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Usimov [2.4K]
3 years ago
15

Consider the plot below describing motion along a straight line with an initial position of 10 m. −4 −3 −2 −1 0 1 2 3 4 5 6 1 2

3 4 5 6 7 8 9 b b b b b time (s) velocity (m/s) What is the acceleration at 1 second
Physics
1 answer:
Ugo [173]3 years ago
4 0

Answer:

+1 m/s^2

Explanation:

Acceleration is given by

a=\frac{\Delta v}{\Delta t}

where

\Delta v is the change in velocity

\Delta t is the time interval in which the change in velocity occurs

To find the acceleration at 1 second, we can take the data at t = 1 s and t = 2. We find:

\Delta t = -3 -(-4) = 1 s

\Delta v = 2 - (1) = +1 m/s

So, the acceleration is

a=\frac{+1}{1}=+1 m/s^2

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3 years ago
A 1200 kg car accelerates from 13 m/s to 17 m/s. find the change in momentum of the car.
mixas84 [53]
P=4800kgm/s
As
p=mΔv
where p is momentum, m is mass and v is velocity
Given values is
m =1200kg
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Now
p=mΔv
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8 0
2 years ago
What are the four steps in the machine cycle?
Eddi Din [679]
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3 years ago
Cierto volumen de gas se encuentra a 60°c de temperatura y 5atm de presión, es calentado hasta 140°c, estado en el cual ocupa un
earnstyle [38]

Answer:

V1 = 2221.33 L

Explanation:

The system is about a ideal gas. Then you can use the equation for ideal gases for a volume V1, temperature T1 and pressure P1:

P_1V_1=nRT_1   (1)

And also for the situation in which the variables T, V and P has changed:

P_2V_2=nRT_2   (1)

R: constant of ideal gases = 0.082 L.atm/mol.K

For both cases (1) and (2) the number of moles are the same. Next, you solve for n in (1) and (2):

n=\frac{P_1V_1}{RT_1}\\\\n=\frac{P_2V_2}{RT_2}

Next, you equal these equations an solve for T2:

\frac{P_1V_1}{RT_1}=\frac{P_2V_2}{RT_2}\\\\V_1=\frac{P_2V_2T_1}{P_1T_2}

Finally you replace the values of P2, V2, T1 and T2:

V_1=\frac{(7atm)(680L)(140\°C)}{(60\°C)(5atm)}=2221.33\ L

Hence, the initial volume of the gas is 2221.33 L

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3 years ago
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The radiation that was emitted is still "visible"

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4 0
3 years ago
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