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2 years ago

A ball dropping from 70 cm took 0.6 seconds to travel a horizontal distance of 34 cm.

Physics

1 answer:

Alenkasestr [34]2 years ago

4 0

Answer:

vₓ = 0.566 m / s, W_total = 9.1 J

Explanation:

This exercise is a parabolic type movement, for the x axis where there is no acceleration

x = v t

vₓ = x / t

vₓ = 0.34 / 0.6

vₓ = 0.566 m / s

the work done is

X axis

In this axis there is no acceleration, therefore the sum of the forces is zero and since the work is the force times the distance, we conclude that the lock in this axis is zero.

W₁ = 0

Y axis

in this axis the force that exists is the force of gravity, that is, the weight of the body

W₂ = Fy y

W₂ = mg and

W₂ = m 9.8 0.70

W₂ = m 9.1

the work is a scalar for which we have to add the quantities obtained

W_total = W₁ + W₂

W_total = 0 + 9.1 m

W_total = 9.1 m

In order to complete the calculation, the mass of the body is needed if we assume that the mass is m = 1

W_total = 9.1 J

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Answer:

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Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft × 4 sorings

K = 100 lb/ft

Amplitude of the microscope $\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}]$

where;

$\epsilon = 0$

$W_n = \sqrt { \frac{k}{m}}$

= $\sqrt { \frac{100*32.2}{200}}$

= 4.0124

replacing them into the above equation and making X the subject of the formula:

$X =$ $Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}$

$X =$ $0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}$

$X =$ $5.676*10^{-3} \ mm$

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = $5.676*10^{-3} \ mm$

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3 years ago

Your Lead Teaching Assistant is initially seated on the top of a hemispherical ice mound of radius R = 30 m. Approximate the ice

saw5 [17]

Answer:

<em>Height = 5.65 km</em>

Explanation:

$3\pi r^2$ is the circumference or we can say measures the boundary of hemisphere of friction-less ice that he is sitting on.

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2 years ago

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sweet-ann [11.9K]

Given :

A box weighing 12,000 N is parked on a 36° slope.

To Find :

What will be the component of the weight parallel to the plane that balances friction.

Solution :

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$F \ cos ( 90 - 36)^o\\\\F\ ( sin\ 36^o)$

Therefore, component of force to balance friction is F sin 36° .

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2 years ago

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The relationship between the distance covered, initial and final speeds, and time can be expressed through the equation,

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2ad = Vf² - Vi²

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The value of a from the equation is 7173.92 km/h².

Second equation,
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2 years ago

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over over over
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