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miskamm [114]
3 years ago
6

How much tension must a rope withstand if used to accelerate a 960-kg car from rest horizontally along a frictionless surface to

1.20 m/s in 5 seconds?
Physics
1 answer:
Svetach [21]3 years ago
7 0
Use a=(dv/dt)         (change in velocity/ change in time)=acceleration
(1.2/5)=acceleration

F=ma (Newton's second law, Force= Mass x Acceleration

=960 x 0.24     F=230.4N If  T<230.4N then the tow rope will hold

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Nikolay [14]

Answer: American sports culture has a much greater appreciation of and emphasis towards collegiate and high school sports. This may be the strongest difference between American sports culture and every other countries' sports culture.

4 0
2 years ago
What is the resistance force when you walk up an inclined plane?<br> Please help quick!
Dominik [7]
Friction
Friction also affects the movement of an object on a slope. Friction is a force that offers resistance to movement when one object is in contact with another. Imagine now that you were on the downside of the object and applying force to keep the object in the same place (not moving)
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3 years ago
Compare a wedge and a screw with an inclined plane
Jlenok [28]

a wedge is a small inclined plane

a screw is an inclined lane wkich goes round a centrral axis  ... like a 'spiral' or helical staircase


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3 0
3 years ago
I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
A 0.7 kg block attached to a spring with force constant 160 Nm is free to move on a frictionless, horizontal surface. The block
Tom [10]

Answer:

+ 24 N

Explanation:

the computation is shown below:

Given that

Mass of the block = m = 0.7 kg

Sprint constant = k = 160 N / m

x = 0.15m

Now the force on the block is

F = kx

= (160) (0.15)

= 24 N

As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction

Therefore the third option is correct

3 0
3 years ago
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