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gregori [183]
3 years ago
9

If a copper tube has current flowing to the right in the presence of a magnetic field going forward ("into the board"), we expec

t the weight on the scale to be lighter than with no current flowing.
True/False
Physics
1 answer:
Citrus2011 [14]3 years ago
6 0

Answer:

False, because current flow has no effect in the weight of a conductor.

Explanation:

Current flow has no effect in the weight of a conductor. Thus, the weight on the scale will be the same when current is flowing and when current is not flowing in the copper tube.

Therefore, If a copper tube has current flowing to the right in the presence of a magnetic field going forward ("into the board"), we expect the weight on the scale to be the same whether current is flowing or not.

You might be interested in
According to newton's second law of motion, how much force will be required to accelerate an object at the same rate if its mass
guajiro [1.7K]

<u>D.</u>

For every action ther is an equal and opposite reaction.

equal in force, opposite in direction

5 0
3 years ago
A 3,000 kg truck moving at +10 m/s hits a 1,000 kg parked car which moves off at +15 m/s What is the velocity of
Rina8888 [55]

Answer:

v₃ = 5 [m/s]

Explanation:

To solve this problem we must use the definition of linear momentum, which tells us that momentum is equal to the product of mass by Velocity.

P = m*v

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

We must also clarify that the momentum is preserved i.e. it is equal before the collision and after the collision

Pbeforecollision = Paftercollision

(m₁*v₁) + (m₂*v₂) = (m₁*v₃) + (m₂*v₄)

where:

m₁ = mass of the truck = 3000 [kg]

v₁ = velocity of the truck = 10 [m/s]

m₂ = mass of the car = 1000 [kg]

v₂ = velocity of the car before the collision = 0 (the car is parked)

v₃ = velocity of the truck after the collision [m/s]

v₄ = velocity of the car after the collision = 15 [m/s]

(3000*10) + (1000*0) = (3000*v₃) + (1000*15)

30000 = 3000*v₃ + 15000

3000*v₃ = 30000 - 15000

3000*v₃ = 15000

v₃ = 5 [m/s]

7 0
3 years ago
Flying against the wind, an airplane travels 2670 km in 3 hours. Flying with the wind, the same plane travels 11,070 km in 9 hou
xxMikexx [17]

Answer:

speed of plane in still air = 1060 km/h

speed of wind = 170 km/h

Explanation:

Let teh speed of plane in still air is vp and the speed of air is va.

Irt travels 2670 km in 3 hours against the wind

So,

vp - va = 2670 / 3 = 890 km/h ..... (1)

It travels 11070 km in 9 hours along the wind.

vp + va = 11070 / 9 = 1230 km/h .... (2)

Adding both the equations

2 vp = 2120

vp = 1060 km/h

and va = 1230 - vp = 1230 - 1060 = 170 km/h

5 0
3 years ago
A skydiver of 75 kg mass has a terminal velocity of 60 m/s. At what speed is the resistive force on the skydiver half that when
ankoles [38]

Answer:

The speed of the resistive force is 42.426 m/s

Explanation:

Given;

mass of skydiver, m = 75 kg

terminal velocity, V_T = 60 \ m/s

The resistive force on the skydiver is known as drag force.

Drag force is directly proportional to square of terminal velocity.

F_D = kV_T^2

Where;

k is a constant

k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}

When the new drag force is half of the original drag force;

F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}=  \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}=  0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s

Therefore, the speed of the resistive force is 42.426 m/s

8 0
3 years ago
Someone please help me!
beks73 [17]

Answer:

current = 8,750

Explanation:

you will need to use the formula i = v/r to get this answer

3 0
3 years ago
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