To measure the wavelength of any longitudnal wave, one should measure distance

A 0.5 kg ball moves in a circle that is 0.4 m in radius at a speed of 4.0 m/s. Calculate the force exerted on the ball.

blsea [12.9K]

Answer:

Explanation:

Given a ball of mass m= 0.5kg

The ball moves in as circle of radius r= 0.4m

Speed of the ball is v = 4m/s

Centripetal force is exerted on ball and it is given as

Fc = m•ac

ac is centripetal acceleration and it is given as

ac = v²/r

Then,

Fc = mv²/r

Fc = 0.5 × 4²/0.4

Fc = 20N.

The force exerted on the ball is 20N

5 0

3 years ago

What is the role of equations in this course?

Karolina [17]

That will depend on which course you're talking about. It will be a minor role in, say, Maritime Law or Comparitive Religion, but a major one in, say, Particle Physics or Linear Algebra.

4 0

3 years ago

A railroad car moves under a grain elevator at a constant speed of 3.20 m/s. Grain drops into the car at the rate of 240 kg/min.

dedylja [7]

Answer:

F = 768 N

Explanation:

It is given that,

Speed of the elevator, v = 3.2 m/s

Grain drops into the car at the rate of 240 kg/min, $\dfrac{dm}{dt}=240\ kg/min = 4\ kg/s$

We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :

$F=\dfrac{dp}{dt}$

$F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}$

Since, the speed is constant,

$F=m\dfrac{dv}{dt}$

$F=v\dfrac{dm}{dt}$

$F=3.2\times 240$

F = 768 N

So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.

5 0

3 years ago

If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c

hichkok12 [17]

Answer:

4

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of Earth

$m_2$ = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

$F_o=\dfrac{Gm_1m_2}{r^2}$

New gravitational force

$F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}$

Dividing the equations

$\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4$

The ratio is $\dfrac{F_n}{F_o}=4$

The new force would be 4 times the old force

7 0

3 years ago

Red shift data shows that galaxies are O expanding O shrinking O moving away O moving closer

Anon25 [30]

Answer:

Should be moving away

Explanation:

Red is a longer wavelength therefore further away. Wavelength is stretched out more and on the red end. I hope this is right. I decided to research and answer since you didn’t have other answers. Are you taking this on edg? I hope I helped!

7 0

3 years ago