Answer:
The magnitude of the average force exerted on the water by the blade is 960 N.
Explanation:
Given that,
The mass of water per second that strikes the blade is, 
Initial speed of the oncoming stream, u = 16 m/s
Final speed of the outgoing water stream, v = -16 m/s
We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :
F = -960 N
So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.
Answer:
Explanation:
area of square loop A = side²
= 8.4² x 10⁻⁴
A = 70.56 x 10⁻⁴ m²
when it is converted into rectangle , length = 14.7 , width = 2.1
area = length x width
= 14.7 x 2.1 x 10⁻⁴
= 30.87 x 10⁻⁴ m²
Let magnetic field be B
Change in flux = magnetic field x change in area
= B x ( 70.56 x 10⁻⁴ - 30.87 x 10⁻⁴ )
= 39.69 x 10⁻⁴ B
rate of change of flux = change in flux / time taken
= 39.69 x 10⁻⁴ B / 6.5 x 10⁻³
= 6.1 x 10⁻¹ B
emf induced = 6.1 x 10⁻¹ B
6.1 x 10⁻¹ B = 14.7 ( given )
B = 2.41 x 10
= 24.1 T
B ) magnetic flux is decreasing , so it needs to be increased as per Lenz's law . Hence current induced will be anticlockwise so that additional magnetic flux is induced out of the page.
Answer:
move the point of the fulcrum
Explanation:
you can do this by moving yourself closer to the pivot point, or moving the sea saw so that your sister has more of the board on her side
To find the number of electrons transferred, we should divide the total charge acquired by the rod
by the charge of a single electron (
), and we find:
Answer:
(a)F= 3.83 * 10^3 N
(b)Altitude=8.20 * 10^5 m
Explanation:
On the launchpad weight = gravitational force between earth and satellite.
W = GMm/R²
where R is the earth radius.
Re-arranging:
WR² / GM = m
m = 4900 * (6.3 * 10^6)² / (6.67 * 10^-11 * 5.97 * 10^24) = 488 kg
The centripetal force (Fc) needed to keep the satellite moving in a circular orbit of radius (r) is:
Fc = mω²r
where ω is the angular velocity in radians/second. The satellite completes 1 revolution, which is 2π radians, in 1.667 hours.
ω = 2π / (1.667 * 60 * 60) = 1.05 * 10^-3 rad/s
When the satellite is in orbit at a distance (r) from the CENTRE of the earth, Fc is provided by the gravitational force between the earth and the satellite:
Fc = GMm/r²
mω²r = GMm / r²
ω²r = GM / r²
r³ = GM/ω² = (6.67 * 10^-11 * 5.97 * 10^24) / (1.05 * 10^-3)²
r³ = 3.612 * 10^20
r = 7.12 * 10^6 m
(a)
F = GMm/r²
F=(6.67 * 10^-11 * 5.97 * 10^24 * 488) / (7.12 * 10^6 )²
F= 3.83 * 10^3 N
(b) Altitude = r - R = (7.12 * 10^6) - (6.3 * 10^6) = 8.20 * 10^5 m
