All of Dina's potential energy Ep is converted into kinetic energy Ek so Ep=Ek, where Ep=m*g*h and Ek=(1/2)*m*v². m is the mass of Dina, h is the height of ski slope, g=9.8 m/s² and v is the maximal velocity.
So we solve for v:
m*g*h=(1/2)*m*v², masses cancel out,
g*h=(1/2)*v², we multiply by 2,
2*g*h=v² and take the square root to get v
√(2*g*h)=v, we plug in the numbers and get:
v=9.9 m/s.
So Dina's maximum velocity on the bottom of the ski slope is v=9.9 m/s.
Answer:
recall that heat absorbed released is given by
Q = mc*(T2 - T1)
where
m = mass (in g)
c = specific heat capacity (in J/g-k)
T = temperature (in C or K)
*note: Q is (+) when heat is absorbed and (-) when heat is released.
substituting,
Q = (480)*(0.97)*(234 - 22)
Q = 98707 J = 98.7 kJ
Explanation:
-- The acceleration of gravity is 9.8 m/s².
So if there's no air resistance, the speed of a falling object
always increases by 9.8 m/s for every second it falls.
Speed = (original speed) + (gravity x falling time)
-- If it has no vertical speed when it started, then at the end
of 3 seconds, its speed is
= (0) + (9.8 m/s² x 3 sec)
Velocity = 29.4 m/s downward .
The diameter of the column of the water as it hits the bucket is 4.04 cm
The equation of continuity occurs in the fluid system and it asserts that the inflow and the outflow of the volume rate at the inlet and at the outlet of the system are equal.
By using the kinematics equation to determine the speed of the water in the bucket and applying the equation of continuity to estimate the diameter of the column, we have the following;
Using the kinematics equation:
From the equation of continuity:
Since diameter = 2r;
∴
The diameter of the column of the water is:
= 2(2.02) cm
= 4.04 cm
Learn more about the equation of continuity here:
brainly.com/question/10822213
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV