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3 years ago

56. What is the range of magnitudes that can be attained from the

Physics

1 answer:

Svetach [21]3 years ago

8 0

Answer:

Explanation:

Given the magnitude of the forces, 7N and 2N, the minimum combining force acting on the forces are when the force's is acting in opposite direction.

Magnitude of the force in opposite direction is 7N - 5N = 2N

The maximum combining force occurs when they act in the same direction. Magnitude of the force in the same direction is 5N+7N = 12N

Hence the range of magnitude requires is 2N≤F≤12N

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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

Lelu [443]

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!!!

5 0

3 years ago

Batman (mass = 96.1 kg) jumps straight down from a bridge into a boat (mass = 458 kg) in which a criminal is fleeing. The veloci

MariettaO [177]

Answer:

The velocity of the boat after the batman lands in it is +9.26 m/s

Explanation:

Applying the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

Note: The collision between the Batman and the boat is an inelastic collision.

m'u'+mu = V(m+m').................... Equation 1

Where m' = mass of the Batman, u' = initial velcoity of the batman, m = mass of the boat, u = initial velocity of the boat, V = common velocity.

make V the subject of equation 1

V = (m'u'+mu)/(m+m')............... Equation 2

Given: m' = 96.1 kg, u' = 0 m/s, m = 458 kg, u = +11.3 m/s.

Substitute these values into equation 2

V = [(96.1×0)+(458×11.2)]/(96.1+458)

V = 5129.6/554.1

V = +9.26 m/s

8 0

3 years ago

What is the force on an electron in a CRT when it’s moving at 2.5 × 105 meters/second perpendicular to a magnetic field of 1.5 t

Firdavs [7]

F = Magnetic Force
B = Magnetic Field
V = Velocity

*The vectors from the photo you get doing the left-hand rule.

The magnetic force is always perpendicular to the magnetic field.

And as told in the statement, the electron is moving perpendicular to a magnetic field, that is, the Velocity forms an 90 degree angle / Right angle with the magnetic field.

The formula to find the Magnetic Force is:

$f = |q| \times v \times b \times sin \: \theta$

Where "q" is the Charge and the sin theta is the angle formed by the Velocity and Magnetic Field, in this case it's 90°. Sin 90° = 1.

$f = |- 1.6 \times {10}^{ - 19} | \times 2.5 \times {10}^{5} \times 1.5 \times 1 \\ f = 6 \times {10}^{ - 19 + 5} \\ f = 6 \times {10}^{ - 14} \: newtons$
Newton (N) = C x m/s x T = (C x m x T)/s