In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
= (5 * 10) Kg m/s
= 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".
¡Hellow!
For this problem, first, lets convert the seconds in hours:
5,4x10³
5400
h = sec / 3600
h = 5400 s / 3600
h = 1,5
Let's recabe information:
d (Distance) = 386 km
t (Time) = 1,5 h
v (Velocity) = ?
For calculate velocity, let's applicate formula:

Reeplace according we information:
386 km = v * 1,5 h
v = 386 km / 1,5 h
v = 257,33 km/h
The velocity of the train is of <u>257,33 kilometers for hour.</u>
<u></u>
Extra:
For convert km/h to m/s, we divide the velocity of km/h for 3,6:
m/s = km/h / 3,6
Let's reeplace:
m/s = 257,33 km/h / 3,6
m/s = 71,48
¿Good Luck?
Here light ray strikes to interface at an angle of 45 degree and then refracts into other medium such that it will bend towards boundary.
So here the angle of incidence will be less than the angle of refraction as light moves towards the boundary after refraction which mean it will bend away from the normal
here it can be said that medium 2 will be rarer then medium 1
So here the possible options are
1. Water
Air
2. Diamond
Air
So in above two options medium 1 is denser and medium 2 is rarer
Answer:
The minimum coefficient of friction is 0.27.
Explanation:
To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.
Centripetal force is written as

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

with ω denoting the angular velocity, which we are given. With that, the above becomes:

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

and so 45 rev/min = 4.71 rad/s.

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.