If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula
Where | A | is the magnitude of the vector and 
is the angle that it forms with the x axis in the opposite direction to the hands of the clock.
In this problem we know the value of Ax and Ay and we need the angle .
Vector A is in the 4th quadrant
So:
So:
So:
= -47.28 ° +360° = 313 °
= 313 °
Option 4.
Answer:
Explanation:
You are looking for the resistance to start with
W = E * E/R
75 = 240 * 240 / R
75 * R = 240 * 240
R = 240 * 240 / 75
R = 57600 / 75
R = 768
Now let's see what happens when you try putting this into 110
W = E^2 / R
W = 120^2 / 768
W = 18.75
So the wattage is rated at 75. 18.75 is a far cry from that. I think they intend you to set up a ratio of
18.75 / 75 = 0.25
This is the long sure way of solving it. The quick way is to realize that the voltage is the only thing that is going to change. 120 * 120 / (240 * 240) = 1/2*1/2 = 1/4 = 0.25
Answer:
a) It takes her 1.43 s to reach a speed of 2.00 m/s.
b) Her deceleration is - 2.50 m/s²
Explanation:
The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:
v = v0 + a · t
Where:
v = velocty.
v0 = initial velocity.
a = acceleration.
t = time.
a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:
v = v0 + a · t
2.00 m/s = 0 m/s + 1.40 m/s² · t
t = 2.00 m/s / 1.40 m/s²
t = 1.43 s
It takes her 1.43 s to reach a speed of 2.00 m/s
b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:
v = v0 + a · t
0 = 2.00 m/s + a · 0.800 s
-2.00 m/s / 0.800 s = a
a = -2.50 m/s²
Her deceleration is - 2.50 m/s²
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.
