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Basile [38]
3 years ago
6

A tire swing hanging from a branch reaches nearly to the ground. How could you estimate the height of the branch using only a st

opwatch?
Physics
1 answer:
AlexFokin [52]3 years ago
4 0

Answer:

Explanation:

With the help of expression of  time period of pendulum we can calculate the height of the branch . The swinging tire can be considered equivalent to swinging bob of a pendulum . Here length of pendulum will be equal to height of branch .

Let it be h . Let the time period of swing of tire be T then from the formula of time period of pendulum

T = 2\pi\sqrt{\frac{l}{g} }  where l is length of pendulum .

here l = h so

T = 2\pi\sqrt{\frac{h}{g} }  

h = \frac{T^2g}{4\pi^2}

If we calculate the time period of swing of tire , we can calculate the height of branch .

The time period of swing of tire can be estimated with the help of a stop watch . Time period is time that the tire will take in going from one extreme point to the other end and then coming back . We can easily estimate it with the help of stop watch .

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A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0
3 years ago
The Earth and the Moon are attracted to each other by universal gravitation. The Earth is much more massive than is the Moon. Do
OverLord2011 [107]

Answer:

Earth attract the Moon with a force that is greater.

Explanation:

According to the law of gravitation, the gravitational force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mathematically, F1 = Gm1m2/r²... 1

Let m1 be the mass of the earth and m2 be that of the moon

If the Earth is much more massive than is the Moon, the new force of attraction between them will become;

F2= G(2m1)m2/r²

F2 = 2Gm1m2/r² ... (2)

Dividing eqn 1 by 2 we have;

F1/F2 = (Gm1m2/r²)÷(2Gm1m2/r²)

F1/F2 = Gm1m2/r²×r²/2Gm1m2

F1/F2 = 1/2

F2=2F1

This shows that that the earth will attract the moon by a force 2times the initial force of the masses(i.e a much greater force)

6 0
3 years ago
Which of the following calculations is equal to 1?
lutik1710 [3]

Answer:

we need to see the answers but probably 1 or -1

5 0
3 years ago
If the 250 kg bumper car that you are riding in hits another bumper car that is sitting still while driving 3.5 m/s, how much fo
Vilka [71]

Answer:

875 N

Explanation:

From this question, you didn't state the time taken for the bumper car to move or to hit the other bumper car. In calculations of force, time is often needed, because

Force = mass * acceleration, while

Acceleration = velocity / time, basically

Force = mass * velocity / time.

We have our mass, we have our velocity, but we haven't time. So, for this calculation, I'd assume our time to be 1s.

Going by the formula I stated, we can then say that

Force = 250 * 3.5 / 1

Force = 875 N

This means the force my bumper car have while moving at 3.5 m/s for an estimated time of 1s is 875 N

3 0
3 years ago
An elevator is being lowered at steadily decreasing speed by a steel cable attached to an electric motor. There is no air resist
shtirl [24]

Answer:

F = W + ma  a> 0

Explanation:

For this exercise let's use Newton's second law

 we assume the upward direction as positive

          F - W = m a

          F = W + ma

          F = m (g + a)

In this case they indicate that the speed is less and less as it goes down, therefore the acceleration must be opposite to the speed, that is, the acceleration is upwards, consequently it is positive

               

We can see that since a> 0 the force F must have greater than the weight of the elevator

6 0
3 years ago
Read 2 more answers
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