Answer:
337k
Explanation:
First, let us find the difference between the given two temperatures.
Difference = 85°C - 21°C
= 64°C
<u>And now we have to write the temperature in kelvins.</u>
To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.
<u>Let us find it now.</u>
64°C + 273 = 337k
Therefore,
64°C ⇒ <u>337k</u>
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
Answer:
The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)
Explanation:
Fundamental frequency = wave velocity/2L
where;
L is the length of the stretched rubber
Wave velocity = 
Frequency (F₁) = 
To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.
Given:
L₂ =2L₁ = 2L
T₂ = 2T₁ = 2T
(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)
F₂ = ![\frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B%5Cfrac%7B2T%7D%7B0.5%28%5Cfrac%7BM%7D%7BL%7D%29%7D%7D%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B4%28%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%29%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B2%7D%7B2%7D%20%5B%5Cfrac%7B%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%7D%7B2%2AL%7D%5D%20%3D%20F_1)
Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).
Option(a) the mass of cart 2 is twice that of the mass of cart 1 is the right answer.
The mass of cart 2 is twice that of the mass of cart 1 is correct about the mass of cart 2.
Let's demonstrate the issue using variables:
Let,
m1=mass of cart 1
m2=mass of cart 2
v1 = velocity of cart 1 before collision
v2 = velocity of cart 2 before collision
v' = velocity of the carts after collision
Using the conservation of momentum for perfectly inelastic collisions:
m1v1 + m2v2 = (m1 + m2)v'
v2 = 0 because it is stationary
v' = 1/3*v1
m1v1 = (m1+m2)(1/3)(v1)
m1 = 1/3*m1 + 1/3*m2
1/3*m2 = m1 - 1/3*m1
1/3*m2 = 2/3*m1
m2 = 2m1
From this we can conclude that the mass of cart 2 is twice that of the mass of cart 1.
To learn more about inelastic collision visit:
brainly.com/question/14521843
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The source of information was biased. It was like walking along a river bank in the country and asking everybody you meet whether they like fishing. Or asking 500 people sitting in the bleachers whether they like baseball.
I'm sure the scientist would have gotten different data if she interviewed 500 teenagers at neighborhood basketball courts, or 500 teenagers at a rock concert.