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3 years ago

What is the average velocity in the time interval 3 to 4 seconds

Physics

2 answers:

AlekseyPX3 years ago

7 0

Since the position doesn't change over that time, it's zero

iVinArrow [24]3 years ago

3 0

Answer: 0 it did not change.

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Question 2 of 10

emmainna [20.7K]

Answer:

Explanation:

I just did it on apex

8 0

3 years ago

A 30 g bullet moving a horizontal velocity of 500 m/s comes to a stop 12 cm within a solid wall. (a) what is the change in the b

Sidana [21]

M = 30 g = 0.03 kg, the mass of the bullet
v = 500 m/s, the velocity of the bullet

By definition, the KE (kinetic energy) of the bullet is
KE = (1/2)*m*v²
= 0.5*(0.03 kg)*(500 m/s)² = 3750 J
Because the bullet comes to rest, the change in mechanical energy is 3750 J.

The work done by the wall to stop the bullet in 12 cm is
W = (1/2)*(F N)*(0.12 m) = 0.06F J

If energy losses in the form of heat or sound waves are ignored, then
W = KE.
That is,
0.06F = 3750
F = 62500 N = 62.5 kN

Answer:
(a) 3750 J
(b) 62.5 kN

7 0

3 years ago

In 2 1/2 hours an airplane travels 1150 km against the wind. It takes 50 min to travel 450 km with the wind. Find the speed of t

Tamiku [17]

Answer:

Velocity of airplane is 500 km/h

Velocity of wind is 40 km/h

Explanation:

$V_a$ = Velocity of airplane in still air

$V_w$ = Velocity of wind

Time taken by plane to travel 1150 km against the wind is 2.5 hours

$V_a-V_w=\frac {1150}{2.5}\\\Rightarrow V_a-V_w=460\quad (1)$

Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours

$V_a+V_w=\frac {450}{50}\times 60\\\Rightarrow V_a-V_w=540\quad (2)$

Subtracting the two equations we get

$V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h$

Applying the value of velocity of wind to the first equation

$V_a-40=460\\\Rightarrow V_a =500\ km/h$

∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h

5 0

3 years ago

A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the

Bond [772]

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

$qvB = m\frac{v^2}{r}$