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N76 [4]
3 years ago
5

Which car has the most kinetic energy? A car of mass 1500 kg with speed 5 m/s or A car of mass 1300 kg with speed 3 m/s or a car

of mass 1200 kg with speed 2 m/s or a car mass 1400 kg with speed 4 m/s
Physics
2 answers:
solmaris [256]3 years ago
5 0
<span>The formula to find the kinetic energy of an object is:

KE = 1/2 MV^2 

So if we plug in the information 
1/2 x 1500 x 5^2
1/2 x 1500 x 25
750 x 25
 Car 1 has a KE of 18750

If needed you can find the other three on your own. It'd be good practice.

The answer is car 1.</span>
AlekseyPX3 years ago
3 0

Answer:

The first car has the most kinetic energy.

Explanation:

Given that,

For first car

Mass of car = 1500 kg

Speed of car = 5 m/s

For second car

Mass of car = 1300 kg

Speed of car = 3 m/s

For third car

Mass of car = 1200 kg

Speed of car = 2 m/s

For fourth car

Mass of car = 1400 kg

Speed of car = 4 m/s

The kinetic energy is defined as,

K.E=\dfrac{1}{2}mv^2

Where, m = mass

v = velocity

The kinetic energy of first car,

K.E =\dfrac{1}{2}\times1500\times(5)^2

K.E=18750\ J

The kinetic energy of second car,

K.E =\dfrac{1}{2}\times1300\times(3)^2

K.E=5850\ J

The kinetic energy of third car,

K.E =\dfrac{1}{2}\times1200\times(2)^2

K.E=2400\ J

The kinetic energy of fourth car,

K.E =\dfrac{1}{2}\times1400\times(4)^2

K.E=11200\ J

Therefore, The first car has 18750 J kinetic energy.

Hence, The first car has the most kinetic energy.

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Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot
almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0 (Ec. 1)

\Sigma F_{y} = N - W = 0 (Ec. 2)

Where:

P - Pushing force, measured in newtons.

T - Tension, measured in newtons.

\mu_{k} - Coefficient of kinetic friction, dimensionless.

N - Normal force, measured in newtons.

W - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

P+T -\mu_{k}\cdot W = 0

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

\mu_{k} =\frac{P+T}{m\cdot g}

If we know that P = 400\,N, T = 290\,N, m = 300\,kg and g = 9.807\,\frac{m}{s^{2}}, then:

\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\mu_{k} = 0.235

The kinetic coefficient of friction of the crate is 0.235.

5 0
3 years ago
In a particular case of an object in front of a spherical mirror with a focal length of +12.0 cm, the magnification is +4.00.(a)
salantis [7]

Answer:

9 cm

-36 cm

Explanation:

u = Object distance

v = Image distance

f = Focal length = 12

m = Magnification = 4

m=-\frac{v}{u}\\\Rightarrow 4=-\frac{v}{u}\\\Rightarrow v=-4u

Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{12}=\frac{1}{u}+\frac{1}{-4u}\\\Rightarrow \frac{1}{12}=\frac{3}{4u}\\\Rightarrow u=9\ cm

Object distance is 9 cm

v=-4\times 9=-36\ cm

Image distance is -36 cm (other side of object)

7 0
3 years ago
Hansel pushes a 2 lb. box 5 feet in 20 seconds. How much work has he done?
Licemer1 [7]
Work = force x distance

You can see time doesn’t matter (if we were talking about power, which is the RATE at which work is performed, that would be a different story).

W = 2 x 5 = 10 foot-pounds of work

Foot-pounds are gross units. Better to work in SI units when you can!

8 0
3 years ago
Ô tô chuyển động trên đường cong có bán kính R = 300(m). Nếu vận tốc ôtô tăng đều từ v1 =
kow [346]

Answer:

Mc ,12,4

Explanation:

7 0
2 years ago
A force on a toy airplane exerted a work of 100 J. What was the change in the kinetic energy?
klemol [59]

Answer:

it probs gos fast then it was it would be easy if u asked us if their was any mass or friton

Explanation:

7 0
3 years ago
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