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Sauron [17]
3 years ago
5

If the magnitude of the magnetic force on a proton is F when it is moving at 18.0 ∘ with respect to the field, what is the magni

tude of the force (in terms of F) when this charge is moving at 33.0 ∘ with respect to the field?
Physics
1 answer:
Lelu [443]3 years ago
5 0

Answer:

The force when θ = 33° is 1.7625 times of the force when θ = 18°

Explanation:

The force on a moving charge through a magnetic field is given by

F = qvB sin θ

q = charge of the moving particle

v = Velocity of the moving charge

B = Magnetic field strength

θ = angle between the magnetic field and the velocity (direction of the motion) of the moving charge

Because qvB are all constant, we can call the expression K.

F = K sinθ

when θ = 18°,

F = K sin 18° = 0.309K

when θ = 33°, let the force be F₁

F₁ = K sin 33° = 0.5446K

(F₁/F) = (0.5446K/0.309K) = 1.7625

F₁ = 1.7625 F

Hope this Helps!!!

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A roller coaster begins at the top of a hillIf it accelerates at the rate of 2m/s ^2 And has a mass of 2000kg, what net force is
Luden [163]

The correct answer to the question is : 4000 N

EXPLANATION:

As per the question, the mass of the roller coaster is given as m = 2000 kg

The acceleration of the roller coaster is given as a = 2m/s2

We are asked to calculate the net force acting on the roller coaster.

From Newton's second laws of motion,we know that net external force acting on a particle is equal to the product of mass with acceleration of the particle.

Mathematically it is written as-

F = ma

Hence,net force acting on the roller coaster is calculated as-

F = 2000Kg×2m/s2

= 4000 N.

Here, Newton (N) is the unit of force.

Hence,net force acting on the roller coaster is 4000 N.

5 0
3 years ago
You are designing an airport for small planes. One kind of airplane that might use this airfield must reach a speed before takeo
ladessa [460]

Explanation:

Given that,

Initial speed of the airfield, u = 0

Final speed, v = 27.8 m/s

Acceleration of the airfield, a=2\ m/s^2

Length of the runway, d = 150 m

Let v' is the speed of the airplane to reach the required speed for takeoff. Finding v' using third equation of motion as :

v'^2-u^2=2ad\\\\v'=\sqrt{2ad} \\\\v'=\sqrt{2\times 2\times 150} \\\\v'=24.49\ m/s

This speed is not enough as the airfield must reach a speed before takeoff of at least 27.8 m/s. Now, the required length of the runways is :

v^2=2ax\\\\x=\dfrac{v^2}{2a}\\\\x=\dfrac{(27.8)^2}{2\times 2}\\\\x=193.21\ m

So, the minimum length of the runways is 193.21 meters.

8 0
3 years ago
4. Which of the following statement is correct regarding velocity and speed of a moving body?
lara31 [8.8K]

Answer:

Explanation:

Hello friend!!!

The correct option is <u><em>(d) Velocity of a moving body is its speed in a given direction.</em></u>

Hope this helps

plz mark as brainliest!!!!!!!!

7 0
3 years ago
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solong [7]

Answer:

Opposite sides are congruent (AB = DC).

Opposite angels are congruent (D = B).

Consecutive angles are supplementary (A + D = 180°).

If one angle is right, then all angles are right.

The diagonals of a parallelogram bisect each other.

Each diagonal of a parallelogram separates it into two congruent triangles.

Explanation: #if you need any queshtions answered within secs/mins hit me up and I gotchu.

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A student practicing for track ran 800 meter in 110 seconds. what was her speed?
adoni [48]
Her speed was 7.27 meters per second
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