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2 years ago

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A car tries to tow a stranded van out of a ditch, and the car applies 300 N and pulls the van for 3 meters in less than a minute

before getting it safely back onto the road. How much work was done by the car on the car? The van on the car?

1 answer:

sergejj [24]2 years ago

8 0

Answer:

Work done, W = 900 Joules

Explanation:

It is given that,

Force applies by the car on the van, F = 300 N

The van pulls a distance of, d = 3 m

We need to find the work was done by the car on the car. We know that the product of force and distance is equal to the work done by the object i.e.

$W=F\times d$

$W=300\ N\times 3\ m$

W = 900 Joules

So, the work done by the car on the van is 900 Joules. Hence, this is the required solution.

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Provide an example of when momentum is conserved and explain your answer you can get 10 PTS if answered with a good explaination

dezoksy [38]

Answer:

$m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s$

Explanation:

<u>Conservation of Momentum </u>

The total momentum of a system of two particles is

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Where m1,m2,v1, and v2 are the respective masses and velocities of the particles at a given time. Then, the two particles collide and change their velocities to v1' and v2'. The final momentum is now

$p'=m_1v_1'+m_2v_2'$

The momentum is conserved if no external forces are acting on the system, thus

$m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$

Let's put some numbers in the problem and say

$m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s$

$(8)(12)+(6)(4)=(8)(-6)+(6)(28)$

$96+24=-48+168$

120=120

It means that when the particles collide, the first mass returns at 6 m/s and the second continues in the same direction at 28 m/s

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You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warb

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Answer:

The angle it subtend on the retina is $\theta_z = 0.44586^o$

Explanation:

From the question we are told that

The length of the warbler is $L = 14cm = \frac{14}{100} = 0.14m$

The distance from the binoculars is $d = 18cm = \frac{18}{100} = 0.18m$

The magnification of the binoculars is $M =8$

Without the 8 X binoculars the angle made with the angular size of the object is mathematically represented as

$\theta = \frac{L}{d}$

$\theta = \frac{0.14}{0.18}$

$= 0.007778 rad$

Now magnification can be represented mathematically as

$M = \frac{\theta _z}{\theta}$

Where $\theta_z$ is the angle the image of the warbler subtend on your retina when the binoculars i.e the binoculars zoom.

So

$\theta_z = M * \theta$

=> $\theta_z =8 * 0.007778$

$= 0.0622222224$

Generally the conversion to degrees can be mathematically evaluated as

$\theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )$

$\theta_z = 0.44586^o$

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