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3 years ago

A car traveling on the highway at 15 m/s accelerates at 3.0 m/s2 for 5.0 s. What is its final velocity?

Physics

1 answer:

Nataly_w [17]3 years ago

3 0

Answer:

The answer to your question is: vf = 30 m/s

Explanation:

Data

vo = 15 m/s

a = 3.0 m/s²

t = 5 s

vf = ?

Formula

vf = vo + at

Substitution

vf = 15 + (3)(5)

vf = 15 + 15

vf = 30 m/s

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6. As the sky diver's fall the______energy is converted into _______.

notka56 [123]

Answer: B

D makes no sense because energy is created so we can remove that. C also makes no sense because if energy is not created how did it get destroyed if there's no energy in the first place. This leaves us with a and B the correct answer is B because energy is created and cannot be destroyed.

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3 years ago

An object of volume 0.0004m^3 and density 6000kg/m^3 is immersed inside a fluid of density 5000kg/m^3. The force exerted by the

FromTheMoon [43]

We have that the density of the fluid is

$\rho_f=178.57 kg/m3$

From the question we are told that

object of volume 0.0004m^3

density 6000kg/m^3

fluid of density 5000kg/m^3.

Force F=0.7N

Generally the equation for the Force is mathematically given as

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$\rho_f=178.57 kg/m3$

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3 years ago

AlekseyPX

I believe it’s tackling but I’m not quite sure

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4 years ago

Read 2 more answers

At time t=0 a 2150 kg rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. This forc

amm1812

Explanation:

Given that,

Mass of the rocket, m = 2150 kg

At time t=0 a rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. The force is given by equation :

$F=At^2$

Here F = 888.93 N when t = 1.25 s

$F=At^2\\\\A=\dfrac{F}{t^2}\\\\A=\dfrac{888.93}{(1.25)^2}\\\\A=568.91\ N/s^2$

The value of A is $568.91\ N/s^2$ .

(a) Let J is the impulse does the engine exert on the rocket during the 4.0 s interval starting 2.00 s after the engine is fired. It is given in terms of force as :

$J=\int\limits {F{\cdot} dt}$

Limits will be from 2 s to 2+ 4 = 6 s

It implies :

$J=\int\limits^6_2 {At^2{\cdot} dt}\\\\J=A\int\limits^6_2 {t^2{\cdot} dt}\\\\J=A\dfrac{t^3}{3}|_2^6\\\\J=568.91\times \dfrac{1}{3}\times (6^3-2^3)\\\\J=39444.42\ Ns$

(b) Impulse is also equal to the change in momentum as :

$J=m\Delta v\\\\\Delta v=\dfrac{J}{m}\\\\\Delta v=\dfrac{39444.42}{2150}\\\\\Delta v=18.34\ m/s$

Hence, this is the required solution.

5 0

3 years ago

A certain unbalanced force gives a 5kg object an acceleration of 15 m/s2. What would be the acceleration if the same force was a

stira [4]

Answer:

a2 = 2.5 m/s2

Explanation:

F1 = m1 a1 We use the same force so F1 = F2

= 5kg × 15m/s2 F2 = m2 a2

= 75N a2 is required

a2 = F2 / m2

= 75N / 30 kg

= 2.5 m/s2

7 0

4 years ago