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Soloha48 [4]
3 years ago
13

Suppose that you toss a rock upward so that it rises and then falls back to the earth. If the acceleration due to gravity is 9.8

m/sec2,
what is the rock’s acceleration at the instant that it reaches the top of its trajectory (where its velocity is momentarily zero)?
Assume that air resistance is negligible.

A)The acceleration of the rock is zero.
B)The rock has an upward acceleration of 19.6 m/s2.
C)The rock has a downward acceleration of 19.6 m/s2.
D)The rock has a downward acceleration of 9.8 m/s2.
E)The rock has an upward acceleration of 9.8 m/s2.
Physics
1 answer:
arsen [322]3 years ago
3 0
"The rock has a downward acceleration of 9.8 m/s2" is the one among the following choices that explains the <span>rock’s acceleration at the instant that it reaches the top of its trajectory (where its velocity is momentarily zero). The correct option among all the options that are given in the question is option "D". </span>
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What is one way to lower gravitational potential energy?
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decrease the height

Explanation:

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This same car gets pulled over for speeding, and goes from 68 m/s to 0 m/s in 14
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the acceleration of the car is -4.9m/s2.

the direction is opposite to the actual direction, since the acceleration is negative.

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3 years ago
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

We know that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, temperature on left side is 1.48 times the temperature on right

6 0
3 years ago
What do you call any two colors of light that combine to form white light?
Sever21 [200]
Hello there!


We called that complementary colors.

As always, it is my pleasure to help students like you!

4 0
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E = mgh +  \frac{1}2} m v^{2} + \frac{1}{2} I \omega^{2} = mgh +  \frac{1}2} m  r ^{2}   \omega ^{2}  + \frac{1}{2} I \omega^{2}

for a solid cylinder:  I =  \frac{1}{2} m r^{2}
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I will look at the case of a hollow cylinder:

E = mgh + I \omega ^{2} = constant \\ \\ I =  \frac{mgh}{  \omega^{2} }

That is as far as i get.


7 0
4 years ago
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