Answer: 17.68 s
Explanation:
This problem is a good example of Vertical motion, where the main equation for this situation is:
(1)
Where:
is the height of the ball when it hits the ground
is the initial height of the ball
is the initial velocity of the ball
is the time when the ball strikes the ground
is the acceleration due to gravity
Having this clear, let's find from (1):
(2)
Rewritting (2):
(3)
This is a quadratic equation (also called equation of the second degree) of the form 
, which can be solved with the following formula:
(4)
Where:
Substituting the known values:
(5)
Solving (5) we find the positive result is:
Answer:
<em>The amount of electric charge transported = 0.192 C</em>
Explanation:
Electric Charge: This is defined as the product of electric current and time in an electric circuit, The S.I unit of electric charge is Coulombs (C)
Q = It..................... Equation 1
Where Q = Electric charge, I = electric current, t = time.
<em>Given:</em> I = 285 mA, t = 674 milliseconds.
<em>Conversion: (i) Convert from 285 mA to A = (285/1000) A = 0.285 A</em>
<em> (ii) convert from 674 milliseconds to seconds = (674/1000) s = 0.674 s </em>
Substituting these values into equation 1
Q = 0.285 × 0.674
<em>Q = 0.192 C</em>
<em>Therefore the amount of electric charge transported = 0.192 C</em>
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Answer:
1. 75N
2. 67,983 J (=67.98 kJ)
Explanation:
1. Work = Force x Distance
we are given that Work = 1,500J and Distance = 20m
hence,
Work = Force x Distance
1,500 = Force x 20
Force = 1,500 ÷ 20 = 75N
2. Potential Energy, PE = mass x gravity x change in height
we are given that mass = 165 kg and change in height = 42m
assuming that gravity, g = 9.81 m/s²
Potential Energy, PE = mass x gravity x change in height
Potential Energy, PE = 165 x 9.81 x 42 = 67,983 J (=67.98 kJ)
Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore,
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore,
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
