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3 years ago

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A harmonic force of maximum value 25 N and frequency of 180 cycles>min acts on a machine of 25 kg mass. Design a support syst

em for the machine (i.e., choose c, k) so that only 10% of the force applied to the machine is transmitted to the base supporting the machine.

1 answer:

kramer3 years ago

3 0

Answer:

registrar y de h2rhrjlekeek tert2h2jwkwlwlwlwlwlwñlwk

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1. Mr. Ure has a mass of 65 kg, due to the fact that he is WAY too skinny! What is the force of Earth's gravity on him?

shepuryov [24]

1.

m = mass of Mr. Ure = 65 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Ure is given as

F = mg

F = 65 x 9.8

F = 637 N

2.

F = force of gravity on car = 3050 N

m = mass of the car = ?

g = acceleration due to gravity = 9.8 m/s²

force of gravity on car is given as

F = mg

3050 = m (9.8)

m = 3050/9.8

m = 311.22 kg

3.

m = mass of Mr. Rees = 90 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Rees is given as

F = mg

F = 90 x 9.8

F = 882 N

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3 years ago

The research of social psychologist Vincent Hsu determined that Americans tend to fulfill their ideal of self-reliance through

dsp73

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3 years ago

SI unit for electrical current

Mashcka [7]

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3 years ago

A ball rolls off the end of a horizontal table that is 4 meters off the ground. It is measured that the ball lands 3 meters away

ElenaW [278]

Answer:

The speed at which the ball rolled off the end of the table is 3.3 m/s

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the table.

The position vector of the ball can be calculated as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

r = position vector.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity.

When the ball reaches the ground, its position will be:

r final = (3, -4)

Then:

3 = x0 + v0x · t

-4 = y0 + v0y · t + 1/2 · g · t²

Since the origin of the frame of reference is located at the edge of the table, x0 and y0 = 0. v0y is also 0 ( see the initial velocity vector in the figure to elucidate why). Then:

3 m = v0x · t

-4 m = 1/2 · g · t²

We can solve for "t" in the equation of the y-component and use it in the equation of the x-component to obtain v0x:

-4 m = 1/2 · g · t²

-4 m = -1/2 · 9.8 m/s² · t²

8 m / 9.8 m/s² = t²

t = 0.9 s

Then:

3 m = v0x · 0.9s

3 m/ 0.9 s = v0x

v0x = 3.3 m/s

The speed at which the ball roll off the end of the table is 3.3 m/s

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A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50

laila [671]

Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

<u>Given:</u>

Height of the window=1.5 m
Time taken by ball to cover the window height=0.15

Now using equation of motion in one dimension we have

$s=ut+\dfrac{at^2}{2}$

Let u be the velocity of the ball when it reaches the top of the window

then

$1.5=0.15u+\dfrac{9.8\times0.15^2}{2}\\u=3.96\ \rm m/s\\$

Now u is the final velocity of the ball with respect to the top of the building

so let t be the time taken for it to reach the top of the window with this velocity

$3.96=gt\\t=0.4\ \rm s\\$

Let h be the height above the top of the window

$h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 0.4^2}{2}\\h=9.71\ \rm m$

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