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abruzzese [7]
3 years ago
6

You have landed on an unknown planet, Newtonia, and want to know what objects weigh there. When you push a certain tool, startin

g from rest, on a frictionless horizontal surface with a 12.0-N force, the tool moves 16.0 m in the first 2.00 s. You next observe that if you release this tool from rest at 10.0 m above the ground, it takes 2.58 s to reach the ground. What does the tool weigh on Newtonia, and what does it weigh on Earth?
Physics
1 answer:
a_sh-v [17]3 years ago
8 0

Answer:

Explanation:

Given

Force applied F=12\ N

time taken t_1=2\ s

Displacement s=16\ m

using

s=ut+\frac{1}{2}at^2

u=initial velocity

s=displacement

t=time

16=0+\frac{1}{2}a(2)^2

a=8\ m/s^2

thus mass of body m=\frac{F}{a}=\frac{12}{8}=1.5\ kg

Next it is released from a height of h=10\ m

time taken to reach ground t_2=2.58\ s

using  s=ut+\frac{1}{2}at^2 in vertical direction

here acceleration is due to acceleration due to gravity(g') of the planet

as it at rest so u=0 here

10=0+\frac{1}{2}(g')(2.58)^2

g'=3.004\ m/s^2

Thus acceleration due to gravity on Newtonian is 3\ m/s^2

Weight of tool on Newtonia W'=mg'

W'=1.5\times 3.004=4.506\ N

Weight on Earth W=1.5\times 9.8=14.7\ N

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Kazeer [188]

Answer:

17.5 m/s²

1.90476 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

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Time taken by the rocket to reach 120 km/h is 1.90476 seconds

Change in the velocity of a rocket is given by the Tsiolkovsky rocket equation

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3 years ago
A Brayton cycle has air into the compressor at 95 kPa, 290 K, and has an efficiency of 50%. The exhaust temperature is 675 K. Fi
motikmotik

Answer:

The specific heat addition is 773.1 kJ/kg

Explanation:

from table A.5 we get the properties of air:

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cp=specific heat at constant pressure=1.004 kJ/kg*K

We calculate the pressure range of the Brayton cycle, as follows

n=1-(1/(P2/P1)^(k-1)/k))

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P2/P1=(1/0.5)^(1.4/0.4)=11.31

the temperature of the air at state 2 is equal to:

P2/P1=(T2/T1)^(k/k-1)

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11.31=(T2/290)^(1.4/(1.4-1))

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The temperature of the air at state 3 is equal to:

P2/P1=(T3/T4)^(k/(k-1))

11.31=(T3/675)^(1.4/(1.4-1))

T3=1350K

The specific heat addition is equal to:

q=Cp*(T3-T2)=1.004*(1350-580)=773.1 kJ/kg

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In this sense, I believe that we simply divide both currents.

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