The question is incomplete! The complete question along with answer and explanation is provided below.
Question:
A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.
What is the change in the potential energy (in Joules) of the mass as it goes up the incline?
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?
Given Information:
Mass = m = 0.5 kg
Horizontal distance = d = 40 cm = 0.4 m
Vertical distance = h = 7 cm = 0.07 m
Normal force = Fn = 1 N
Required Information:
Potential energy = PE = ?
Work done = W = ?
Answer:
Potential energy = 0.343 Joules
Work done = 0.39 N.m
Explanation:
The potential energy is given by
PE = mgh
where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.
PE = 0.5*9.8*0.07
PE = 0.343 Joules
As you can see in the attached image
sinθ = opposite/hypotenuse
sinθ = 0.07/0.4
θ = sin⁻¹(0.07/0.4)
θ = 10.078°
The horizontal component of the normal force is given by
Fx = Fncos(θ)
Fx = 1*cos(10.078)
Fx = 0.984 N
Work done is given by
W = Fxd
where d is the horizontal distance
W = 0.984*0.4
W = 0.39 N.m
Answer:
114.26
Explanation:
a)Formula for per unit impedance for change of base is
Zpu2= Zpu1×(kV1/kV2)²×(kVA2/kVA1)
Zpu2: New per unit impedance
Zpu1: given per unit impedance
kV1: give base voltage
kV2: New bas votlage
kVA1: given bas power
kVA2: new base power
In the question
Zpu2=??
Zpu1= 0.3
kV2=24kV
kV1= 13.8 kV
kVA2= 1MVA ×1000= 1000 kVA
kVA1=500kVA
Zpu2= 0.3(13.8/24)²×(1000/500)
Zpu2= 0.198
b) to find ohmic impedance we will first calculate base value of impedance(Zbase). So,
Zbase= kV²/MVA
Zbase= 13.8²/(500/1000)
Zbase=380.88
Now that we have base value of impedance, Zbase, we can calculate actual ohmic value of impedance(Zactual) by using the following formula:
Zpu=Zactual/Zbase
0.3= Zactual/380.88
Zactual= 114.26 ohms
Answer:
Explanation:
The cannonball goes a horizontal distance of 275 m . It travels a vertical distance of 100 m
Time taken to cover vertical distance = t ,
Initial velocity u = 0
distance s = 100 m
acceleration a = 9.8 m /s²
s = ut + 1/2 g t²
100 = .5 x 9.8 x t²
t = 4.51 s
During this time it travels horizontally also uniformly so
horizontal velocity Vx = horizontal displacement / time
= 275 / 4.51 = 60.97 m /s
Vertical velocity Vy
Vy = u + gt
= 0 + 9.8 x 4.51
= 44.2 m /s
Resultant velocity
V = √ ( 44.2² + 60.97² )
= √ ( 1953.64 + 3717.34 )
= 75.3 m /s
Angle with horizontal Ф
TanФ = Vy / Vx
= 44.2 / 60.97
= .725
Ф = 36⁰ .
Answer:
the pressure will decrease by 1/2
Explanation:
PV=nRT
P=(nRT)/(V)
nRT are all constant so they will equal 1
V is 2
P=1/2
It often requires STANDARD reaction time
