Given :
Magnitude of vector, v = 15.3 m
Angle of vector from horizon, 
.
To Find :
The x - component of the given vector.
Solution :
We know, a vector is given by :
Now, the x - component is :
Hence, this is the required solution.
The answer is "the same than the mercury in the bigger tube".
If one barometer tube has twice the cross-sectional area of another, mercury in the smaller tube will rise the same than the mercury in the bigger tube.
The mercury will rise to the point where the column of mercury has the same weight as the force exerted by the atmosphere.
The force exerted by the atmosphere is pressure * cross-sectional area
Anf the weight of the column of mercury, W, will be:
W = m* g
where m = density * volume, and volume = cross-sectional area * height
=> W = density * cross-sectional area * height
Then, you make W = F and get:
density * cross-sectional area * height = P * cross-sectional area
The term cress-sectional area appears on both sides so it gets cancelled, and the height of the column of mercury does not depend on the cross-sectional area of the barometer.
Answer:
Part A) the angular acceleration is α= 44.347 rad/s²
Part B) the angular velocity is 195.13 rad/s
Part C) the angular velocity is 345.913 rad/s
Part D ) the time is t= 7.652 s
Explanation:
Part A) since angular acceleration is related with angular acceleration through:
α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²
Part B) since angular acceleration is related
since
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s
since
ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s
Part C) at t=0
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s
ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s
Part D ) since the radial acceleration is related with the velocity through
ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s
therefore
v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s
t= 7.652 s
Answer: Input Resistance (Rin) = 500 ohms , overall voltage Gain (GV) = 2 volts / volts , ID = 4/9 = 0.44 A
Explanation:
Answer:
A:- 50 J
B:- 500 J
Explanation:
a) Given that a 25 N force is applied to move the box. Also the floor is having friction surface.
So in order to move the box, the floor should have friction of atleast 25 N.
∴ friction = 25 N
Work done = force * displacement of box
Given, the box is moved 2 m across the floor
So, Work done = friction * 2 m
= 25 * 2
= 50 J
b) Given, the box is having weight of 250 N weight
Gravitational force is acting on the box which is equal to (mass * gravity)
∴ Force = 250 N
The box is lifted 2 m above the floor.
So, displacement = 2 m
Work done = force * displacement of box
Work done = 250 * 2
= 500 J
