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3 years ago

A cheetah runs with a constant speed of 8 m/s. How far has the cheetah gone after 25 seconds?

Physics

2 answers:

Radda [10]3 years ago

7 0

Multiply 25 * 8 = 200
So the cheetah has gone 200 miles in 25 seconds

Rom4ik [11]3 years ago

5 0

200 meters.

In one second, a cheetah runs 8m/s, so multiply 8 by 25.

8 x 25 = 200

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An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 23.0 days on average to co

Gelneren [198K]

Answer:

See it in the pic

Explanation:

See it in the pic

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3 years ago

You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner

omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

$Q_1 = Q_2$

$V_1A_1=V_2A_2$

Our values are given as,

$A_1=\frac{1}{2}^2*\pi=0.785 in^2$

$A_2=\frac{5}{8}^2*\pi=1.227 in^2$

Re-arrange the equation to find the first ratio of rates we have:

$\frac{V_1}{V_2}=\frac{A_2}{A_1}$

$\frac{V_1}{V_2}=\frac{1.227}{0.785}$

$\frac{V_1}{V_2}=1.56$

The second ratio of rates is

$\frac{V2}{V1}=\frac{A_1}{A2}$

$\frac{V2}{V1}=\frac{0.785}{1.227}$

$\frac{V2}{V1}=0.640$

3 0

3 years ago

Phil is riding a scooter and pushes off the ground with his foot. this causes him to accelerate at 12 m /s. Phil weighs 600 N. h

Dvinal [7]

Answer:

734.16 kg m/ $s^{2}$

Explanation:

The problem is asking for the Force of pushing off the ground.

The formula of Force is: F = mass x acceleration

Given = Mass: 600 newtons (N)

Acceleration: 12 m/ $s^{2}$

We have to convert the mass into kg first. Remember that 1 kg is equal to 9.80665 newtons.

Let x be the mass in newtons.

Let's convert: $\frac{1 kg}{9.80665 N}$ x $\frac{x}{600 N}$ = $\frac{600}{9.80665}$ = 61.18 kg

Phil's weight is 61.18 kg

Let's go back to finding the force.

F = m x a

F = 61.18 kg x 12 m/ $s^{2}$

F = 734.16 kg m/ $s^{2}$

7 0

2 years ago

Why does light change its velocity (or slows down) when it encounters a glass lens?

Rufina [12.5K]

Explanation:

what happens is that light slows down when it passes from the less dense air into the denser glass or water. This slowing down of the ray of light also causes the ray of light to change direction. It is the change in the speed of the light that causes refraction.

6 0

3 years ago

An object is propelled straight up from ground level with an initial velocity of 48 feet per second. Its height at time t is mod

Ket [755]

Answer:

Explanation:

For a. its max height and when it occurs. First the max height. That's a y-dimension thing, and in the y-dimension we have this info:

v₀ = 48 ft/s

a = -32 ft/s/s

v = 0 (the max height of an object occurs when the final velocity of the object is 0). Use the following equation for this part of the problem:

v² = v₀² + 2aΔx and filling in:

$0=48^2+2(-32)$ Δx and

0 = 2300 - 64Δx and

-2300 = -64Δs so

Δx = 36 feet.

Now for the time it takes to get to this max height. Final velocity is still 0 here, but the equation is a different one for this part of the problem. Use:

v = v₀ + at and filling in:

0 = 48 - 32t and

-48 = -32t so

t = 1.5 sec. That's part a. Onto part b:

The object hits the ground when its displacement, Δx, is 0. Use this equation for this problem:

Δx = v₀t + $\frac{1}{2}at^2$ and filling in:

$0=48t+\frac{1}{2}(-32)t^2$ and

$0=48t-16t^2$ and

0 = 16t(3 - t) so

t = 0 and t = 3. t = 0 is before the object is propelled, so it makes sense that at 0 seconds, the object was still on the ground, right? Then at 3 seconds, it's back on the ground. (Isn't math just perfectly, beautifully sensible!?) Now onto part c:

We are looking for the time interval when the object is >32 feet. So we use the same equation we just used, but with an inequality instead of an equals sign:

$48t+\frac{1}{2}(-32)t^2 >32$ and get everything on one side and factor it again:

$-16t^2+48t-32>0$ and we find that

1 < t < 2 so the time interval is between 1 and 2 seconds that the object is over 32 feet in the air.

8 0

3 years ago