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3 years ago

6

if a 10kg object is lifted a distance of 10m from the rest what was the final velocity of the object?

1 answer:

Anastasy [175]3 years ago

5 0

I think it will be 100 kg

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Now assume that the frictional force f is not at its maximum value. What is the relation between the torque Ï„ applied to each w

leva [86]

Answer:

a)The direction the frictional force will acts is in the positive x direction.

Explanation:

a)The direction the frictional force will acts is in the positive x direction

b)in the horizontal direction, the total force F(total) is equal to 4times the frictional force in the wheel.

F(total)=4f

''f'' is taken as the frictional force.

c)4times the normal force on each wheel minus the acceleration equals zero i.e 4N(wheel)-a=0

=4N(wheel)-mg=0

d) torque is the force that tends to bend rotation

ζ=rf

but acceleration=4×frictional force

cross multiply

f=ζ/r

f=ma/4

ma/4=ζ/r

a=4ζ/r

5 0

3 years ago

Read 2 more answers

camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Usin

son4ous [18]

The question is incomplete. Here is the complete question.

The image below was taken with a camera that can shoot anywhere between one and two frames per second. A continuous series of photos was combined for this image, so the cars you see are in fact the same car, but photographed at differene times.

Let's assume that the camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive?

Answer: v = 6.5 m/s

Explanation: The question asks for velocity of the car. Velocity is given by:

$v=\frac{\Delta x}{\Delta t}$

The camera took 7 pictures of the car and knowing its length is 5.3, the car's displacement was:

Δx = 7(5.3)

Δx = 37.1 m

The camera delivers 1.3 frames per second and it was taken 7 photos, so time the car drove was:

1.3 frames = 1 s

7 frames = Δt

Δt = 5.4 s

Then, the car was driving:

$v=\frac{37.1}{5.4}$

v = 6.87 m/s

The car drove at, approximately, a velocity of 6.87 m/s

7 0

3 years ago

light spring of force constant k = 158 N/m rests vertically on the bottom of a large beaker of water (Figure a). A 4.36-kg block

Fantom [35]

Answer:

0.146 m

Explanation:

f = -KΔL according to Hooke's law

volume of water displaced = mass / density of block since a body will displace equal volume of its own

weight of water displaced = mass of water × acceleration due to gravity

and mass of water = volume of water / density of water

weight of water displaced = Vw × dw × g = mg (dw / dblock)

net force = mg - mg (dw / dblock) = 42.728 - 65.74 = -23.00

it will be balanced by a restoring force of 23 N

ΔL = F / k = 23 / 158 = 0.146 m

4 0

2 years ago

What does atomic motion mean?

vaieri [72.5K]

Atomic Motion definition: Atomic motion is the continual movement of atoms and molecules that are contained within everything in the universe.

5 0

3 years ago

A 0.5 m diameter wagon wheel consists of a thin rim having a mass of 7 kg and six spokes, each with a mass of 1.2 kg. 1.2 kg 7 k

Arte-miy333 [17]

Explanation:

It is given that,

Mass of the rim of wheel, m₁ = 7 kg

Mass of one spoke, m₂ = 1.2 kg

Diameter of the wagon, d = 0.5 m

Radius of the wagon, r = 0.25 m

Let I is the the moment of inertia of the wagon wheel for rotation about its axis.

We know that the moment of inertia of the ring is given by :

$I_1=m_1r^2$

$I_1=7\times (0.25)^2=0.437\ kgm^2$

The moment of inertia of the rod about one end is given by :

$I_2=\dfrac{m_2l^2}{3}$

l = r

$I_2=\dfrac{m_2r^2}{3}$

$I_2=\dfrac{1.2\times (0.25)^2}{3}=0.025\ kgm^2$

For 6 spokes, $I_2=0.025\times 6=0.15\ kgm^2$

So, the net moment of inertia of the wagon is :

$I=I_1+I_2$

$I=0.437+0.15=0.587\ kgm^2$

So, the moment of inertia of the wagon wheel for rotation about its axis is $0.587\ kgm^2$ . Hence, this is the required solution.

4 0

3 years ago