xercise 15.29: At a distance of 6.00×1012 m from a star, the intensity of the radiation from the star is 14.7 W/m2. Assuming tha
1 answer:
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Answer:
(a) Tangential velocity will be 38.648 m/sec
(b) Acceleration will be
Explanation:
We have given radius r = 11.2 m
Angular speed
(a) We have to find the tangential velocity
We know that tangential velocity is given by
(b) We know that acceleration is given by
Answer:
d= 7.32 mm
Explanation:
Given that
E= 110 GPa
σ = 240 MPa
P= 6640 N
L= 370 mm
ΔL = 0.53
Area A= πr²
We know that elongation due to load given as
A= 42.14 mm²
πr² = 42.14 mm²
r=3.66 mm
diameter ,d= 2r
d= 7.32 mm
Answer:
The mass of the meter stick is
Explanation:
Here since the meter stick is in equilibrium position its net torque and net force should be equal to zero
Since at the begging the meter stick is balanced at center of the meter stick that means its center of mass should be present at
Now lets consider the later case where stick is balanced by two 5.12 g coins .
Here torque due to two coins =
Torque due to weight of meter stick = =
where m = mass of the meter stick
Here .
Upon equating we will be getting mass of the meter stick =
Answer:
29,7 m
Explanation:
We need to devide the problem in two parts:
A) Energy
B) MRUV
<u>Energy:</u>
Since no friction between pint (1) and (2), then the energy conservatets:
Energy = constant ----> Ek(cinética) + Ep(potencial) = constant
Ek1 + Ep1 = Ek2 + Ep2
Ek1 = 0 ; because V1 is zero (the ball is "dropped")
Ep1 = m*g*H1
Ep2= m*g*H2
Then:
Ek2 = m*g*(H1-H2)
By definition of cinetic energy:
m*(V2)²/2 = m*g*(H1-H2) --->
Replaced values: V2 = 14,0 m/s
<u>MRUV:</u>
The decomposition of the velocity (V2), gives a for the horizontal component:
V2x = V2*cos(α)
Then the traveled distance is:
X = V2*cos(α)*t.... but what time?
The time what takes the ball hit the ground.
Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²
In the vertical axis:
Y3 = 0 ; Y2 = H2 = 2 m
Reeplacing:
-2 = 14*t + (1/2)*(-9,81)*t²
solving the ecuation, the only positive solution is:
t = 2,99 sec ≈ 3 sec
Then, for the distance:
X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m
