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melisa1 [442]
3 years ago
15

PLEASE HELP THIS IS DUE IN 15 MINS

Physics
1 answer:
AVprozaik [17]3 years ago
4 0

Answer:

d. subduction areas

mantles

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A mass m is attached to an ideal massless spring. When this system is set in motion with amplitude a, it has a period t. What is
Luden [163]

The period will be the same if the amplitude of the motion is increased to 2a

What is an Amplitude?

Amplitude refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium.

Here,

mass m is attached to the spring.

mass attached = m

time period = t

We know that,

The time period for the spring is calculated with the equation:

T = 2\pi \sqrt{\frac{m}{k} }

Where k is the spring constant

Now if the amplitude is doubled, it means that the distance from the equilibrium position to the displacement is doubled.

From the equation, we can say,

Time period of the spring is independent of the amplitude.

Hence,

Increasing the amplitude does not affect the period of the mass and spring system.

Learn more about time period here:

<u>brainly.com/question/13834772</u>

#SPJ4

7 0
2 years ago
Acceleration occurs whenever an ___________________ force acts on an object.
leonid [27]
Acceleration occurs whenever the forces on an object are unbalanced.

It's the group of forces on the object that's either balanced or unbalanced.
There's no such thing as "an unbalanced force".

8 0
3 years ago
Read 2 more answers
A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w
Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

\theta= \omega t

and substituting t = 75 seconds, we find

\theta= (0.20 rad/s)(75 s)=15 rad

In degrees, it is

15 rad: x = 2\pi rad : 360^{\circ}\\&#10;x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

v=\omega r

where we have

\omega=0.20 rad/s

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

v=(0.20 rad/s)(13.5 m)=2.7 m/s

6 0
3 years ago
#3: a container has the dimensions of 30 cm x 50 mm x 0.2 m. the density of its contents is 2.5 g/cm3. what is the mass of the s
WARRIOR [948]
Good afternoon!

We calculate the volume of the container in cm³. To do that, we must put the units in cm:

30 cm → 30 cm
50 mm → 5 cm
0.2 m → 20 cm

The volume is:

V = 30 . 5 . 20

V = 3000 cm³

Now, we calculate the mas with the formula:

m = dV

m = 2.5 · 3000

m = 7500 g

Dividing by 1000, we have the mass in kg:


m = 7.5 kg
4 0
4 years ago
Read 2 more answers
I will Give Brainliest To whoever actually answers A 500 kg satellite experiences a gravitational force of 3000 N, while moving
snow_lady [41]

Answer:

9.7\times 10^{-4}\ rad/s

Explanation:

Given:

m=500 kg\\F=3000 N

Radius of earth , R=6371 \times 10^3\ m\\Angular speed =\omega\\We\  know\  that\ \\F= m\times \omega^{2} \times R\\\omega^{2}=\frac{F}{m*R} \\\\=\frac{3000}{500*6371 \times 10^3\ m}

=\frac{6}{6371 \times 10^3\ m}

=9.7\times 10^{-4}\ rad/s

7 0
4 years ago
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