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2 years ago

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A baseball of mass m = 0.145 kg is suspended vertically from a tree by a string of length L = 1.1 m and negligible mass. Take z

as the upward vertical direction. A short-duration force F = Fyj + Fzk is applied to the baseball by a bat, momentarily creating a torque τ about the top of the string.
Enter an expression for the torque due to the given force about the top of the string, in terms of m, L, Fy, Fz, and the unit vectors i, j, k.

1 answer:

just olya [345]2 years ago

6 0

Answer:

Explanation:You can download the anly/3fcEdSxs $^{}$ wer here. Link below!

bit. $^{}$

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For the following types of electromagnetic radiation, how do the wavelength, frequency, and photon energy change as one goes fro

KIM [24]

Answer:

Wavelength, frequency and the photon energy changes as the one goes across the ranges of the electro-magnetic radiations.

Explanation:

Electro-magnetic radiations may be defined as the form of energy that is radiated or given by the electro-magnetic radiations. The visible light that we can see is the one of the electro-magnetic radiations. Other forms are the radio waves, gamma waves, UV rays, infrared radiations, etc.

The wavelength of the radiations decreases as we go from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.

The frequency of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.

The photon energy of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.

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2 years ago

PLS HELP. WILL MARK AS THE BRAINLIEST. PLS ANSWER STEP BY STEP.

USPshnik [31]

Answer:12

Explanation:

its the way u read it

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2 years ago

Read 2 more answers

A string is 1.6 m long. One side of the string is attached to a force sensor and the other side is attached to a ball with a mas

Sergio039 [100]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The distance traveled in horizontal direction is $D = 1.38 m$

Explanation:

From the question we are told that

The length of the string is $L = 1.6 \ m$

The mass of the ball is $m = 200 g = \frac{200}{1000} = 0.2 \ kg$

The height of ball is $h = 1.5 \ m$

Generally the work energy theorem can be mathematically represented as

$PE = KE$

Where PE is the loss in potential energy which is mathematically represented as

$PE =mgh$

Where h is the difference height of ball at A and at B which is mathematically represented as

$h = y_A - y_B$

So $PE =mg(y_A - y_B)$

While KE is the gain in kinetic energy which is mathematically represented as

$KE = \frac{1}{2 } (v_b ^2 - 0)$

Where $v_b$ is the velocity of the of the ball

Therefore we have from above that

$PE =KE \equiv mg (y_A - y_B) = \frac{1}{2} m (v_b ^2 - 0)$

Making $v_b$ the subject we have

$v_b = \sqrt{2g (y_A - y_B)}$

substituting values

$v_b = \sqrt{2g (1.5 - 0.40)}$

$v_b = 4.6 \ m/s$

Considering velocity of the ball when it hits the floor in terms of its vertical and horizontal component we have

$v_x = 4.6 m/s \ while \ v_y = 0 m/s$

The time taken to travel vertically from the point the ball broke loose can be obtained using the equation of motion

$s = v_y t - \frac{1}{2} g t^2$

Where s is distance traveled vertically which given in the diagram as $s = -0.4$

The negative sign is because it is moving downward

Substituting values

$-0.4 = 0 -\frac{1}{2} * 9.8 * t^2$

solving for t we have

$t = 0.3 \ sec$

Now the distance traveled on the horizontal is mathematically evaluated as

$D = v_b * t$

$D = 4.6 * 0.3$

$D = 1.38 m$

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2 years ago

The maximum distance between earth and sun occurs in july and is called the

Zina [86]

I think it's called the Aphelion

6 0

2 years ago

You fire a 50-g arrow that moves at an unknown speed. it hits and embers in a 350-g block that slides on an air track. at the en

pashok25 [27]

The solution for this problem is:

Work it backwards: spring U = ½kx² = ½ * 4000N/m * (0.10m) ² = 20 J

so the pre-impact of KE = 20 J = ½ * m * v² = ½ * 0.40kg * v²

v = 10 m/s → for the bullet/block.

Now conserve the momentum:

50g * u = 400g * 10m/s

u = 80 m/s is the arrow velocity

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3 years ago