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3 years ago

8

The electromagnetic interaction _______.A. applies only to charges at rest B. applies only to charges in motion C. is responsibl

e for sliding friction and contact forces D. all of the above E. none of the above

1 answer:

Bond [772]3 years ago

8 0

Answer:

C. is responsible for sliding friction and contact forces

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What task requires the most work, lifting a 12-kg sack 2 meters or lifting a 25-kg sack 1 meter?

MrMuchimi

Multiply the masses by the respective distances:

(12 kg) (2 m) = 24 J

(25 kg) (1 m) = 25 J

so the heavier bag takes more work to lift, and (b) is the answer.

(d) is technically correct if the sacks are carrying different contents whose masses are not equal, but since we don't know what's inside each sack, assume 12 kg and 25 kg are the masses of each sack *and* their contents.

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The bond enthalpy value for a carbon to hydrogen bond is 413 kJ. What does this mean?

AlladinOne [14]

Breaking bond requires energy. The bond between the carbon and hydrogen is broken when the energy is absorbed. The enthalpy is defined to be the energy taken to break the one mole of the stated carbon and hydrogen bond. Thus a should be the correct answer

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3 years ago

Read 2 more answers

Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude

aivan3 [116]

Answer:

$E=6.8Kv/m$

Explanation:

From the question we are told that

Distance b/w plate $d=10cm=>0.1m$

P_1 Potential at 7.35 $V=533v$

Generally the equation for electric field at a distance is mathematically given as

$E=\frac{v}{d}$

$E=\frac{533}{7.85*10^-^2}$

$E=6789.808917$

$E=6.8*10^3$

$E=6.8Kv/m$

7 0

3 years ago

A car with a mass of 1380 Kg is traveling at 23 m/s to the north. A truck with a mass of 1625 Kg is traveling at 26 m/s to the s

trasher [3.6K]

Answer: -3.49 m/s (to the south)

Explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum $p_{i}$ must be equal to the final momentum $p_{f}$ , and taking into account this is aninelastic collision:

Before the collision:

$p_{i}=mV_{o}+MU_{o}$ (1)

After the collision:

$p_{f}=(m+M)V_{f}$ (2)

Where:

$m=1380 kg$ is the mass of the car

$V_{o}=23 m/s$ is the velocity of the car, directed to the north

$M=1625 kg$ is the mass of the truck

$U_{o}=-26 m/s$ is the velocity of the truck, directed to the south

$V_{f}$ is the final velocity of both the car and the truck

$p_{i}=p_{f}$ (3)

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1380 kg)(23 m/s)+(1625 kg)(-26 m/s)}{1380 kg+1625 kg}$ (6)

Finally:

$V_{f}=-3.49 m/s$ The negative sign indicates the direction of the velocity is to the south

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3 years ago

A boy pushes a cart with a constant velocity of 0.5m/s by applying a force of 60 N. What is the total frictional force acting on

Nutka1998 [239]

Answer:

60N

Explanation:

in this case the minimum amount of force required must be equal to the friction Force. i.e <u>Newton</u><u>'s</u><u> </u><u>first</u><u> </u><u>law</u><u> of</u><u> </u><u>mot</u><u>ion</u><u>.</u>

therefore the maximum amount of frictional force is equal to the applied force which is 60N.

because of the net force acting on the object is zero the object is in constant motion . i.e equal and opposite force must be applied so that the object is in constant velocity therefore the total frictional force must be 60N

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3 years ago