Answer:
The shearing stress is 10208.3333 Pa
The shearing strain is 0.25
The shear modulus is 40833.3332 Pa
Explanation:
Given:
Block of gelatin of 120 mm x 120 mm by 40 mm
F = force = 49 N
Displacement = 10 mm
Questions: Find the shear modulus, Sm = ?, shearing stress, Ss = ?, shearing strain, SS = ?
The shearing stress is defined as the force applied to the block over the projected area, first, it is necessary to calculate the area:
A = 40*120 = 4800 mm² = 0.0048 m²
The shearing stress:

The shearing strain is defined as the tangent of the displacement that the block over its length:

Finally, the shear modulus is the division of the shearing stress over the shearing strain:

Answer:
166 666 666.7 years
Explanation:
We start the question by making the units uniform. We are told that the continents move at 3 cm/year = 0.03 m/year.
We are also told that the continents are now 5000 km = 5 000 000 m apart
So to calculate the time it took for them to be this far apart
t = distance/speed
t = 5 000 000 m/(0.03 m/year) = 166 666 666.7 years
Answer:
The force applied to the surface is 9 kilo Newton.
Explanation:
While jumping on the surface the player applies the force that is equal to its weight on the surface.
The mass of the player is given as 90 kg.
Force applied by the player = weight of the player
Force applied by the player = m × g
Where m is the mass of the player and g is acceleration due to gravity
Force applied by the player = 90 × 9.8
Force applied by the player = 882 Newton
Expressing your answer to one significant figure, we get
Force applied by the player =0. 9 kilo Newton
The force applied to the surface is 0.9 kilo Newton.
The new oscillation frequency of the pendulum clock is 1.14 rad/s.
The given parameters;
- <em>Mass of the pendulum, = M </em>
- <em>Length of the pendulum, = L</em>
- <em>Initial angular speed, </em>
<em> = 1 rad/s</em>
The moment of inertia of the rod about the end is given as;

The moment of inertia of the rod between the middle and the end is calculated as;
![I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} = \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}](https://tex.z-dn.net/?f=I_f%20%3D%20%5Cint%5Climits%5EL_%7BL%2F2%7D%20%7Br%5E2%5Cfrac%7BM%7D%7BL%7D%20%7D%20%5C%2C%20dr%20%3D%20%5Cfrac%7BM%7D%7B3L%7D%20%5Br%5E3%5D%5EL_%7BL%2F2%7D%20%3D%20%20%5Cfrac%7BM%7D%7B3L%7D%20%5BL%5E3%20-%20%5Cfrac%7BL%5E3%7D%7B8%7D%20%5D%20%3D%20%5Cfrac%7BM%7D%7B3L%7D%20%5B%5Cfrac%7B7L%5E3%7D%7B8%7D%20%5D%3D%20%5Cfrac%7B7ML%5E2%7D%7B24%7D)
Apply the principle of conservation of angular momentum as shown below;

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.
Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129