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2 years ago

1 A ball is rolling along the floor with a constant velocity of 8 m/s. How far will it have gone after

Physics

1 answer:

Dennis_Churaev [7]2 years ago

3 0

Answer:

448 meters

Explanation:

every second it moves 8 meters, so all you have to do is multiply 56x8 or 8x56 either way it is the same thing and you will get the same answer

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How to find id in this app.

AysviL [449]

Go and click to the invitation bar and you can find an option written as " search friends " . Then it's easy to find that unknown user if you're pretty fond with his/her username and DP ( display picture ).

6 0

3 years ago

Cavendish used the attraction between large and small lead balls to measure an experimental value for the gravitational constant

irina1246 [14]

The gravitational constant was experimentally measured by W Cavendish using the attraction between big and small lead balls. is true

The correct answer is true

<h3>How do you define gravitational constant?</h3>

the strength of gravity. a factor in use in Newton's gravity law to relate the strength of the gravitational pull between two bodies with their masses and distance from one another. 6.67259 X 10-11 newtons per square kilogram is roughly the gravitational constant. G is its identifier.

<h3> where is the strongest gravity is?</h3>

The gravitational pull of the earth is greatest near sea level, normally, and weakens as you get further from the center, such as to the summit of Mt. Everest. Because the obloid earth was slightly wider, but only by a minor ratio, the gravity just at poles is stronger than that at the equator.

To know more about gravitational constant visit:

brainly.com/question/858421
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8 0

1 year ago

The average speed of a greyhound bus from lansing to detroit is 104.5 km/h. on the return trip from detroit to lansing the avera

Nitella [24]

The average speed of the bus from Lansing to Detroit is
$v_1 = 104.5 km/h$
while the average speed of the bus from Detroit to Lansing is
$v_2 =56.2 km/h$

The distance covered by the bus in the two trips is the same (the distance between the two cities), therefore, the average speed of the round trip can be calculated as the mean of the two speeds:
$v= \frac{v_1+v_2}{2}= \frac{104.5+56.2}{2}=80.35 km/h$

7 0

3 years ago

Light travels from medium X into medium Y. Medium Y has a higher index of refraction. Consider each statement below:(i) The ligh

ahrayia [7]

Answer:

i) FALSE, ii) TRUE, iii) FALSE, iv) FALSE

Explanation:

When light (electromagnetic radiation) travels through a material medium, its speed is less than the speed of light in a vacuum. If we define the index of parts

n = c / v

where v is the speed of light in the material medium.

The direction of the ray can be determined by the law of refraction

n₁ sin θ₁ = n₂ sin θ₂

Let's apply these equations to our case where

nₓ <n_y

i) The expression of the refractive index

nₓ < n_y

$\frac{c}{v_x} = \frac{c}{n_y}$

v_y <vₓ

therefore the expression is FALSE

ii) If we use the law of refraction, the light, when passing from a medium with a lower start to another with a higher index, must approach the normal one, away from what would be the continuation of the path of the incident ray

so the expression is TRUE

iii) The speed of light is constant in all material media

the statement is FALSE

iv) light approaches normal

Let me clarify that the normal is a line perpendicular to the surface at the point of contact, not the direction of Io.

the statement of FALSE

8 0

3 years ago

A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i

NARA [144]

Answer:

The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.

Explanation:

In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.

Momentum is Mass X velocity.

5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.

The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.

Next possible options are in kinetic Energy

Initial Kinetic energy = $\frac{1}{2}mu^2$

Final kinetic energy = $\frac{1}{2}mv^2$

Change in kinetic energy = Final Kinetic Energy - Initial Kinetic Energy

Change in kinetic energy of 5kg ball = $\frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2$

Since the 5-kg ball has initial velocity (u), the magnitude of the change in velocity will be reduced.

Change in kinetic energy of 10kg ball:

the ball is initially at rest, therefore the initial velocity (u) will be zero (0)

Δ K.E = $\frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2 = \frac{1}{2}m(v-0)^2 = \frac{1}{2}mv^2$

From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.

Hence, The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball

4 0

3 years ago