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grin007 [14]
3 years ago
8

A 5,000 kg truck moving at 8 m/s has the same momentum as a 2,500 kg car. What is the velocity of the car?

Physics
1 answer:
rodikova [14]3 years ago
5 0

Answer:

16 m/s.

Explanation:

The following data were obtained from the question:

Mass of truck = 5000 Kg

Velocity of truck = 8 m/s

Mass of car = 2500 kg

Velocity of car =..?

Next, we shall determine the momentum of the truck. This can be obtained as follow:

Mass of truck = 5000 Kg

Velocity of truck = 8 m/s

Momentum of truck =.?

Momentum = mass × velocity

Momentum = 5000 × 8

Momentum of the truck = 40000 Kg.m/s

Finally, we shall determine the velocity of the car as follow:

From the question given above, we were told that the car and truck has the same momentum.

This implies that:

Momentum of the truck = momentum of car = 40000 Kg.m/s

Thus, the velocity of the car can be obtained as shown below:

Mass of car = 2500 kg

Momentum of the car = 40000 Kg.m/s

Velocity of car =..?

Momentum = mass × velocity

40000 = 2500 × velocity

Divide both side by 2500

Velocity = 40000/2500

Velocity = 16 m/s

Therefore, the velocity of the car is 16 m/s.

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mihalych1998 [28]

Answer:

   a = w² r

Explanation:

In this exercise, indicate that the wheel has angular velocity w, the worm experiences the same angular velocity if it does not move, and has an acceleration towards the center of the circle, according to Newton's second law, called the centripetal acceleration.

           a = v² / r

angular and linear variables are related

           v = w r

we substitute

          a = w² r

where r is the radius of the wheel

4 0
3 years ago
an object weighting 100g is thrown upwards from the ground at a speed of 100 m/s.where will the potential energy of the object b
Kay [80]

Answer:

333.3 m

Explanation:

Given

m =100g\ =\  0.1kg\\v = 100 m/s\\g = 10 m/s ^2

Potential energy =\frac{2}{3}\  of\  Kinetic\  energy......Equation(1)

We know that

Potential energy=mgh

Kinetic energy =\frac{1}{2} mv^{2}

Now From the Equation(1)

mgh=\frac{2}{3}*\frac{1}{2} mv^{2}\\  gh=\frac{v^{2} }{3} \\10 * h=\ \frac{10000}{3}\\ h=\ \frac{1000}{3} \\h=333.3\  m

3 0
2 years ago
Newer telephone circuits, built during the last decade, offer higher quality because they were built using analog transmission.
hammer [34]

Answer:

Well, newer telephone circuits built during the last decade are based on the digital transmission, not on the analog transmission. So it's the digital transmission circuit that has made the higher quality. Digital circuits converts the voice signals into the binary codes which is then translated again into the voice signal at the receiving end.

The answer is false.

Explanation:

6 0
3 years ago
An automobile moves on a level horizontal road in a circle of radius 30 m. The coefficient of
blondinia [14]

Answer:

v = 12.12 m/s      

Explanation:

It is given that,

Radius of circle, r = 30 m

The coefficient friction between tires and road is 0.5,

The centripetal force is balanced by the force of friction such that,

v = 12.12 m/s

So, the maximum speed with which this car can round this curve is 12.12 m/s. Hence, this is the required solution.

5 0
3 years ago
A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

3 0
3 years ago
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