Answer:
0.3376 mm
Explanation:
The computation of the spacing in mm between the slits is shown below:
As we know that
where,
= wavelength
L = distance from the scrren
= spanning distance
As there are 11 bright fingers seen so m would be
= 11 - 1
= 10
Now placing these values to the above formula
So, the spacing is
= 0.3376 mm
We simply applied the above formula.
Answer:
The magnitude of the static frictional force is 1200 N
Explanation:
given information :
radius, r = 0.380 m
applied-torque, τ1 = 456 N
The car has a constant velocity, thus the acceleration is zero
α = 0
Στ = I α
τ1 - τ2 = I α
τ2 = counter-torque
τ1 - τ2 = 0
τ1 = τ2
r x 
= τ1
= the static frictional force (N)
= τ1 /r
= 456 N/0.380 m
= 1200 N
Answer:
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Answers
Related Questions
How many non-square numbers lie between the squares of 12 and 13?
Answer
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Hint: Here, we can see that 12 and 13 are consecutive numbers. So, all numbers between squares of 12 and 13 are non-square numbers. Therefore, first find squares of 12 and 13 and then subtract square of 12 from square of 13, we get numbers of non-square numbers. At the last subtract 1 from the result obtained as both extremes numbers are not included.
Complete step-by-step answer:
In these types of questions, a simple concept of numbers should be known that is between squares of two consecutive numbers all numbers are non-square numbers. Also one tricky point should remember that whenever we find the difference between two numbers we get a number of numbers between them including anyone of the extreme numbers. So we subtract 1 to exclude both extreme numbers.
Square of 12 = 122=144 and square of 12 = 132=169
As 12 and 13 are consecutive numbers so all numbers between their squares will be non-square numbers.
Therefore, 169 – 144 = 25
Total number of numbers between 169 and 144 (i.e., excluding 144 and 169) = 25 – 1 = 24.
Explanation:
Brian least po please
Answer:
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
Explanation:
Given that
Yield strength ,Sy= 240 MPa
Tensile strength = 310 MPa
Elastic modulus ,E= 110 GPa
L=380 mm
ΔL = 1.9 mm
Lets find strain:
Case 1 :
Strain due to elongation (testing)
ε = ΔL/L
ε = 1.9/380
ε = 0.005
Case 2 :
Strain due to yielding
ε '=0.0021
Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.
For computation of load strain due to testing should be less than the strain due to yielding.
