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3 years ago

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In the great shopping cart race, two students push on shopping carts. A having twice the mass of B, with the same force applied

over the same distance. Which shopping cart has more KE by the time it gets to the end of the course?
A. cart A
B. cart B
C. neither, they have the same KE

1 answer:

Lesechka [4]3 years ago

6 0

K = 1/2 m x v^2

m = mass on the cart

V = velocity imparted to the cart

KA = 1/2 mA x vA^2.......................(1)

KB = 1/2 mB x vB^2........................(2)

Diving equation 1 by equation 2, we get -

KA/KB = mA/mB

= 2

KA = 2 x KB

Option A is correct

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Answer:

The total energy of the composite system is 7.8 J.

Explanation:

Given that,

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Suppose, the entire track is friction less. a bullet with a m₁ = 30 g mass is fired horizontally into a block of wood with m₂ = 5.29 kg mass. the acceleration of gravity is 9.8 m/s.

Calculate the total energy of the composite system at any time after the collision.

We need to calculate the total energy of the composite system

Total energy of the system at any time = Potential energy of the system at the stopping point

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$E=(m+M)gh$

Put the value in to the formula

$E=(30\times10^{-3}+5.29)\times 9.8\times0.15$

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Scientists are experimenting with a kind of gun that may eventually be used to fire payloads directly into orbit. In one test, t

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Answer:

$t=0.038s$

Explanation:

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Final speed vf= 9.3×10³ m/s

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Solution

From Newtons second law we know that

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$=m\frac{v_{f}-v_{i}}{t}$

Rearrange this equation to find time t

So

$t=m\frac{v_{f}-v_{i}}{F}$

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3 years ago

A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?a.What is the speed of the mountain lio

Arisa [49]

Answer:

What is the speed of the mountain lion as it leaves the ground?

9.98m/s

At what angle does it leave the ground?

50.16°

Explanation:

This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.

To solve this, you will need use two kinematic equations and SOHCAHTOA:

$d = v_it + \dfrac{1}{2}at^{2}\\\\vf = vi + at$

With these formulas, we can derive formulas for everything you need:

Things you need to remember:

A projectile at an angle has a x-component (horizontal movement) and y-component (vertical movement), which is the reason why it creates an angle.

Treat them separately.

At maximum height, the vertical final velocity is always 0 m/s going up. And initial vertical velocity is 0 m/s going down.

Horizontal movement is not influenced by gravity.
acceleration due to gravity (a) on Earth is constant at 9.8m/s

First we need to take your given:

10.0 m long (horizontal) and maximum height of 3.0m (vertical).

$d_x=10.0m\\d_y=3.0m$

What your problem is looking for is the initial velocity and the angle it left the ground.

Vi = ? Θ =?

Vi here is the diagonal movement and do solve this, we need both the horizontal velocity and the vertical velocity.

Let's deal with the vertical components first:

We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.

$d_y=V_i_yt+\dfrac{1}{2}at^{2}$

Since it is at maximum height at this point, we can assume that the lion is already making its way down so the initial vertical velocity would be 0 m/s. So we can reduce the formula:

$d_y=0+\dfrac{1}{2}at^{2}$

$d_y=\dfrac{1}{2}at^{2}$

From here we can derive the formula of time:

$t=\sqrt{\dfrac{2d_y}{a}}$

Now we just plug in what we know:

$t=\sqrt{\dfrac{(2)(3.0m}{9.8m/s^2}}\\t=0.782s$

Now that we know the time it takes to get from the highest point to the ground. The time going up is equal to the time going down, so we can use this time to solve for the intial scenario of going up.

$vf_y=vi_y+at$

Remember that going up the vertical final velocity is 0m/s, and remember that gravity is always moving downwards so it is negative.

$0m/s=vi_y+-9.8m/s^{2}(0.782s)\\-vi_y=-9.8m/s^{2}(0.782s)\\-vi_y=-7.66m/s\\vi_y=7.66m/s$

So we have our first initial vertical velocity:

Viy = 7.66m/s

Next we solve for the horizontal velocity. We use the same kinematic formula but replace it with x components. Remember that gravity has no influence horizontally so a = 0:

$d_x=V_i_xt+\dfrac{1}{2}0m/s^{2}(t^{2})\\d_x=V_i_xt$

But horizontally, it considers the time of flight, from the time it was released and the time it hits the ground. Also, like mentioned earlier the time going up is the same as going down, so if we combine them the total time in flight will be twice the time.

T= 2t

T = 2 (0.782s)

<em>T = 1.564s</em>

<em>So we use this in our formula:</em>

<em> $d_x=V_i_xT\\\\10.0m=Vi_x(1.564s)\\\\\dfrac{10.0m}{1.564s}=V_i_x\\\\6.39m/s=V_i_x$ </em>

Vix=6.39m/s

Now we have the horizontal and the vertical component, we can solve for the diagonal initial velocity, or the velocity the mountain lion leapt and the angle, by creating a right triangles, using vectors (see attached)

To get the diagonal, you just use the Pythagorean theorem:

c²=a²+b²

Using it in the context of our problem:

$Vi^{2}=Viy^2+Vix^2\\Vi^2=(7.66m/s)^2+(6.39m/s)^2\\\sqrt{Vi}=\sqrt{(7.66m/s)^2+(6.39m/s)^2}\\\\Vi=9.98m/s$

The lion leapt at 9.98m/s

Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:

$Tan\theta=\dfrac{O}{A}\\\\Tan\theta=\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{7.66m/s}{6.39m/s}\\\\\theta=50.17$

The lion leapt at an angle of 50.16°.

6 0

3 years ago

1) Archeologists have found fossil remains of some of the first land

dimaraw [331]

Answer: Carbon 14 and Uranium 238 are not used together to determine fossil ages.

Explanation:

Carbon 14 with a half life of 5,700 years can only be used to date fossils of approximately 50,000 years. Most fossils are thought to be much older than 50,000 years. Also most fossils no longer contain any Carbon. The fossilized remains have been mineralized where the original organic material has been replaced and turned into stones containing no carbon.

Uranium 238 has a half life of 4.5 billion years. Uranium can be used to date the age of the earth. If 50% of pure uranium' is left in a sample the sample is assumed to be 4.5 billion years old.( This is assuming that the original sample was 100% uranium and no Uranium 238 has been eroded or lost in 4.5 billion years old. If a fossil has only 25 % of the Uranium 238 the sample has an estimated age of 3.2 Billion years. This would be the estimated age of the earliest life or formation of fossils.

Note no fossils contain Uranium 238. Uranium 238 is only found in igneous or volcanic rocks. So no fossils can be dated directly using U 238.

Because of the huge differences in the half lives of Carbon 14 and Uranium238 they cannot be used together. Carbon 14 can only be used to date fossils of a very recent age. Uranium 238 can only be used to date volcanic rocks of a very old age.

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