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2 years ago

1. Calculate the work done by a 47 N force pushing a pencil 0.26 m. 47x .26 = 12J Solve it

Physics

1 answer:

ziro4ka [17]2 years ago

7 0

Answer:

I need help with one to

Explanation:

Im trying to ask someone but no one knows :(

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In an adiabatc process, what happens when gases in a system are compressed?

likoan [24]

Answer:work is done, and temperature increases

Explanation:

In an adiabatic process, when gases are compressed, work is done on the liquid and the temperature increases

8 0

3 years ago

Do you think that rubbing two pieces of woods together would have more or less friction than wood rubbing against smooth metal ?

salantis [7]

Answer: More friction

Explanation: The wood would probably splinter, causing even more friction, while the metal would probably smooth the wood out.

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3 years ago

Two point charges, with charge magnitudes q and ????, are placed a distance r apart. In this arrangement, each point charge expe

sammy [17]

Answer:

1) Q ’= 8 Q , 2) q ’= 16 q , 3) r ’= ¾ r

Explanation:

For this exercise we will use Coulomb's law

F = k q Q / r²

It asks us to calculate the change of any of the parameters so that the force is always F

Original values

q, Q, r

Scenario 1

q ’= 2q

r ’= 4r

F = k q ’Q’ / r’²

we substitute

F = k 2q Q ’/ (4r)²

F = k 2q Q '/ 16r²

we substitute the value of F

k q Q / r² = k q Q '/ 8r²

Q ’= 8 Q

Scenario 2

Q ’= Q

r ’= 4r

we substitute

F = k q ’Q / 16r²

k q Q / r² = k q’ Q / 16 r²

q ’= 16 q

Scenario 3

q ’= 3/2 q

Q ’= ⅜ Q

we substitute

k q Q r² = k (3/2 q) (⅜ Q) / r’²

r’² = 9/16 r²

r ’= ¾ r

6 0

3 years ago

A train, traveling at a constant speed of 25.8 m/s, comes to an incline with a constant slope. While going up the incline, the t

zhannawk [14.2K]

The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:

Explanation:

So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.

So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then

v = 25.8 + (-1.66×8.3)

v =12.022 m/s.

So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

8 0

3 years ago

A mobile starts with a speed of 250m / s and begins to decelerate at a rate of 3m / s². How fast is it after 45s?

Korvikt [17]

$\large{ \underline{ \underline{ \bf{ \purple{Given}}}}}$

Speed of the mobile = 250 m/s
It starts decelerating at a rate of 3 m/s²
Time travelled = 45s

$\large{ \underline{ \underline{ \bf{ \green{To \: find}}}}}$

Velocity of mobile after 45 seconds

$\large{ \underline{ \underline{ \red{ \bf{Now, \: What \: to \: do?}}}}}$

We can solve the above question using the three equations of motion which are:-

v = u + at
s = ut + 1/2 at²
v² = u² + 2as

So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

$\large{ \bf{ \underline{ \underline{ \orange{Solution:}}}}}$

We are provided with,

u = 250 m/s
a = -3 m/s²
t = 45 s

By using 1st equation of motion,

⇛ v = u + at

⇛ v = 250 + (-3)45

⇛ v = 250 - 135 m/s

⇛ v = 115 m/s

✤ Final velocity of mobile = 115 m/s

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4 0

3 years ago