Answer:
The value of acceleration due to gravity is greater in terai than in mountain. In terai region the radius of earth is less as it lies close to the centre of the earth. Thus, the value of g is more in terai region.
Answer:
Explanation:
cSep 20, 2010
well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.
Anonymous
Sep 20, 2010
First you need to solve for time by using
d=(1/2)(a)(t^2)+(vi)t
1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s
t=.45 sec
Then you find the horizontal distance traveled by using
v=d/t
1.3m/s=d/.54sec
d=.585m
Then you need to find the time of player B by using
d=(1/2)(a)(t^2)+(vi)t
1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0
t=.61 sec
Finally to find player Bs initial horizontal velocity you use the horizontal equation
v=d/t
v=.585m/.61 sec
so v=.959m/s
Answer:
Incomplete questions
Let assume we are asked to find
Calculate the induced emf in the coil at any time, let say t=2
And induced current
Explanation:
Flux is given as
Φ=NAB
Where
N is number of turn, N=1
A=area=πr²
Since r=2cm=0.02
A=π(0.02)²=0.001257m²
B=magnetic field
B(t)=Bo•e−t/τ,
Where Bo=3T
τ=0.5s
B(t)=3e(−t/0.5)
B(t)=3exp(-2t)
Therefore
Φ=NAB
Φ=0.001257×3•exp(-2t)
Φ=0.00377exp(-2t)
Now,
Induce emf is given as
E= - dΦ/dt
E= - 0.00377×-2 exp(-2t)
E=0.00754exp(-2t)
At t=2
E=0.00754exp(-4)
E=0.000138V
E=0.138mV
b. Induce current
From ohms laws
V=iR
Given that R=0.6Ω
i=V/R
i=0.000138/0.6
i=0.00023A
i=0.23mA
Ok so we know:
The time (t) is 18seconds
The acceleration (a) is 2.2m/s2
The displacement (r) is 660
Using the equation
With 'u' being the initial velocity we want, we get:
So:
So:
So the original/initial velocity was 16.8666 or 16.87 m/s
Hope this helped
·The acceleration of gravity is proportional to
1 / (the square of the distance from the center) .
When we're on the surface, we're 1 radius from the center of the Earth,
and the acceleration of gravity is 9.8 m/s² .
The boy's weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)
= 490 newtons .
At the distance of 5 radii from the center (4 radii altitude from the surface),
the acceleration of gravity is
(9.8 m/s²) · (1/5)² = 0.39 m/s² .
The boy's weight is (mass) · (gravity) = (50kg) · (0.39 m/s²)
= 19.6 newtons .
Just as we expected, his weight at that distance is
(19.6 / 490) = 0.04 = 1/25 = 1/5² of his weight on the surface.
