Answer:
The maximum velocity is 0.489 m/s
Explanation:
Maximum velocity (v) = angular velocity (w) × radius (r)
w = 33.33 rpm = 33.33×0.1047 = 3.4897 rad/s
r = 14 cm = 14/100 = 0.14 m
v = 3.4897×0.14 = 0.489 m/s
Explanation:
As per the problem,
When q > 0 then -q is a negative charge . Since, change in potential energy (
) increases.
or, 
> 0
or, 
Therefore, both positive and negative charge will move from 
to 
and as 
so both of them move through a negative potential difference.
Thus, we can conclude that the true statements are as follows.
- The positively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).
- The negatively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).
Explanation:
given,
mass of one planet (m1)=2*10^23 kg
mass of another planet (m2)=5*10^22kg
distance between them(d)=3*10^16m
gravitational constant(G)=6.67*10^-11Nm^2kg^-2
gravitational force between them(F)=?
we know,
F=Gm1m2/d^2
or, F=6.67*10^-11*2*10^23*5*10^22/(3*10^16)^2
or, F=6.67*2*5*10^-11+23+22/3*3*10^32
or, F=66.7*10^34/9*10^32
or, F=7.41*10^34-32
•°• F=7.41*10^2
thus, the gravitational force between them is 7.14*10^2
Answer:
Explanation:
Initial kinetic energy of M = 1/2 M vi²
let final velocity be vf
v² = u² + 2a s
vf² = vi² + 2 (F / M) x D
Kinetic energy
= 1/2 Mvf²
= 1/2 M ( vi² + 2 (F / M) x D
1/2 M vi² + FD
Ratio with initial value
1/2 M vi² + FD) / 1/2 M vi²
RK = 1 + FD / 2 M vi²
There is one mistake in the question.The Correct question is here
A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.
Answer:
y(1s) - y(1/2s) = - 3.675 m
The cat falls 3.675 m between time 1/2 s and 1 s.
Explanation:
Given data
time=1/2 sec to 1 sec
v(t)=-9.8t m/s
To find
Distance
Solution
As the acceleration as first derivative of velocity with respect to time
So
acceleration(-g)= dv/dt
Solve it
dv = a dt
dv = -g dt
v - v₀ = -gt
v= dy/dt
dy = v dt
dy = ( v₀ - gt ) dt
y(1s) - y(1/2s) = ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]
y(1s) - y(1/2s) = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]
y1s - y1/2s = ( - 4.9 m/s² ) ( 3/4 s² )
y(1s) - y(1/2s) = - 3.675 m
The cat falls 3.675 m between time 1/2 s and 1 s.
