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3 years ago

Help me out guysI'm stuck in this

Physics

1 answer:

Juli2301 [7.4K]3 years ago

6 0

Answer: The answer will be R1 = B

Explanation:

Cause Here x component of A = A

X component of 2B = 2BcosФ

X component of Resultant = 0

I.e. A + 2BcosФ = 0

cosФ = -A/2B

also R1 = √(A² + B² + 2ABcosФ)

=√ [A² + B² + 2AB(-A/2B)]

=√A² + B² -A²

R1 = √B²

R1 = B so R1 = B is correct.

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Gamma rays, X-rays, most ultraviolet rays, and some infrared are absorbed by the atmosphere but do not reach the Earth's surface

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3 years ago

A solenoid with an inductance of 8 mH is connected in series with a resistance of 5 Ω and an EMF forming a series RL circuit. A

monitta

Answer:

induced EMF = 240 V

and by the lenz's law direction of induced EMF is opposite to the applied EMF

Explanation:

given data

inductance = 8 mH

resistance = 5 Ω

current = 4.0 A

time t = 0

current grow = 4.0 A to 10.0 A

to find out

value and the direction of the induced EMF

solution

we get here induced EMF of induction is express as

E = - L $\frac{dI}{dt}$ ...................1

so E = - L $\frac{I2 - I1}{dt}$

put here value we get

E = - 8 × $10^{-3}$ $\frac{10 - 4}{0.2*10^{-3}}$

E = -40 × 6

E = -240

take magnitude

induced EMF = 240 V

and by the lenz's law we get direction of induced EMF is opposite to the applied EMF

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3 years ago

A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with rig

aleksandr82 [10.1K]

Answer:

The rate of change of the shadow length of a person is 2.692 ft/s

Solution:

As per the question:

Height of a person, H = 20 ft

Height of a person, h = 7 ft

Rate = 5 ft/s

Now,

From Fig.1:

b = person's distance from the lamp post

a = shadow length

Also, from the similarity of the triangles, we can write:

$\frac{a + b}{20} = \frac{a}{7}$

$a = \frac{7}{13}b$

Differentiating the above eqn w.r.t t:

$\frac{da}{dt} = \frac{7}{13}.\frac{db}{dt}$

Now, we know that:

Rate = $\frac{db}{dt} = 5\ ft/s$

Thus

$\frac{da}{dt} = \frac{7}{13}.\times 5 = 2.692\ ft/s$

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3 years ago

1. To calulate the molarity of a solution, you need to know the moles of the solute and the?

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1. Volume of the solution (B)

2. Celery (D)

3. Hydroxide ions in solution (A)

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3 years ago

A basketball weighing 0.63 kg is dropped from a height of 6.0 meters onto a court. Use the conservation of energy equation to de

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Answer:

The velocity of the ball at a height of 2.0 meters above the court is approximately 8.85 m/s

Explanation:

The given parameters of the ball are;

The mass of the ball, m = 0.63 kg

The height from which the ball is dropped, h₁ = 6.0 meters

The height at which the velocity of the ball is sought, h₂ = 2 meters

The initial potential energy of the ball, P.E. = m·g·h₁ = 0.63 × 9.8 × 6.0 = 37.044

The initial potential energy of the ball, P.E.₁ = 37.044 J

The potential energy of the ball, when the ball is at 2 meters above the court, P.E.₂ = m·g·h₂ = 0.63 × 9.8 × 2.0 = 12.348

The potential energy of the ball, when the ball is at 2 meters above the court, P.E.₂ = 12.348 J

From M.E> = P.E. + K.E.

Where;

M.E = The total mechanical energy of the ball = Constant

P.E. = The potential energy of the ball

K.E. = The kinetic energy of the ball

By the conservation of energy principle, we have;

The potential energy lost by the ball = The kinetic energy gained by the ball

The potential energy lost by the ball = P.E.₁ - P.E.₂ = 37.044 - 12.348 = 24.696

The potential energy lost by the ball = 24.696 J

The kinetic energy gained by the ball = 1/2·m·v² = 1/2×0.63×v²

Where;

v = The velocity of the ball

∴ The potential energy lost by the ball at 2.0 meters above the court = 24.696 J = The kinetic energy gained by the ball at 2.0 meters above the court = 1/2×0.63×v²

24.696 J = 1/2×0.63 kg ×v²

v² = 24.696 J / (1/2×0.63 kg) = 78.4 m²/s²

∴ v = √(78.4 m²/s²) = 8.85437744847 m/s

The velocity of the ball at a height of 2.0 meters above the court, v ≈ 8.85 m/s.

7 0

3 years ago

Help me out guysI'm stuck in this ​

Help me out guysI'm stuck in this