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topjm [15]
3 years ago
8

A ball is thrown straight up with a velocity of 50 m/s.(use g = -10 m/s^2) a) What is the velocity in m/s after 2 seconds?

Physics
1 answer:
Andrei [34K]3 years ago
3 0

Answer:

(a) 30 m/sec

(b) -50 m/sec

Explanation:

We have given initial velocity of ball u = 50 m/sec

Acceleration due to gravity g=-10m/sec^2

(a) Time t = 2 sec

Now according to first equation of v = u-gt

So v=50-10×2=30 m/sec

(b) Time t = 10 sec

Now according to first equation of motion

So final velocity v = u-gt = 50-10×10 =-50 m/sec

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1)the car's engine power is 44000W. Explain this number in a physical sense
Ratling [72]

Answer:

1) It expresses the rate (top speed) at which it can move with time.

2) P = 20 W

3) h = 18 km

Explanation:

1) Power is the rate of transfer of energy.

⇒ Power = \frac{Energy(or workdone)}{Time}

i.e P = \frac{E}{t}

Thus a car's engine power is 44000W implies that the engine of the car can propel the car at this rate. This expresses the rate (top speed) at which it can move with time.

2) m = 400g = 0.4 kg

    t = 20 s

h = 100m

g = 10 m/s^{2}

P = \frac{mgh}{t}

  = \frac{0.4*10*100}{20}

  = \frac{400}{20}

P = 20 W

3) u = 600 m/s

   g = 10 m/s^{2}

From the third equation of free fall,

V^{2} = U^{2} - 2gh

V is the final velocity, U is the initial velocity, h is the height.

0 = (600)^{2} - 2 x 10 x h

0 = 360000 - 20h

20h = 360000

h = \frac{360000}{20}

  = 18000

h = 18 km

The maximum height of the bullet would be 18 km.

3 0
3 years ago
1/A ball is dropped from the top of a building. After 3 seconds, its speed is measured to be 29.4 m/s. Calculate the acceleratio
AleksAgata [21]

Answer:

9.8

Explanation:

7 0
3 years ago
3. Sailors use a capstan (shown below) to raise and lower the anchor. It takes two sailors located 1 meter from the center to tu
mr_godi [17]

Answer:

B. 2 meters.

Explanation:

To rotate the capstan a certain amount of torque is required, and if each sailor applies a force F at a distance D from the center, then for two sailors the total torque will be

\tau = 2FD\\;

therefore,  for one sailor to apply the same torque it must be that the torque \tau_2 he applies must be equal to the torque that the two sailors applied:

FD_2 =2FD

which gives

D_2 = 2D.

and since D = 1\:meter,

\boxed{D_2 = 2\: meters}

which is choice B.

4 0
3 years ago
He more gentle the
Alex787 [66]
I don’t really know the answer cause I need more information about the question
4 0
3 years ago
two teams are playing tug of war. team a pulls to the right with a force of 450n .team b pulls to the left with a force of 415 n
Alex_Xolod [135]

Explanation:

It is given that, two teams are playing tug of war.

Force applied by Team A, F_A=450\ N

Force applied by Team B, F_B=415\ N

We need to find the net force acting on the rope. It is equal to :

F_{net}=F_A-F_B

F_{net}=450-415

F_{net}=35\ N

So, the net force acting on the rope is 35 N and it is acting toward right. Hence, this is the required solution.

4 0
3 years ago
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