1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Andrej [43]
3 years ago
8

A desktop computer and monitor together draw about 0.6 A of current. They

Physics
2 answers:
liubo4ka [24]3 years ago
6 0

Answer:

200 ohms

Explanation:

Masteriza [31]3 years ago
4 0

Answer:

200

Explanation:

You might be interested in
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 student is measuring the volumes of nectar produced by a flowering plant for an experiment. He measures nectar from 50 flowers
pantera1 [17]

The question is incomplete, the complete question is;

A student is measuring the volumes of nectar produced by a flowering plant for an experiment. He measures nectar from 50 flowers using a graduated cylinder that measures to the nearest millilitre (mL). Which statement describes a change that can help improve the results of his experiment?

A.) His measurements will be more precise if he takes measurements from an additional 100 flowers. B.) His measurements will be more accurate if he uses a graduated cylinder that measures to the nearest tenth of a mL. C.) His measurements will be more precise if he uses a graduated cylinder that measures to the nearest tenth of a mL. D.) His measurements will be more accurate if he takes measurements from an additional 100 flowers.

Answer:

His measurements will be more accurate if he uses a graduated cylinder that measures to the nearest tenth of a mL.

Explanation:

In the measurements of volume using most graduated cylinders, the cylinders are calibrated to the nearest tenth owing to the uncertainty in the measurement of volume.

Hence if a cylinder has measures to the nearest milliliter(mL), then he can improve his experiment by using a graduated cylinder that measures to the nearest tenth of a mL

6 0
3 years ago
What happens to the force needed to stretch a rubber band when putting it on a stack of papers?
zmey [24]

Answer:the force is needed to stretch it farther.

Explanation: the definition of force is push or pull so imagine stretching a rubber band your pulling the band which means you increase the force.

4 0
3 years ago
Read 2 more answers
Please help quick please
Ede4ka [16]
I can’t see the picture what do you need help with
3 0
3 years ago
Two forces act on a 41-kg object. One force has magnitude 65 N directed 59° clockwise from the positive x-axis, and the other ha
Genrish500 [490]

Answer:

a = 1.41 m/s²

Explanation:

Given that

mass ,m= 41 kg

F₁ = 65 N , θ = 59°

F₂ = 35 N ,θ = 32°

The component of Force F₁

F₁x= F₁cos59° i

F₁x= 65 x cos59° i = 33.47 i

F₁y= - F₁ sin 59° j

F₁y= - 65 x sin 59° j = - 55.71 j

The component of Force F₂

F₂x= F₂ sin 32° i

F₂x= 35 x sin 32° i = 18.54 i

F₂y=  F₂ cos 32° j

F₂y=  35 x cos 32° j =  29.68 j

The total force F

F= 33.47 i +   18.54 i - 55.71 j +   29.68 j

F= 52.01 i - 26.03 j

The magnitude of the force F

F=\sqrt{52.01^2+26.03 ^2}\ N

F=58.16 N

We know that

F= m a

a= Acceleration

m=mass

58.16 = 41 x a

a = 1.41 m/s²

6 0
4 years ago
Other questions:
  • A BB gun is discharged at a cardboard box of mass m2 = 0.65 kg on a frictionless surface. The BB has a mass of m1 = 0.012 kg and
    13·1 answer
  • Which best describes what is made of matter?
    13·2 answers
  • An engine absorbs 2150 J as heat from a hot reservoir and gives off 750 J as heat to a cold reservoir during each cycle. What is
    8·1 answer
  • The water table is the
    13·1 answer
  • 19. The particles in an object move very quickly and are spread apart. Then they move slower and get slower and get closer toget
    9·1 answer
  • If a string has a density of 1.2 g/m and a 329 g weight is attached, what is the expected wave velocity of the string?
    6·1 answer
  • If no external forces act on a moving object it will
    6·2 answers
  • 1) How is the temperature of a gas related to the kinetic energy of its particles?
    6·1 answer
  • HELPPPPPPPP PLEASWEW
    8·1 answer
  • The first law of thermodynamics is just another form of the _____.
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!