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MariettaO [177]
3 years ago
15

State Newton’s third law of motion.

Physics
1 answer:
bagirrra123 [75]3 years ago
7 0

Answer:every force has an equal and oppisite reaction. meeaning if you punch a wall wwith 50lbs of force the wall pushes back on your hand with 50 lbs of force

Explanation:

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tester [92]
Great question the answer is -25x.
3 0
2 years ago
All of the following measure mass except
mixer [17]
A centimeter cannot measure mass because mass is measured by grams. 

Answer: Centimeter 
5 0
3 years ago
Read 2 more answers
A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at lif
Serggg [28]

Answer:

Height of the rocket be one minute after liftoff is 40.1382 km.

Explanation:

v(t)=-gt-v_e\times \ln \frac{m-rt}{m}

v = velocity of rocket at time t

g = Acceleration due to gravity =9.8 m/s^2

v_e = Constant velocity relative to the rocket = 2,900m/s.

m = Initial mass of the rocket at liftoff = 29000 kg

r = Rate at which fuel is consumed = 170 kg/s

Velocity of the rocket after 1 minute of the liftoff =v

t = 1 minute = 60 seconds'

Substituting all the given values in in the given equation:

v(60)=-9.8 m/s^2\times 60 s-2,900m/s\times \ln (\frac{29,000 kg-170 kg/s\times 60 s}{2,9000 kg})

v(60) = 668.97 m/s

Height of the rocket = h

Velocity=\frac{Displacement}{time}

668.97 m/s=\frac{h}{60 s}

h=668.97 m/s\times 60 s=40,138.2 m = 40.1382 km

Height of the rocket be one minute after liftoff is 40.1382 km.

4 0
3 years ago
If a net force of 250 N causes an object to accelerate at 20m/s^2 what must its mass be?
aleksandr82 [10.1K]
The answer is 12.5 kg because 250N / 20m/s^2

I hope that helped
7 0
3 years ago
Observe yourself breathing and count the number of times you inhale per second. During each breath you probably inhale 0.66 L of
Pavel [41]

To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

PV=nRT

Where,

P = Pressure

V = Volume

T = Temperature

n = amount of substance

R = Ideal gas constant

We start by calculating the volume of inhaled O_2 for it:

V = 21\% * 0.66L

V = 0.1386L

Our values are given as

P = 1atm

T=293K R = 0.083145kJ*mol^{-1}K^{-1}

Using the equation to find n, we have:

PV=nRT

n = \frac{PV}{RT}

n = \frac{(1)(0.1386)}{(0.0821)(293)}

n = 5.761*10^{-3}mol

Number of molecules would be found through Avogadro number, then

\#Molecules = 5.761*10^{-3}*6.022*10^{23}

\#Molecules = 3.469*10^{21} molecules

7 0
3 years ago
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