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Bond [772]
3 years ago
10

A cheetah can run at 30 m/s but only for about 12s. how far will it run in that time

Physics
2 answers:
Mashutka [201]3 years ago
6 0
About 360 meters I hope this helps 
Anika [276]3 years ago
3 0

Answer: 360 meters

Explanation: A cheetah can run at 30m/s, this means that in each second the animal can travel a distance of 30m.

If the cheeta runs for 12 seconds at that speed, then it travels 12 times that distance.

The distance that a cheetah that runs at 30m/s by 12 seconds is:

D = 30m/s*12 = 360 m

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Exactly one pound of bread dough is placed in a baking tin. the dough is cooked in an oven at 350 °f releasing a wonderful aroma
Sladkaya [172]
The mass of the baked loaf will be less than the original dough. In making dough for bread, we have ingredients that are liquid such as water, melted butter, food flavoring, etc. All of this liquid ingredients mixed on the dough will definitely turn into vapor. This vapor is responsible for releasing of the aroma of the freshly baked bread.
7 0
3 years ago
Find the unit vector in the direction of vector B=4i+2j-5k​
solmaris [256]

Answer:

\frac{4}{\sqrt{45} } i\,+\frac{2}{\sqrt{45} } j-\frac{5}{\sqrt{45} } k

Explanation:

for the unit vector, we need to divide the given vector by its norm, because it should be in the SAME direction as the original vector, but of magnitude "1".

We notice that the norm of the given vector is:

\sqrt{4^2+2^2+(-5)^2} =\sqrt{45}

Then, the unit vector becomes:

\frac{4}{\sqrt{45} } i\,+\frac{2}{\sqrt{45} } j-\frac{5}{\sqrt{45} } k

7 0
3 years ago
What is a stable and unstable equilibrium
Citrus2011 [14]

Answer:

here

Explanation:

Equilibrium is a state of a system which does not change. ... An equilibrium is considered stable (for simplicity we will consider asymptotic stability only) if the system always returns to it after small disturbances. If the system moves away from the equilibrium after small disturbances, then the equilibrium is unstable.

4 0
3 years ago
PLEASE HELP!!!!
defon

Answer:

v = 2.45 m/s

Explanation:

first we find the time taken during this motion by considering the vertical motion only and applying second equation of motion:

h = Vi t + (1/2)gt²

where,

h = height of cliff = 15 m

Vi = Initial Vertical Velocity = 0 m/s

t = time taken = ?

g = 9.8 m/s²

Therefore,

15 m = (0 m/s) t + (1/2)(9.8 m/s²)t²

t² = (15 m)/(4.9 m/s²)

t = √3.06 s²

t = 1.75 s

Now, we consider the horizontal motion. Since, we neglect air friction effects. Therefore, the horizontal motion has uniform velocity. Therefore,

s = vt

where,

s = horizontal distance covered = 4.3 m

v = original horizontal velocity = ?

Therefore,

4.3 m = v(1.75 s)

v = 4.3 m/1.75 s

<u>v = 2.45 m/s</u>

8 0
3 years ago
An arrow is launched from a bow with an initial horizontal velocity of 40
Ivan

Answer:

V = (Vx^2 + Vy^2)^1/2 = (40^2 + 62^2)^1/2

V = 73.8 m/s

tan theta = Vy / Vx = 62/40 = 1.55

theta = 57.2 deg

4 0
2 years ago
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