Answer:
<em>a) Fvt cosθ</em>
<em>b) Fv cosθ</em>
<em></em>
Explanation:
Each horse exerts a force = F
the rope is inclined at an angle = θ
speed of each horse = v
a) In time t, the distance traveled d = speed x time
i.e d = v x t = vt
also, the resultant force = F cosθ
Work done W = force x distance
W = F cosθ x vt = <em>Fvt cosθ</em>
<em></em>
b) Power provided by the horse P = force x speed
P = F cosθ x v
P = <em>Fv cosθ</em>
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is 
Explanation:
From the question we are told that
The diameter of the wire is 
The radius of the wire is 
The resistivity of aluminum is 
The electric field change is mathematically defied as

Generally the charge is mathematically represented as

Where A is the area which is mathematically represented as

So

Therefore

substituting values
![Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5Cint%5Climits%5E%7Bt%7D_%7B0%7D%20%7B%20%5B%200.0004t%5E2%20-%200.0001t%20%2B0.0004%5D%20%7D%20%5C%2C%20dt)
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%20t%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
From the question we are told that t = 5 sec
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%205%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004%285%29%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20%285%29%5E2%7D%7B2%7D%20%2B0.0004%285%29%5D%20%7D)

Answer: Mass of the planet, M= 8.53 x 10^8kg
Explanation:
Given Radius = 2.0 x 106m
Period T = 7h 11m
Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.
This is represented by the equation
T^2 = ( 4π^2/GM) R^3
Where T is the period in seconds
T = (7h x 60m + 11m)(60 sec)
= 25860 sec
G represents the gravitational constant
= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet
Making M the subject of the formula,
M = (4π^2/G)*R^3/T^2
M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2
Therefore Mass of the planet, M= 8.53 x 10^8kg
Answer:
This is because the rubbing releases negative charges, called electrons, which can build up on one object to produce a static charge. For example, when you shuffle your feet across a carpet, electrons can transfer onto you, building up a static charge on your skin.
Explanation:
This is because the rubbing releases negative charges
Answer:
The cannonball fly horizontally before it strikes the ground, S = 323.25 m
Explanation:
Given data,
The height of the cliff, h = 80 m
The horizontal velocity of the cannonball, Vₓ = 80 m/s
The range of the cannon ball with initial vertical velocity is zero is given by the formula,


S = 323.25 m
Hence, the cannonball fly horizontally before it strikes the ground, S = 323.25 m