To answer the two questions, we need to know two important equations involving centripetal movement:
v = ωr (ω represents angular velocity <u>in radians</u>)
a = 
Let's apply the first equation to question a:
v = ωr
v = ((1800*2π) / 60) * 0.26
Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.
Back to the equation:
v = ((1800*2π)/60) * 0.26
v = (1800*2(3.14)/60) * 0.26
v = (11304/60) * 0.26
v = 188.4 * 0.26
v = 48.984
v = 49 (m/s)
Now that we know the linear velocity, we can find the centripetal acceleration:
a =
a = 
a = 9234.6 (m/
)
Wow! That's fast!
<u>We now have our answers for a and b:</u>
a. 49 (m/s)
b. 9.2 * 
(m/)
If you have any questions on how I got to these answers, just ask!
- breezyツ
Answer:
60 cm
Explanation:
We are given;
- Focal length of a concave mirror as 30.0 cm
- Object distance is 15.0 cm
We are required to determine the radius of curvature.
We need to know that the radius of a curvature is the radius of a circle from which the curved mirror is part.
We also need to know that the radius of curvature is twice the focal length of a curved mirror.
Therefore;
Radius of curvature = 2 × Focal length
Therefore;
Radius of curvature = 2 × 30 cm
= 60 cm
<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
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
