Answer:
a = 17.68 m/s²
Explanation:
given,
length of the string, L = 0.8 m
angle made with vertical, θ = 61°
time to complete 1 rev, t = 1.25 s
radial acceleration = ?
first we have to calculate the radius of the circle
R = L sin θ
R = 0.8 x sin 61°
R = 0.7 m
now, calculating at the angular velocity


ω = 5.026 rad/s
now, radial acceleration
a = r ω²
a = 0.7 x 5.026²
a = 17.68 m/s²
hence, the radial acceleration of the ball is equal to 17.68 rad/s²
For a standing wave on a string, the wavelength is equal to twice the length of the string:

In our problem, L=50.0 cm=0.50 m, therefore the wavelength of the wave is

And the speed of the wave is given by the product between the frequency and the wavelength of the wave:
B) not work ,because the water would freeze
Answer:
T= 38.38 N
Explanation:
Here
mass of can = m = 3 kg
g= 9.8 m/sec2
angle θ = 40°
From figure we see the vertical and horizontal component of tension force T
If the can is to slip - then horizontal component of tension force should become equal to force of friction.
First we find force of friction
Fs= μ R
where
μ = 0.76
R = weight of can = mg = 3 × 9.8 = 29.4 N
Now horizontal component of tension
Tx= T cos 40 = T× 0.7660 N
==>T× 0.7660 = 29.4
==> T= 38.38 N
Answer:
a.Distance = 150 m
b. Displacement = 50 m
Time lapsed = 5 seconds
Explanation:
a. Distance is the change in the position of an object.
The distance covered by the car = 100 + 50
= 150 m
b. Since displacement is a vector quantity,
Displacement of the car = 100 - 50
= 50 m due east
c. Time elapsed is the time taken for the motion of the car starting from when its starts to when it stops.
Thus, the time elapsed = 4 + 1
= 5 seconds