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4 years ago

11

Calculate the average velocity of a dancer who moves 5 m toward the left of the stage over the course of 15 s. ** Velocity = dis

placement/time Question 1 options: A. 3 m/s B. 1/3 m/s C. 1/3 m/s west D. 3 m/s west

1 answer:

Alex73 [517]4 years ago

3 0

Answer:

B

Explanation:

Velocity=disp/time

V=5m/15s

V=1/3 m/s

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A sound wave has a wavelength of 15 meters with a frequency of 2.5 Hz. What would the velocity be for this situation in

rodikova [14]

Answer:

6m/s

Explanation:

V = frequency * wavelength

15 * 2.5 = 6m/s

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2 years ago

An arrow strikes a target moving at 75 m/s and embeds itself 15 cm into the target. If the arrow stopped with constant accelerat

zhenek [66]

Answer:

Answer:

4 ms

Explanation:

initial velocity, u = 75 m/s

final velocity, v = 0

distance, s = 15 cm = 0.15 m

Let the acceleration is a and the time taken is t.

Use third equation of motion

v² = u² + 2 a s

0 = 75 x 75 - 2 a x 0.15

a = - 18750 m/s^2

Use first equation of motion

v = u + at

0 = 75 - 18750 x t

t = 4 x 10^-3 s

t = 4 ms

thus, the time taken is 4 ms.

Explanation:

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3 years ago

Q. At what point in a waterfall do the drops of water contain the most kinetic energy ?

antoniya [11.8K]

Answer:

2.When they reach the bottom of the fall

Explanation:

The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).

Thus, the correct option is "2" When they reach the bottom of the fall

7 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:

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

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

$T=\frac{2\pi r}{v}$

Re-arranging for r

$r=\frac{Tv}{2\pi}$

And substituting into the previous equation

$qvB = m \frac{Tv^3}{2\pi}$

Solving for T,

$T=\frac{2\pi q B}{m v^2}$

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

4 0

3 years ago

An assault rifle fires an eight-shot burst in 0.40 s. Each bullet has a mass of 7.5 g and a speed of 300 m/s as it leaves the gu

myrzilka [38]

Answer:

The average recoil force on the gun during that 0.40 s burst is 45 N.

Explanation:

Mass of each bullet, m = 7.5 g = 0.0075 kg

Speed of the bullet, v = 300 m/s

Time, t = 0.4 s

The change in momentum of an object is equal to impulse delivered. So,

$F\times t=mv\\\\F=\dfrac{mv}{t}$

For 8 shot burst, average recoil force on the gun is :

$F=\dfrac{8mv}{t}\\\\F=\dfrac{7.5}{1000}\cdot\dfrac{300}{0.4}\cdot8\\\\F=45\ N$

So, the average recoil force on the gun during that 0.40 s burst is 45 N.

5 0

3 years ago