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
<u>Answer:</u>
Chipmunk's average acceleration during the 2.07 s time interval = 0.1256
<u>Explanation:</u>
We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
In this case we need to find acceleration value, we have initial velocity = 1.47 m/s, final velocity = 1.73 m/s and time taken = 2.07 seconds.
Substituting
1.73 = 1.47 + a * 2.07
a = 0.1256
Chipmunk's average acceleration during the 2.07 s time interval = 0.1256
Answer:
28 m/s
Explanation:
vf^2=vi^2+2a(delta y)
=sqrt(2* -9.8 m/s^2* -40 m)
=28 m/s
Explanation:
It is given that,
Mass of the object, m = 0.8 g = 0.0008 kg
Electric field, E = 534 N/C
Distance, s = 12 m
Time, t = 1.2 s
We need to find the acceleration of the object. It can be solved as :
m a = q E.......(1)
m = mass of electron
a = acceleration
q = charge on electron
"a" can be calculated using second equation of motion as :
a = 16.67 m/s²
Now put the value of a in equation (1) as :
q = 0.0000249 C
or
Hence, this is the required solution.