that

$m_1 = 5 kg$

$v_1 = 2 m/s$

$m_2 = 6 * 10^{24} kg$

from above equation velocity of earth will be

$v_2 = \frac{m_1 v_1}{m_2}$

$v_2 = \frac{5*2}{6 * 10^{24}}$

$v_2 = 1.67 * 10^{-24} m/s$

so above will be the recoil speed of earth

3 0

3 years ago

A soccer ball is kicked from the ground with an initial speed of 20.1 m/s at an upward angle of 47.8˚. A player 53.0 m away in t

e-lub [12.9K]

Answer:3.95 m/s

Explanation:

Given

Initial velocity(u) $=20.1 m/s$

launch angle $=47.8^{\circ}$

Player is 53 m away

Range of projectile

$R=\frac{u^2sin2\theta }{g}$

$R=\frac{20.1^2\times sin95.6}{9.8}$

R=41.02 m

so he need to run 53-41.02=11.98 m

Time of flight of projectile $=\frac{2usin\theta }{g}$

T=3.03 s

thus average speed of boy $s_{avg}=\frac{11.98}{3.03}=3.95 m/s$

5 0

3 years ago