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2 years ago

7

a 15-kg block is on a frictionless ramp that is inclined at 20° above the horizontal. it is connected by a very light string ove

r an ideal pulley at the top edge of the ramp to a hanging 19-kg block, as shown in the figure. the string pulls on the 15-kg block parallel to the surface of the ramp. find the magnitude of the acceleration of the 19-kg block after the system is gently released?

1 answer:

saw5 [17]2 years ago

5 0

Hi there!

Since the string is light and there is no friction in the pulley, the acceleration of the system is equal to the acceleration of both blocks.

We can begin by summing the forces of each block:

Block on incline:

- Force of gravity (in the negative direction away from the acceleration)

- Force of Tension

∑F = -M₁gsinФ + T

Block hanging:

- Force of gravity (Positive, in direction of acceleration)

- Force of Tension (Negative, opposite from acceleration)

∑F = M₂g - T

Sum both of these net forces for each block:

∑Fт = -M₁gsinФ + T - T + M₂g

∑Fт = -M₁gsinФ + M₂g

Divide by the mass to solve for acceleration:

$a = \frac{-M_1gsin\theta+M_2g}{M_1+M_2}$

Plug in the given values:

$a = \frac{-(15)(9.81)(sin20)+19(9.81)}{15+19} = 4.002 m/s^2$

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The maximum emf generated around the perimeter of a cell in a field is given by

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For maximum emf cos(ωt) = 1

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ω = 2π(72.5)

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Emf = BAω

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Explanation:

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Using Snell's law as:

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3 years ago

an empty boat floats in water with 10% of its volume submerged. and when it is loaded with 1200kg , the volume submerged will in

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Let volume of empty boat be = 100% = 1V

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boat submerging.

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By comparison

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1 year ago

The wheel having a mass of 100 kg and a radius of gyration about the z axis of kz=300mm, rests on the smooth horizontal plane.a.

pickupchik [31]

Answer:

a) 20 rad/s

b) 6 m/s

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From Newton's second law of motion, F = m × a

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m is mass of the body

a is the acceleration of the body

By substituting the values we get, a = 2 m/s²

As acceleration is constant, we can use the below formula for calculating the final velocity of the object

v = u + a × t

Where v is the final velocity

u is the initial velocity

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t is the time taken

u = 0 (∵ it starts from rest)

By substituting the values we get

v = 0 + 2 × 3 = 6 m/s

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Torque acting on the wheel about axis of rotation = 300 mm × 200 N =

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Assuming that the mass of spokes of the wheel to be negligible,

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Then,

60 = 9 × (angular acceleration)

∴ angular acceleration ≈ 6·67 rad/s²

As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed

$w_{f}$ = $w_{i}$ + α × t

Where

$w_{f}$ is the final angular velocity

$w_{i}$ is the initial angular velocity

α is the angular acceleration

t is the time taken

$w_{i}$ is 0 (∵ initially it starts from rest)

By substituting the values we get

$w_{f}$ = 6·67 × 3 = 20 rad/s

∴ Angular velocity of the wheel after three seconds = 20 rad/s

3 0

3 years ago