Answer:
Given that
T= 0.43 s
Radius of the ball path's , r=2.1 m
a)
We know that
f= 1/T
Here f= frequency
T= Time period
Now by putting the values
f= 1/T
T= 0.43 s
f= 1/0.43
f=2.32 Hz
b)
We know that
V= ω r
ω = 2 π f
ω=Angular speed
V= Linear speed
ω = 2 π f=ω = 2 x π x 2.32 =14.60 rad/s
V= ω r= 14.60 x 2.1 = 30.66 m/s
c)
Acceleration ,a
a =ω ² r
a= 14.6 ² x 2.1 = 447.63 m/s²
We know that g = 10 m/s²
So
a= a/g= 447.63/10 = 44.7 g m/s²
a= 44.7 g m/s²
Ok so what we need to find is how large is the potential difference between points A and B.
Now you make a statement about what this potential difference would do in this problem
We now know that the points from A and B relies to transport a difference by using points A to B per unit change.
V
=
W/
q
=
50/
10
=
5
V=
q
W
=
10
50
=5 V (5 volt).
"<span>The strength and duration of the wind controls size and frequency" is the one statement among the following choices given in the question that is true about surface waves. The correct option among all the options that are given in the question is the first option. I hope that the answer has come to your desired help.</span>
If you mean 6.9 miles the it is 11.1045 kilometers. You can round it to the tenth place to get 11.1
Answer:
The stretched length is 0.18816 in.
Explanation:
Given that,
Distance = 0.5 in
length = 8 ft
Tensile stress = 20 kpsi
We need to calculate the stretched length
Using formula of length
Where, = Tensile stress
L = length
E = young's modulus of aluminum
Put the value into the formula
Hence, The stretched length is 0.18816 in.