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Maru [420]
3 years ago
8

Which is not a way to conserve existing energy resources?

Physics
2 answers:
zimovet [89]3 years ago
8 0
For the answer to the question above, the answer is simple.<span> among</span> the choices given the only way on not to conserve energy is by using the available fossil fuel.
I hope my answer helped you. feel free to ask more questions. Have a nice day!
Ghella [55]3 years ago
5 0
Use available fossil fuel
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What is the velocity of an object that has a momentum of 4000 kg-m/s and a mass of 115 kg? Round to the nearest hundredth.
insens350 [35]
The answer is C. You divide 4000 kg/s by 115 kg.
5 0
3 years ago
An object is thrown straight down with an initial speed of 4 m/s from a window which is 8 m above the ground. Calculate: a) The
frosja888 [35]

Answer:

0.41s

8.01m/s

Explanation:

Using the formula; v = u + at

Where;

u = initial velocity (m/s)

v = final velocity (m/s)

t = time (s)

a = acceleration (m/s²)

According to the provided information, u = 4m/s, s = 8m,

V = u + at

4 = 0 + 9.8t

4 = 9.8t

t = 0.41s

b) v = u + at

v = 4 + 9.8(0.41)

v = 4 + 4.018

v = 8.018m/s

8 0
3 years ago
Help with all of them plsss
MakcuM [25]
I):final velocity=initial velocity+acceleration due to gravity*time of travel

Therefore,
v=0+9.8*3.1
=30.38 m/s

If by floor you meant water bed then i think there isn't enough info in the question.
8 0
3 years ago
A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction
yawa3891 [41]

Answer:

F'=\dfrac{F}{4}

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

F=\dfrac{mv^2}{r}

v = 2v

F=\dfrac{m(2v)^2}{r}

F=4\dfrac{m(v)^2}{r}..........(1)

For the slower car on the road,

F'=\dfrac{mv^2}{r}............(2)

Equation (1) becomes,

F=4F'

F'=\dfrac{F}{4}

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

8 0
3 years ago
A system of ideal gas at 22°C undergoes an ischoric process with an internal energy decrease of 4.30 × 10 3 4.30×103 J to a fina
Komok [63]

Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

Given that,

Temperature = 22°C

Internal energy U=4.30\times10^{3}\ J

Final temperature = 16°C

We need to calculate the approximate change in entropy

Using formula of the entropy

\Delta S=\dfrac{\Delta U}{T}

Where, \Delta U = internal energy

T = average temperature

Put the value in to the formula

\Delta S=\dfrac{-4.30\times10^{3}}{\dfrac{22+273+16+273}{2}}

\Delta S=-14.72\ J/K

Hence, The approximate change in entropy is -14.72 J/K.

5 0
3 years ago
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