Answer:
d=1.49×1011m
Explanation:
Velocity is defined as the rate of travel, and can be found using the distance formula.
velocity=distancetime
Rearranging this formula we can solve for distance given velocity and time of travel.
d=vt
We are given velocity and time, and so can solve for distance, but if we plug in the values given;
d=(3.00×108m/s)(8.3minutes)
We can see that the units do not match up. Since seconds are the SI unit for time, we will need to convert 8.3 minutes to seconds.
t=(8.3minutes)(60seconds/minute)=(498s)
Now our units work out and we can solve for distance.
= 15.85
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Answer:
1.6 x 10^-5 T
Explanation:
i = 4 A
r = 0.05 m
The magnetic field due to long wire at a distance r is given by

B = 10^-7 x 2 x 4 / 0.05
B = 1.6 x 10^-5 T
Answer:
53
Explanation:
Because there are 1609.34 meters in a mile. 1609.34÷30=53.64 but because you put one at the beginning of the mile it will stay 53 and not round up to 54
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:

where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:

where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:

- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:

- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

- Replacing by the givens in (5), we can solve for μsmín, as follows:
