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tamaranim1 [39]

1 year ago

6

A child’s toy rake is held so that its resistance length is 0.85 meters. If the mechanical advantage is 0.43, what is the effort

distance? plz

1 answer:

mart [117]1 year ago

7 0

Answer:

1.28

Explanation:

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To make the jump, Neo and Morpheus have pushed against their respective launch points with their legs applying a _____ to the la

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Force, newtons 3rd law of motion stated for every action there is an equal and opposite reaction

7 0

1 year ago

A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

Kinetic energy K.E = $\frac{1}{2}mv^2$ .

From above we know that -

Kinetic energy at the bottom of the cliff = potential energy at a height h

$\frac{1}{2}mv^2=\ mgh$

⇒ $v^2=\ 2gh$

⇒ $h=\ \frac{v^2}{2g}$

⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m

5 0

1 year ago

The image shows one complete cycle of a mass on a spring in simple harmonic motion. An illustration of a mass on a vertical spri

Alja [10]

Answer:

D. "The net force is zero, so the acceleration is zero"

Explanation:

edge 2020

6 0

2 years ago

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that

mars1129 [50]

Answer:

The maximum transverse speed of the bead is 0.4 m/s

Explanation:

As we know that the Amplitude of the travelling wave is

A = 3.65 mm

Now the speed of the travelling wave is

$v_x = 13.5 m/s$

now we know that distance of first antinode from one end is 27.5 cm

so length of the loop of the standing wave is given as

$\frac{\lambda}{4} = 27.5 cm$

$\lambda = 110 cm$

now we have

$N = \frac{2L}{\lambda}$

$N = \frac{2(1.65)}{1.10}$

$N = 3$

now we have

$R = 2A sin(kx)$

$R = 2(3.65) sin(\frac{2\pi}{1.10}x)$

$R = 7.3 sin(1.82 \pi x)$

now at x = 13.8 cm

$R = 7.3 sin(1.82 \pi (0.138))$

$R = 5.18 mm$

now we have

$f = \frac{v}{\lambda}$

$f = \frac{13.5}{1.1}$

$f = 12.27 Hz$

now maximum speed is given as

$v_y = R\omega$

$v_y = (5.18 \times 10^{-3})(2\pi(12.27))$

$v_y = 0.4 m/s$

4 0

1 year ago

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

Assoli18 [71]

Answer:

The tension in the cable when the craft was being lowered to the seafloor is 4700 N.

Explanation:

Given that,

When the craft was stationary, the tension in the cable was 6500 N.

When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N.

The drag force of 1800 N will act in the upward direction. As it was lowered or raised at a steady rate, so its acceleration is 0. As a result, net force is 0. So,

T + F = W

Here, T is tension

F = 1800 N

W = 6500 N

Tension becomes :

$T=W-F\\\\T=6500-1800\\\\T=4700\ N$

So, the tension in the cable when the craft was being lowered to the seafloor is 4700 N.

7 0

1 year ago