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3 years ago

I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air.

Physics

1 answer:

drek231 [11]3 years ago

8 0

Answer:

Explanation:

I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .

As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.

If I am sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.

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A girl walked 200 m from her home to her friend’s home, then she walked 1 km to a shop. She then walked 800 m back to her home.

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Answer:

200 meters+1000 meters+800 meters=2000 meters or 2km

Explanation:

1km=1000m

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3 years ago

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3 years ago

The henry's law constant for n2 is 6. 2×10−4matm at 25∘c. what pressure of nitrogen is needed to maintain a n2 concentration of

hodyreva [135]

The pressure of nitrogen which is needed to maintain a N2 concentration of 0. 53 m is 3.2 × 10^(4).

<h3>What is pressure? </h3>

It is defined as the continuous physical force applied on or against an object by something which is in contact with it.

It is also defined as the force per unit area.

<h3>What is henry's law? </h3>

The henry law constant is thr ratio of the partial pressure of compound in air to the concentration of compound in water at given temperature.

C= kp

where,

C is the concentration of compound = 0.53m

k is the henry constant = 6. 2×10−4matm

p is the pressure of compound

By substituting all the value we get,

C = 6. 2×10−4 × p

0.53 = 6. 2×10−4 × p

p = 0.53/6. 2×10−4

p = 3.2 × 10^(4)

Thus we find that the pressure needed to maintain a N2 concentration of 0. 53 m is 3.2 × 10^(4).

learn more about Henry's law:

brainly.com/question/16222358

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2 years ago

Which combination of cell increases both electric current and electromotive force?

inn [45]

Answer:

In parallel combination, the cells are connected in such a way that the potential difference of the circuit is same as the cells used in the circuit but the time of current flowing increases

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1 year ago

A student on a skateboard is moving at a speed of 1.40 m/s at the start of a 2.15 m high and 12.4 m long incline. The total mass

saw5 [17]

Answer:

W = 609.97J

Explanation:

In order to calculate the work done by the student, you take into account that the total work is equal to the change of the kinetic energy of the student, as follow:

$W_T=\Delta K\\\\W+W_g-W_f=\frac{1}{2}m(v^2-v_o^2)\\\\W+Mgsin\alpha d-F_fd=\frac{1}{2}m(v^2-v_o^2)$ (1)

The work done by the friction force is negative because it is against the motion of the student.

W: work done by the student = ?

Wf: work done by the friction force

Wg: work done by the gravitational force

Ff: total friction force = 41.0N

m: mass of the skateboard = 53.0kg

d: distance traveled by the student

v: final speed of the student = 6.90m/s

vo: initial speed of the student = 1.40m/s

α: angle of the incline

You first calculate the distance d, with the Pythagoras' theorem

$d=\sqrt{(2.15m)^2+(12.4m)^2}=12.58m$

Furthermore, the angle α is:

$\alpha=tan^{-1}(\frac{2.15m}{12.4m})=9.83\°$

Then, you solve the equation (1) for W and replace the values of all parameters:

$W=\frac{1}{2}m(v^2-v_o^2)+F_fd-Mgsin\alpha d\\\\W=\frac{1}{2}(53.0kg)((6.90m/s)^2-(1.40m/s)^2)+(41.0N)(12.58m)\\-(53.0kg)(9.8m/s^2)sin(9.83\°)(12.58m)\\\\W=609.97J$

The work done by the student is 609.97J

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4 years ago