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3 years ago

14

How much tension must a rope withstand if it is used to accelerate a 960 kg car horizontally along a frictionless track at 1.20

m/s²?

2 answers:

raketka [301]3 years ago

5 0

Answer:

<em>The tension of the rope is 1152 N</em>

Explanation:

<u>Force</u>

Newton's second law states that the acceleration of an object is directly proportional to the net force and inversely related to its mass.

The law can be written as:

F = m.a

The rope must apply a force F to accelerate the car of m=960 Kg at $a=1.2~ m/s^2$ . Given the track is frictionless, all the force applied produces acceleration, thus:

$F = 960~Kg\cdot 1.2~ m/s^2$

F = 1152 N

The tension of the rope is 1152 N

EleoNora [17]3 years ago

4 0

T=ma
T=960*1.2
T=1152 N

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3 years ago

An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi

Arturiano [62]

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

$V_x'=0.8c\\V=0.6c\\m=4*10^5kg$

Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

$KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}$

So, to $V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016$ find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
We have an expression from Lorents transformation for relativistic law of addition of velocities as,

$V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V$

Substituting values, we get,

$V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c$

Thus, the KE will be,

$KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J$

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

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3 0

1 year ago

Calculate a rate of cooling down of air from 80 C to 5C Show calculation. Give an answer in cubic meters per minute and cfm.

antoniya [11.8K]

Explanation:

Given that,

Rate of cooling of air

Initial temperature= 80°C

Final temperature = 5°C

We need to calculate

Using newton's law of cooling

$\dfrac{dT}{dt}=c(T-T_{0})$

$\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})$

Where, $dT=T_{1}-T_{2}$

Here, $T =\dfrac{T_{1}+T_{2}}{2}$

$T_{0}$ = 25°C (surrounding temperature)

dt = 1 minute

$\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})$

Put the value into the formula

$\dfrac{80-5}{1}=c(\dfrac{85}{2}-25)$

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Hence, This is the required answer.

3 0

2 years ago

A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal

slamgirl [31]

Answer:

18.89cm

Explanation:

As we know that the person is standing 5m in front of the camera

$d_0=5m=500cm$

The focal length of the lens =50cm

f=50 cm

By Lens formula we have:

$\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm$

By the formula of magnification

$\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm$

The height of the image formed is 18.89cm.

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2 years ago

Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav

umka2103 [35]

Answer:

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Explanation:

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$=2.0 \ m/s$

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ $a=\frac{v_f^2-v_i^2}{2s}$

On substituting the given values, we get

⇒ $=\frac{(2.0)^2-(4.0)^2}{2\times 18.2}$

⇒ $=\frac{4-8}{37}$

⇒ $=\frac{-4}{37}$

⇒ $=0.108 \ m/s^2$

As we know,

⇒ $f=ma$

and,

⇒ $\mu_rmg=ma$

⇒ $\mu_r=\frac{a}{g}$

On substituting the values, we get

⇒ $=\frac{0.108}{9.8}$

⇒ $=0.011$

7 0

3 years ago