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Vaselesa [24]
3 years ago
10

As the force applied to an object increases, the acceleration of that object also

Physics
1 answer:
Ganezh [65]3 years ago
3 0

Answer:

C. posotive

I dont have any idea to explain my answer.. all I know is that is the aswer

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A force of 30 N is applied tangentially to the rim of a solid disk of radius 0.14 m. The disk rotates about an axis through its
sammy [17]

Answer: 3.27kg

Explanation:

Inertia can be said to be the resistance of an object to a change in its motion. This includes a change in its direction. An object will stay still or keep moving at the same speed and in a straight line, except it is acted upon by an unbalanced external force.

Given F = 30N

r = 0.14m

a = 130rad/s²

Then, T = Fr

T = 30*0.14

T = 4.2Nm

Also, T = inertia * acceleration

Inertia = 4.2/a

Inertia = 4.2/130

Inertia = 0.032

Also, inertia = mr²/2

0.032 = m * (0.14²)/2

0.032 = m * 0.0098

m = 0.032/0.0098

m = 3.27kg

4 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.
3 0
3 years ago
A 15.0 m long steel rod expands when its temperature rises from 34.0 degrees C to 50.0 degrees C. What is the change in the beam
azamat
0.00288................
8 0
4 years ago
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A measure of the quantity of matter is<br> a. density.<br> c. force.<br> b. weight.<br> d. mass.
ohaa [14]
A measure of the quantity of matter is called mass. D.

This is how much of matter is contained in an object. It is different from weight, which is the pull of gravity on an object.  
6 0
4 years ago
In a collision between two objects having unequal masses, how does magnitude of the impulse imparted to the lighter object by th
olasank [31]
<span>Both objects receive the same impulse.</span>
3 0
3 years ago
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