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inessss [21]
2 years ago
11

A charge Q experiences no net force at a particular point in space. Which of the following situations described below must ALWAY

S be true?
-There are no other charges nearby.

-If there are other charges nearby, they must all have the same sign as Q.

-If there are other charges nearby, they must all have the opposite sign of Q.

-If there are other charges nearby, the total positive charge must equal the total negative charge.

-None of the Above
Physics
1 answer:
laila [671]2 years ago
8 0

Answer:

-There are no other charges nearby.

Explanation:

-There are no other charges nearby.

If there is no net charge in nearby space then the force on this charge will be ZERO

-If there are other charges nearby, they must all have the same sign as Q.

There there is nearby charge of same sign then it will have repulsion force on Q

-If there are other charges nearby, they must all have the opposite sign of Q.

if there is nearby charge of opposite sign then the force must be attraction force.

-If there are other charges nearby, the total positive charge must equal the total negative charge.

If there exist two type of charges nearby then there may exist either attraction or repulsion force on it

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Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a
Naily [24]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

m_1v_1 + m_2v_2 = 0

final they both comes to rest so here we can say that final momentum must be zero

now we have

34 v + 54 (3 m/s) = 0

v = -4.76 m/s

8 0
1 year 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
1 year ago
A solution is a special mixture where you cannot see the individual substances that are mixed. You look at two bowls of ice crea
maxonik [38]
The two bols of ice creams are composed of vanilla the same subtances
7 0
1 year ago
A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text
antoniya [11.8K]

Answer:

The energy of the system is 15 J.

Explanation:

Given that,

Energy E = 2.5 J

Amplitude = 10 cm

We need to calculate the spring constant

Using formula of mechanical energy of the system

E=\dfrac{1}{2}kA^2

Put the value into the formula

2.5=\dfrac{1}{2}k\times(10\times10^{-2})^2

k=\dfrac{2.5\times2}{(10\times10^{-2})^2}

k=500\ N/m

If the block is replaced by a block with twice the mass of the original block

Amplitude = 6 cm

We need to calculate the energy

Using formula of mechanical energy

E=\dfrac{1}{2}kA^2

Put the value into the formula

E=\dfrac{1}{2}\times500\times(6\times10^{-2})

E=15\ J

Hence, The energy of the system is 15 J.

8 0
2 years ago
A boat is headed with a velocity of 18 meters/second toward the last with respect to the water in the river. if the river is Flo
scoray [572]
In this case, the two vectors are in the same direction, so they simply add:

<span> total motion = 18m/s + 2.5m/s = 20.5m/s to the west </span>
8 0
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