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
Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :

m is the mass of object


m = 0.25 kg


k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.
Answer:
Explanation:
The explanation is given in the attached document.
Answer:
Explanation:
Given
Minute hand length =16 cm
Time at a quarter after the hour to half past i.e. 1 hr 45 min
Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270


(c)For the next half hour
Effectively it has covered 2 revolution and a quarter

angle turned 
(f)Hour after that
After an hour it again comes back to its original position thus displacement is same =25.136
Angle turned will also be same i.e. 
Answer:
correct option is c. 31.0°C
Explanation:
given data
copper of mass = 100 g
electric drill = 1/2"
power = 40.0 W
time = 30 s
C copper = 387 J/kgC
to find out
copper's increase in temperature
solution
we get here energy that is express a s
energy = 40 W × 30 s
energy = 1200 Watt seconds
and heat acquired by drill is here as
heat acquired = 100 × T × 387
here temperature rise in copper mass as
temperature rise in copper mass =
× T × 387
temperature rise in copper mass = 38.7 × T Watt second
we know that all the energy from the drill heats the copper
so we can say
38.7 × T = 1200
T = 31°C
so correct option is c. 31.0°C