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gayaneshka [121]
3 years ago
8

A 20 N unbalanced force causes an object to accelerate at 1.5 m/s2. What is the mass of the object?

Physics
1 answer:
kicyunya [14]3 years ago
4 0

As per Newton's II law we know that

F = ma

here we know that

F = 20 N

a = 1.5 m/s^2

now the mass of the object will be given as

m = \frac{F}{a}

m = \frac{20}{1.5}

m = 13.3 kg

so mass of the object will be 13.3 kg

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direction, speed, and shape of a body

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A proton moves eastward in the plane of Earth's magnetic field so that the distance from the ground remains constant. What speed
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Answer:

Speed is 1.962*10^-3m/s

Explanation:

Given that

Mass of proton==1.67*10^-27

Acceleration due to gravity (g)=9.81m/s^2

Angle of inclination A=90°

Charge (q)=1.6*10^-19

Magnetic field (B)=5*10^-5T

From lorentz force

It can be seen that

F=w =mg=qvBsinA

We're Velocity (v)=mg /qBsinA

V=1.67*10-27*9.81/1.6*10^-19*5*10^-5*sin90

V=1.962*10^-3m/s

7 0
3 years ago
Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is
ad-work [718]

Answer:

T_{out}=27.0000077 ºC

Explanation:

First, let's write the energy balance over the duct:

H_{out}=H_{in}+Q

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

m*Cp(T_{out}-T_{in})=Q

So, let's isolate T_{out}:

T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}

The Cp of the air at 27ºC is 1007\frac{J}{kgK} (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are T_{out} and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

P=2*H+2*A=2*0.004m+2*0.016m=0.04m

Surface area:

A=P*L=0.04m*1m=0.04m^2

Then, the heat Q is:

600\frac{W}{m^2} *0.04m^2=24W

Finally, find the exit temperature:

T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077

T_{out}=27.0000077 ºC

The temperature change so little because:

  • The mass flow is so big compared to the heat flux.
  • The transfer area is so little, a bigger length would be required.
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3 years ago
If energy is continually added to a gas, what will it likely become?
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The gas will remain a gas and the energy will be transformed into heat raising the temperature
(Some very rare exceptions might occur)
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Question 8 of 10
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Answer:

B. Manipulated Variable

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