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Crazy boy [7]
3 years ago
6

A proton traveling to the right enters a region of uniform magnetic field that points into the screen. When the proton enters th

is region, it will be..??
deflected toward bottom of the screen
deflected out of the plane of screen
deflected toward top of the screen
deflected into plane of screen
unaffected in its direction of motion
Physics
1 answer:
Natali5045456 [20]3 years ago
7 0

Answer:

deflected toward top of the screen

Explanation:

We can find the direction of the force acting on the proton by using the right-hand rule. In fact, we have:

- Index finger: direction of motion of the proton --> to the right

- Middle finger: direction of the magnetic field --> into the screen

- Thumb: it gives the direction of the force --> therefore, it will be upward

And since the charge is positive (it is a proton), we don't have to revers the direction: so, the force on the proton points upward.

Therefore, the correct answer will be

deflected toward top of the screen

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A rock has a mass of 25kg. The acceleration due to gravity is 10m/s and the height of the rock is 42 m. Find the potential energ
alisha [4.7K]

Answer:

 The potential energy of the rock = 10.5 kN

Explanation:

 Mass of rock = 25 kg

 Acceleration due to gravity = 10 m/s²

 Height = 42 m

 Potential energy, PE = mgh, where m is the mass, g is acceleration due to gravity and h is the height.

 PE = 25 x 10 x 42 = 10500 N = 10.5 kN

 The potential energy of the rock = 10.5 kN

7 0
2 years ago
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A force of 5.00 N to the left causes a 1.35 kg book to have a net acceleration of 0.76 m/s2 to the left. What is the frictional
Nuetrik [128]

Answer:

4.0 N

Explanation:

Sum the forces in the x direction:

∑F = ma

F − Fr = ma

Fr = F − ma

Fr = 5.00 N − (1.35 kg) (0.76 m/s²)

Fr = 4.0 N

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3 years ago
What is the function of the Cytoplasm?
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3 years ago
A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.
jeka57 [31]

The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

  • The body is gaining the speed, which means there is a change in kinetic energy.
  • The change in kinetic energy is equal to the work done.
  • The friction force is the product of coefficient of the friction and normal force.
  • The friction force for the given case is zero. Thus, the normal force must be equal to the zero.

Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

brainly.com/question/10941832

7 0
2 years ago
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An archer draws her bow and stores 34.8 J of elastic potential energy in the bow. She releases the 63 g arrow, giving it an init
elena-14-01-66 [18.8K]

Answer:

Approximately 71\%.

Explanation:

The formula for the kinetic energy \rm KE of an object is:

\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2,

where

  • m is the mass of that object, and
  • v is the speed of that object.

Important: Joule (\rm J) is the standard unit for energy. The formula for \rm KE requires two inputs: mass and speed. The standard unit of mass is \rm kg while the standard unit for speed is \rm m \cdot s^{-1}. If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,

Convert the unit of the arrow's mass to standard unit:

m = 63\; \rm g = 0.063\; \rm kg.

Initial \rm KE of this arrow:

\begin{aligned}\mathrm{KE} &= \frac{1}{2} \, m \cdot v^2 \\ &= \frac{1}{2}\times 0.063\; \rm kg \times \left(\rm 28 \; m \cdot s^{-1}\right)^2 \\ &\approx 24.696\; \rm J\end{aligned}.

That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:

\displaystyle \frac{24.696\; \rm J}{34.8\; \rm J} \times 100\% \approx 71\%.

8 0
3 years ago
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