All categories

3 years ago

As the mass oscillates on a spring, when is displacement at its maximum and velocity equal to zero?

Physics

2 answers:

Ann [662]3 years ago

6 0

Answer:

option (c)

Explanation:

The velocity of the oscillating body is zero when it is at its maximum displacement position from its mean position.

So, when the spring is fully stretched, means at maximum displacement the velocity is zero.

Mazyrski [523]3 years ago

3 0

C. when it is fully stretched

You might be interested in

If the length of a pendulum is doubled,what will be the change in its time period??

Phantasy [73]

Explanation:

see the attachment.. hope it will help you

6 0

3 years ago

The Kelly family and the Garcia family each used their sprinklers last summer. The water output rate for the Kelly family's spri

elena55 [62]

Answer:

30 hours

15 hours.

Explanation:

Let the Kelly used sprinkler for t hours and Gracia used the sprinkler for the rest ie (45 - t) hours .

Water output by Kelly = (35 x t) L

Water output by Gracia = 20 ( 45 - t ) L

Total output = 35t + 20(45 - t)

So, 35 t + 20 (45 - t ) = 1350

35 t + 900 - 20 t = 1350

15 t = 450

t = 30 hours

So Kelly used sprinkler for 30 hours and Gracia used it for 45 - 30 = 15 hours.

3 0

3 years ago

A 782-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli

Vadim26 [7]

Answer:

a) v = 5.59x10³ m/s

b) T = 4 h

c) F = 1.92x10³ N

Explanation:

a) We can find the satellite's orbital speed by equating the centripetal force and the gravitation force as follows:

$F_{c} = F_{G}$

$\frac{mv^{2}}{r + h} = \frac{GMm}{(r + h)^{2}}$

$v = \sqrt{\frac{gr^{2}}{r+h}$

Where:

g is the gravity = 9.81 m/s²

r: is the Earth's radius = 6371 km

h: is the satellite's height = r = 6371 km

$v = \sqrt{\frac{gr^{2}}{2r}} = \sqrt{\frac{gr}{2}} = \sqrt{\frac{9.81 m/s^{2}*6.371 \cdot 10^{6} m}{2}} = 5.59 \cdot 10^{3} m/s$

b) The period of its revolution is:

$T = \frac{2\pi}{\omega} = \frac{2\pi (r + h)}{v} = \frac{2\pi (2*6.371 \cdot 10^{6} m)}{5.59 \cdot 10^{3} m/s} = 14322.07 s = 4 h$

c) The gravitational force acting on it is given by:

$F = \frac{GMm}{(r + h)^{2}}$

Where:

M is the Earth's mass = 5.97x10²⁴ kg

m is the satellite's mass = 782 kg

G is the gravitational constant = 6.67x10⁻¹¹ Nm²kg⁻²

$F = \frac{GMm}{(r + h)^{2}} = \frac{6.67 \cdot 10^{-11} Nm^{2}kg^{-2}*5.97 \cdot 10^{24} kg*782 kg}{(2*6.371 \cdot 10^{6} m)^{2}} = 1.92 \cdot 10^{3} N$

I hope it helps you!

3 0

4 years ago

Assume you have an ice cube and also a small rock that is the same size and shape as the ice cube. Predict what would happen if

Step2247 [10]

Answer: The ice cube would float on top of the water and the rock would sink to the bottom.

Explanation: The ice cube has a smaller density than the rock which allows the ice cube to float but makes the rock sink to the bottom of the glass of water.

7 0

3 years ago

A raging bull of mass 700kg runs at 10 m/s. How much kinetic energy does it have?

Illusion [34]

Answer: 35000 J or 35kJ

Explanation:

Use equation for kinetic energy : Ek=mV²/2

m=700 kg

V=10m/s

-------------------

Ek= 700kg*100m²7s²/2

Ek=35000 J=35kJ

6 0

4 years ago