What is the net force in this image?

Where can volcanoes form? A. Where there are cracks in the crust B. Along fault lines C. Where the crust is thin and can be rupt

schepotkina [342]

The answer is all of the above

8 0

3 years ago

Read 2 more answers

Juliana was late to physical science class and missed the beginning of the notes, including the title. These are the notes she t

olga_2 [115]

The correct option to the question is Matter.

Matter makes up everything. matter can be solid, liquid, or gas. matter is made up of atoms, or tiny particles that are the smallest unit of matter.

Moreover, Matter can be described as,

Matter is anything that has occupies space (has mass and volume).

For more information visit:

brainly.com/question/13280491

8 0

2 years ago

A mass weighing 14 pounds stretches a spring 2 feet. The mass is attached to a dashpot device that offers a damping force numeri

Elodia [21]

Answer:

The motion is over-damped when λ^2 - w^2 > 0 or when $b^{2}$ > 0.86

The motion is critically when λ^2 - w^2 = 0 or when $b^{2}$ = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when $b^{2}$ < 0.86

Explanation:

Using the newton second law

k is the spring constante

b positive damping constant

m mass attached

$m\frac{d^{2} x}{dt^{2}} = - kx - b\frac{dx}{dt}$

x(t) is the displacement from the equilibrium position

$\frac{d^{2} x}{dt^{2}} +\frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0$

Converting units of weights in units of mass (equation of motion)

$m = \frac{W}{g} = \frac{14}{32} = 0.43 slug$

From hook's law we can calculate the spring constant k

$k = \frac{W}{s} = \frac{14}{2} = 7 lb/ft$

If we put m and k into the DE, we get

$\frac{d^{2} x}{dt^{2}} +\frac{b}{0.43}\frac{dx}{dt} + 16.28x = 0$

Denoting the constants

2λ = $\frac{b}{m}$ = $\frac{b}{0.43}$

λ = b/0.215

$w^{2} = \frac{k}{m} = 16.28$

λ^2 - w^2 = $\frac{b^{2} }{0.046} - 16.28$

This way,

The motion is over-damped when λ^2 - w^2 > 0 or when $b^{2}$ > 0.86

The motion is critically when λ^2 - w^2 = 0 or when $b^{2}$ = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when $b^{2}$ < 0.86

3 0

3 years ago

A grandfather clock depends on the period of a pendulum to keep correct time.(ii) Suppose a grandfather clock is calibrated corr

ANEK [815]

The grandfather clock will now run slow (Option A).

<h3>What is Time Period of an oscillation?</h3>

The time period of an oscillation refers to the time taken by an object to complete one oscillation.
It is the inverse of frequency of oscillation; denoted by "T".

Now,

$\sqrt{\frac{L}{g}}$ , where L is the length and g is the gravitational constant, is the formula for a pendulum's period.
The period will increase as one climbs a very tall mountain because g will slightly decrease.
Due to this and the previous issue, the clock runs slowly and it seems that one second is longer than it actually is.

Hence, the grandfather clock will now run slow (Option A).

To learn more about the time period of an oscillation, refer to the link: brainly.com/question/26449711

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6 0

1 year ago

A child is riding a merry-go-round that is turning at 7.18 rpm. if the child is standing 4.65 m from the center of the merry-go-

Alisiya [41]

First we need to convert the angular speed from rpm to rad/s. Keeping in mind that
$1 rev= 2 \pi rad$
$1 min = 60 s$
the angular speed is
$\omega = 7.18 \frac{rev}{min} \cdot \frac{2 \pi}{60} = 0.75 rad/s$

And so now we can calculate the tangential speed of the child, which is the angular speed times the distance of the child from the center of the motion:
$v= \omega r = (0.75 rad/s)(4.65 m)=3.50 m/s$

3 0

3 years ago