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3 years ago

__________ is a force acting through distance. A. Acceleration B. Work C. Friction D. Power

Physics

2 answers:

Anni [7]3 years ago

4 0

__________ is a force acting through distance.

B. Work

Hope this answers your question!

Tatiana [17]3 years ago

3 0

The answer is probably going the be C. Friction

It is like tranferring energy to another object

Brainliest if it is correct

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A spaceship is headed toward Alpha Centauri at 0.999c. According to us, the distance to Alpha Centauri is about 4 light-years. H

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Answer:

According to the travellers, Alpha Centauri is c) very slightly less than 4 light-years

Explanation:

For a stationary observer, Alpha Centauri is 4 light-years away but for an observer who is travelling close to the speed of light, Alpha Centauri is very slightly less than 4 light-years. The following expression explains why:

v = d / t

where

v is the speed of the spaceship
d is the distance
t is the time

Therefore,

d = v × t

d = (0.999 c)(4 light-years)

d = 3.996 light-years

This distance is very slightly less than 4 light-years.

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3 years ago

Use the following table of a school bus during morning pickups to calculate its average speed between 0h and 2.34h.

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Answer:

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3 years ago

According to the Law of the Conservation of Matter, if you dissolve 25 grams of sugar into 150 grams of water, the mixture shoul

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19grams becauce if u

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3 years ago

You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The

Sliva [168]

Answer:

Explanation:

Given

Height of ceiling is $h=3.6\ m$

Initial speed of Putty $u=9.5\ m/s$

Speed of Putty just before it strike the ceiling is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-9.5^2=2\times (-9.8)\times 3.6$

$v^2=19.69$

$v=4.43\ m/s$

time taken by putty to reach the ceiling

$v=u+at$

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$t=\frac{5.07}{9.8}$

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4 years ago

A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.

kondaur [170]

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

$F_k = \mu_k N$ = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

3 0

3 years ago