The work done on the filled bucket in raising out of the hole is 2, 925 Joules
<h3>How to determine the work done</h3>
Using the formula:
Work done = force * distance
Note that force = mass * acceleration
F = mg + ma
F = 4. 5 * 10 + 28 * 10
F = 45 + 280
F = 325 Newton
Distance = 9m
Substitute into formula
Work done = 325 * 9
Work done = 2, 925 Joules
Therefore, the work done is 2, 925 Joules
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Explanation:
Given that,
Angle by the normal to the slip α= 60°
Angle by the slip direction with the tensile axis β= 35°
Shear stress = 6.2 MPa
Applied stress = 12 MPa
We need to calculate the shear stress applied at the slip plane
Using formula of shear stress
![\tau=\sigma\cos\alpha\cos\beta](https://tex.z-dn.net/?f=%5Ctau%3D%5Csigma%5Ccos%5Calpha%5Ccos%5Cbeta)
Put the value into the formula
![\tau=12\cos60\times\cos35](https://tex.z-dn.net/?f=%5Ctau%3D12%5Ccos60%5Ctimes%5Ccos35)
![\tau=4.91\ MPa](https://tex.z-dn.net/?f=%5Ctau%3D4.91%5C%20MPa)
Since, the shear stress applied at the slip plane is less than the critical resolved shear stress
So, The crystal will not yield.
Now, We need to calculate the applied stress necessary for the crystal to yield
Using formula of stress
![\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}](https://tex.z-dn.net/?f=%5Csigma%3D%5Cdfrac%7B%5Ctau_%7Bc%7D%7D%7B%5Ccos%5Calpha%5Ccos%5Cbeta%7D)
Put the value into the formula
![\sigma=\dfrac{6.2}{\cos60\cos35}](https://tex.z-dn.net/?f=%5Csigma%3D%5Cdfrac%7B6.2%7D%7B%5Ccos60%5Ccos35%7D)
![\sigma=15.13\ MPa](https://tex.z-dn.net/?f=%5Csigma%3D15.13%5C%20MPa)
Hence, This is the required solution.
The universe is made up of baryonic matter.(neutrons,electrons,protons)
the greater the <u>mass</u> of an object the more force is needed to cause acceleration
Answer:
- tension: 19.3 N
- acceleration: 3.36 m/s^2
Explanation:
<u>Given</u>
mass A = 2.0 kg
mass B = 3.0 kg
θ = 40°
<u>Find</u>
The tension in the string
The acceleration of the masses
<u>Solution</u>
Mass A is being pulled down the inclined plane by a force due to gravity of ...
F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N
Mass B is being pulled downward by gravity with a force of ...
F = mg = (3 kg)(9.8 m/s^2) = 29.4 N
The tension in the string, T, is such that the net force on each mass results in the same acceleration:
F/m = a = F/m
(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)
T = (2(29.4) +3(12.5986))/5 = 19.3192 N
__
Then the acceleration of B is ...
a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2
The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.