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3 years ago

If a cheetah goes from 10 m/s to 30 m/s in 5 seconds, what was her acceleration?

Physics

2 answers:

grandymaker [24]3 years ago

5 0

Explanation:

Y CHETOS XDXDXX

CDXCVHH

Ierofanga [76]3 years ago

3 0

Yea i i didn’t see her last name but yet bbbsnsnsnsnddx i was talking abt it to see if he could see her what she was going on with lol but i yea i she said she was like he didn’t do that too but he said it ain’t no big do but it was a weird thing

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HELP ME PLEASE *50 points*

aev [14]

Their is no document for us to look at , can you add it so i can help you

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3 years ago

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Acceleration = change of velocity divided by time interval = Δv/Δt.

MariettaO [177]

Answer:

a=2.378 m/s^2

Explanation:

a=Δv/Δt------eq(1)

Δv=Vf-Vi=120 km/h-0 km/h=120 km/h

or Δv=33.3 m/sec

or time=t=14s

putting values in eq(1)

a=33.3/14

a=2.378 m/s^2

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3 years ago

According to Newton's Law of Cooling, if a body with temperature T 1 is placed in surroundings with temperature T 0, different f

Mila [183]

We can substitute the given values into the equation for T, given the surrounding temperature T0 = 0, initial temperature T1 = 140, constant k = -0.0815, and time t = 15 minutes.

T = 0 + (140 - 0)e^(-0.0815*15) = 140e^(-1.2225) = 41.23°F

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3 years ago

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Someone please helpppp!

Vlad1618 [11]

I believe it’s A, i could be wrong tho 3

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3 years ago

A Carnot engine operates between temperature levels of 600 K and 300 K. It drives a Carnot refrigerator, which provides cooling

KATRIN_1 [288]

Explanation:

Formula for maximum efficiency of a Carnot refrigerator is as follows.

$\frac{W}{Q_{H_{1}}} = \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}$ ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

$\frac{W}{Q_{C_{2}}} = \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}}$ ...... (2)

Now, equating both equations (1) and (2) as follows.

$Q_{C_{2}} \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}}$ = $Q_{H_{1}} \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}$

$\gamma = \frac{Q_{C_{2}}}{Q_{H_{1}}}$

= $\frac{T_{C_{2}}}{T_{H_{1}}} (\frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{2}} - T_{C_{2}}})$

= $\frac{250}{600} (\frac{(600 - 300)K}{300 K - 250 K})$

= 2.5

Thus, we can conclude that the ratio of heat extracted by the refrigerator ("cooling load") to the heat delivered to the engine ("heating load") is 2.5.

4 0

3 years ago