Answer:
YFy = 0 = Ffsinθ + Fncosθ - Fw
Explanation:
From the base of the vector Fn, draw a vertical line. the small angle between this line and Fn is also theta. The component of Fn in the vertical direction is Fncos(theta).
Take a moment to picture extreme cases. Sine is 0 at 0 and 1 at 90. Cosine is 1 at 0 and 0 at 90.
Tilt the incline so that the box is on a flat surface. How much of the gravitational force is along the x direction of the floor.
a) 2.75 s
The vertical position of the ball at time t is given by the equation
where
h = 4 m is the initial height of the ball
u = 12 m/s is the initial velocity of the ball (upward)
g = 9.8 m/s^2 is the acceleration of gravity (downward)
We can find the time t at which the ball reaches the ground by substituting y=0 into the equation:
This is a second-order equation. By solving it for t, we find:
t = -0.30 s
t = 2.75 s
The first solution is negative, so we discard it; the second solution, t = 2.75 s, is the one we are looking for.
b) -15.0 m/s (downward)
The final velocity of the ball can be calculated by using the equation:
where
u = 12 m/s is the initial (upward) velocity
g = 9.8 m/s^2 is the acceleration of gravity (downward)
t is the time
By subsisuting t = 2.75 s, we find the velocity of the ball as it reaches the ground:
And the negative sign means the direction is downward.
Answer:
W = 2352 J
Explanation:
Given that:
- mass of the bucket, M = 10 kg
- velocity of pulling the bucket, v = 3
- height of the platform, h = 30 m
- rate of loss of water-mass, m =
Here, according to the given situation the bucket moves at the rate,
The mass varies with the time as,
Consider the time interval between t and t + ∆t. During this time the bucket moves a distance
∆x = 3∆t meters
So, during this interval change in work done,
∆W = m.g∆x
<u>For work calculation:</u>
The answer is 1.52.
N = C (speed of light) / actual speed
n = index of refraction