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3 years ago

What Initial Velocity would you have to give the Marble to make it hit the ground in 5 s

Physics

2 answers:

Paraphin [41]3 years ago

5 0

Answer:

-49 m/s

Explanation:

You would use the formula vf = vi + at, where vf stands for final velocity, vi stands for initial velocity, a stands for acceleration, and t stands for time. Assuming you're dropping the marble on Earth, the acceleration would be -9.8m/s², and the final velocity would be 0 m/s, as it stops after hitting the ground. (0 = vi + (9.8)(5))

Yakvenalex [24]3 years ago

4 0

Answer:

Traveling 381m in 5 seconds means the average velocity is: 381m/5s = 76.2 m/s.

Explanation:

Letting x be the initial velocity and v the final velocity, we have:

12(x+v)=76.2m/s

and (since v=x+50),

12(2x+50)=76.2m/s

x+25=76.2m/s

x=51.2m/s

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When designing a roller coaster what are the biggest considerations?

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If you are designing a roller coaster that goes upside down, you may consider of course seat belts or something that goes around you to keep yourself safe.

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2 years ago

A drowsy cat spots a flowerpot that sails first up and then down past an open window. the pot was in view for a total of 0.49 s,

Alika [10]

For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. 

An initial condition has been supplied by the problem:

s=1.80m when t=0.245s

This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2)

to solve for v, the initial velocity of the pot as it enters the cat's view through the window. Substituting and solving (note that gravitational acceleration is negative since this is opposite our coordinate orientation):

(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2

v=8.549m/s

Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

(v final)=(v initial)+(a)(t)

to solve for the time it will take for the pot to reach the apex of its flight. Because (v final)=0, this equation will look like

0=(v)+(a)(t)

Substituting and solving for t:

0=(8.549m/s)+(-9.81m/s^2)(t)

t=0.8714s

Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2)

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

s=3.725m

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window:

3.725m – 1.8m=1.925m

Answer:

1.925m

3 0

2 years ago

Matter that emits no light at any wavelength is called

lorasvet [3.4K]

Matter that emits no light at any wavelength is called DARK MATTER.

4 0

2 years ago

Devise an exponential decay function that fits the given data, then answer the accompanying questions. Be sure to identify the

7nadin3 [17]

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

$P=P_0e^{-kh}$

where,

$P_0$ =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

$P=\dfrac{1}{3}P_0$

$\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}$

Now

$P=\dfrac{1}{2}P_0$

$ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft$

The altitude will be 22145.27733 ft

$P=0.02P_0$

$0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft$

The elevation is 124984.76055 ft

6 0

2 years ago

Can we use momentum to see how fast the earth is going?

Kisachek [45]

Yes, if we know the Earth's mass

Explanation:

The momentum of an object is a vector quantity given by the equation

$p=mv$

where

m is the mass of the object

v is its velocity

In this case, we are asked if we can find the velocity of the Earth by starting from its momentum. Indeed, we can. In fact, we can rewrite the equation above as

$v=\frac{p}{m}$

Therefore, if we know the momentum of the Earth (p) and we know its mass as well (m), we can solve the formula to find the Earth's velocity.

Learn more about momentum:

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3 years ago