Answer:
a) γ =0.055556
b) t = 0.4 MPa
Explanation:
Given:
- The dimensions of rubber block : 18 x 21 x 25
- A load was applied at upper frame P = 420 N
- The rubber deflects dx = 1 mm downwards
Find:
(a) average shear strain in the rubber mounts
(b) average shear stress in the rubber mounts.
Solution:
- For average shear strain we have the definition:
γ = dx / y
Where,
γ: The shear strain.
dx : Deflection along the shear force
y : The length perpendicular to deflection.
- From given data we have dx = 1mm, and the dimension of block perpendicular to deflection is the a dimension. Hence, dx = 0.001 and y =0.018 m:
γ = 0.001 / 0.018 = 0.055556
- The average shear stress along the mating flat surface. We have from definition:
t = F_shear / Area
- Where, F_shear: The shear force on each rubber block is P/2.
Hence,
t = (P/2) / b*c
Plug values in:
t = (420/2) / (0.021*0.025)
t = 0.4 MPa
i. 1,350 calories
j. 992 calories
k. 235.6 calories
2. Calories of energy into water = (amount of water, in grams) x (change in the water's temperature, in °C)
3.a. 840 calories
3.b. 840 calories
4. 80°C
Note that by the law of inertia, it's the nature of objects to continually be in motion. For something to be still, there has to be a presence of a force.
They mention "no outside forces" so you know that a kid isn't kicking the object or a train is hitting it, or anything like that. Still, there are internal forces working, and I drew it out for you :) Hope this helps!
<span>Rebecca is learning to play cello and is highly motivated. She is confident that she can learn cello easily and with great ability because she already knows how to play violin. From Mischel’s perspective, Rebecca’s attitude is best explained by the expectancies variable. </span>Expectancy<span> is the belief that increased effort will lead to increased performance </span>
Answer:
442.5 rad
Explanation:
w₀ = initial angular velocity of the disk = 7.0 rad/s
α = Constant angular acceleration = 3.0 rad/s²
t = time period of rotation of the disk = 15 s
θ = angular displacement of the point on the rim
Angular displacement of the point on the rim is given as
θ = w₀ t + (0.5) α t²
inserting the values
θ = (7.0) (15) + (0.5) (3.0) (15)²
θ = 442.5 rad