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3 years ago

9

A car changes velocity at a constant acceleration of 2.5m/s to reach 43.7m/s in 2.7 s how fast was the car moving when it began

to accelerate?

1 answer:

Montano1993 [528]3 years ago

8 0

The formula we can use in this case is:

v = v0 + a t

where v is final velocity, v0 is initial velocity, a is acceleration and t is time

So finding for v0:

v0 = v – a t

v0 = 43.7 – (2.5) 2.7

v0 = 36.95 m/s

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mafiozo [28]

In a mixture, there will be a solute and solvent. The hydrogen and oxygen are still two different atoms that just mixed together. But in a compound, the hydrogen and oxygen have a bond, making a new and bigger molecule. A mixture can easily be separated by physical means but a compound isn't.

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3 years ago

The formula can be used to convert temperatures between degrees Fahrenheit () and degrees Celsius (). How many degrees are in th

Jlenok [28]

Answer:

-30°C

Explanation:

F-32/180 =C-0/100

or, -22-32/180=C/100

or, -54/180*100=C

or, -0.3*100=C

therefore, C= -30

-22°F = -30°C

PLEASE MARK ME AS BRAINLIEST!

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2 years ago

If a small sports car collides head-on with a massive truck, which vehicle experiences the greater impact force? Which vehicle e

garri49 [273]

Answer:

Small sports car.

Explanation:

Lets take

mass of the small car = m

mass of the truck = M

As we know that when car collide with the massive truck then due to change in the moment of the car both car as well as truck will feel force.We also know that from Third law of Newton's ,it states that every action have it reaction with same magnitude but in the opposite direction.

Therefore

F = m a

a=Acceleration of the car

$a=\dfrac{F}{m}$

F= M a'

a'=Acceleration of the massive truck

$a'=\dfrac{F}{M}$

Here given that M > m that is why a > a'

Therefore car will experiences more acceleration.

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3 years ago

A particular HeNe laser beam at 633 nm has a spot size of 0.8 mm. Assuming a

Kobotan [32]

Answer:

Explanation:

A

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3 years ago

A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.

Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

$\Delta K = \Delta E_{p} = q\Delta V$

<u>Where</u>:

K: is the kinetic energy

$E_{p}$ : is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV

$\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J$

Now, the kinetic energy of the particle as it moves through point B is:

$\Delta K = K_{f} - K_{i}$

$K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J$

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!

8 0

3 years ago