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3 years ago

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A screw Jack whose pitch is 2mm is used to raise a bus of mass 900kg through height of 20cm. The length of the Tommy of the Tomm

y of the jack is 40cm. Calculate the velocity ratio of the machine

1 answer:

Nikitich [7]3 years ago

8 0

R= 40cm —> R= 0.4 m

P= 2mm —> P= 0.002 m

$V.R = \frac{2πR}{p} = \frac{2 \times 3.14 \times 0.4}{0.002} = 1256$

I think this is the answer

I hope I helped you

Good luck ^_^

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If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite d

Yuliya22 [10]

Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:

A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answer:

ΔP=20 kg.m/s

Explanation:

Given data

Mass m=0.2 kg

Initial speed Vi=-44.5m/s

Final speed Vf=55.5 m/s

Required

Change in momentum ΔP

Solution

First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:

$v_{i}=-44.5m/s\\v_{f}=55.5m/s$

Now we need to find the initial momentum

So

$P_{1}=m*v_{i}$

Substitute the given values

$P_{1}=(0.2kg)(-44.5m/s)\\P_{1}=-8.9kg.m/s$

Now for final momentum

$P_{2}=mv_{f}\\P_{2}=(0.2kg)(55.5m/s)\\P_{2}=11.1kg.m/s$

So the change in momentum is given as:

ΔP=P₂-P₁

$=[(11.1kg.m/s)-(-8.9kg.m/s)]\\=20kg.m/s$

ΔP=20 kg.m/s

3 0

3 years ago

A boat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, space, start fraction, m, divided by, s,

zepelin [54]

Answer:

$d=42m(-\hat{i})$

Explanation:

For this problem, we need to apply the formulas of constant accelerated motion.

To obtain the boat displacement we need to calculate the displacement because of the river flow and the displacement done because of the boat motor.

for the river:

$d_r=v*t\\d_r=5m/s*6s\\d_r=30m(\hat{i})$

for the boat:

$x=\frac{1}{2}*a*t^2\\\\x=\frac{1}{2}*4.0m/s*(6s)^2\\\\\\x=72m(-\hat{i})$

So the final displacement is given by:

$d=dr+x\\d=30m-72m\\d=42m(-\hat{i})$

8 0

3 years ago

Read 2 more answers

What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a

Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

v = r ω

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}$

$I = \dfrac{m(2gh - v^2)r^2}{v^2}$

$I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}$

$I =mr^2(0.287)$

I = 0.287 MR²

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3 years ago

Two loudspeakers emit identical sound waves along the x-axis. The sound at a point on the axis has maximum intensity when the sp

EleoNora [17]

The concept required to solve this problem is related to the wavelength.

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The waves are in phase, then the first distance is

$\Delta x_1 = 20cm$

And out of the phase when

$\Delta x_2 = 30cm$

Thus the wavelength is

$\Delta x_2-\Delta x_1 = \frac{\lambda}{2}$

Here,

$\lambda =$ Wavelength

If we rearrange the equation to find it, we will have

$\lambda = 2 (\Delta x_2-\Delta x_1 )$

$\lambda= 2(30-20)$

$\lambda = 20cm$

Therefore the wavelength of the sound is 20cm.

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The answer is 3+5+4 = 10+3 so then you have to add the number to the part of the equation and you will get the answer of five

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2 years ago