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Gekata [30.6K]
3 years ago
9

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.
Physics
1 answer:
just olya [345]3 years ago
6 0

Answer:

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

bit.^{}

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The photo shows a skateboarder pushing her foot against the ground as she rides down a hill.
Alinara [238K]

Answer:

yes

Explanation:

please Mark my answer in brainlist and follow me

4 0
2 years ago
Your office has a 0.025 m^3 cylindrical container of drinking water. The radius of the container is about 13 cm. Required:a. Whe
uranmaximum [27]

Answer:

<em>a) 105935.7 Pa</em>

<em>b) 103630.35 Pa</em>

Explanation:

The volume of the container = 0.025 m^3

The radius of the container = 13 cm = 0.13 m

We have to find the height of the tank

From the equation for finding the volume of the cylinder,

V = \pi r^2h

where

V is the volume of the cylinder

h is the height of the cylinder

substituting values, we have

0.025 = 3.142 x 0.13^2 x h

0.025 = 0.0531h

h = 0.025/0.0531 = 0.47 m

Pressure at the bottom of the tank P = ρgh

where

ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the depth of water which is equal to the height of the tank

substituting values, we have

P = 1000 x 9.81 x 0.47 = 4610.7 Pa

atmospheric pressure = 101325 Pa

therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = <em>105935.7 Pa</em>

b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m

pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa

This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = <em>103630.35 Pa</em>

7 0
3 years ago
3. What is the gravitational force between a 70 kg physics student and her 1 kg textbook, at a distance of 1 meter? (This number
Rina8888 [55]

ANSWER

\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}

EXPLANATION

Parameters given:

Mass of the student, M = 70 kg

Mass of the textbook, m = 1 kg

Distance, r = 1 m

To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

F=\frac{GMm}{r^2}

where G = gravitational constant

Therefore, the gravitational force acting between the student and the textbook is:

\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\  \\ F=4.67*10^{-9}\text{ }N \end{gathered}

That is the answer.

6 0
1 year ago
PLEASE HELP ME,, I WOULD BE SO HAPPY
Juliette [100K]

Answer:

Energy is force times distance. For your problem, no matter how long you push, the wall still goes nowhere, so there is no obvious energy transfer. so in conclusion, you actually didn't do anything :(

Explanation:

6 0
3 years ago
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The phase change in which a substance changes from a gas directly to a solid is
vlabodo [156]
<span>Condensation is the change of the substance from liquid to solid phase. Example of this is the formation of ice. Vaporization is the change of substance from liquid to gas phase. Example of this is the boiling of water. Deposition is the change of a substance from gas to solid phase. Example of this is the formation of ice on a winter day. Sublimation is the change of a substance from solid to gas phase. Example of this is dry ice. The answer is letter C.</span>
7 0
3 years ago
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