Answer
i'm not 100% sure but 1764
Explanation:
Work done = gravitational potential energy
Gravitational potential energy = mass(kg) × height(m) × gravitational field strength(N/kg)
We can assume that the student is on earth so the gravitational field strength is 9.8N/kg
So work done = 60 × 3×9.8
=1764
(if you need help calculating power but if you do just divide your answer by 12 and you will get 147)
You must observe the object twice.
-- Look at it the first time, and make a mark where it is.
-- After some time has passed, look at the object again, and
make another mark at the place where it is.
-- At your convenience, take out your ruler, and measure the
distance between the two marks.
What you'll have is the object's "displacement" during that period
of time ... the distance between the start-point and end-point.
Technically, you won't know the actual distance it has traveled
during that time, because you don't know the route it took.
Answer:
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
Explanation:
Given;
wheel rotates from rest with constant angular acceleration.
Initial angular speed v = 0
Time t = 2.50
Distance x = 8 rev
Applying equation of motion;
x = vt +0.5at^2 ........1
Since v = 0
x = 0.5at^2
making a the subject of formula;
a = x/0.5t^2 = 2x/t^2
a = angular acceleration
t = time taken
x = angular distance
Substituting the values;
a = 2(8)/2.5^2
a = 2.56 rev/s^2
velocity at t = 2.50
v1 = a×t = 2.56×2.50 = 6.4 rev/s
Through the next 5 second;
t2 = 5 seconds
a2 = 2.56 rev/s^2
v2 = 6.4 rev/s
From equation 1;
x = vt +0.5at^2
Substituting the values;
x2 = 6.4(5) + 0.5×2.56×5^2
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
The electric field at the surface of the cylinder is 51428V/m
Given data:
• The length of the charge is l= 7m.
• The charge is q = 2μC..
• The radius the cylinder is r = 10 cm
Since the filament length is so large as compared to the cylinder length that the infinite line of charge can be assumed.
The expression to calculate the electric field is given as,
E=2kλ/r
Here, λ is the linear charge density.
Substitute the values in the above equation,
E = (2×9×109N⋅m^2/C^2×2×10^−6C)/0.1m×7m
E = 51428N/C×(V/m)/(N/C)
=51428V/m
An electric charge is the property of matter where it has more or fewer electrons than protons in its atoms. Electrons carry a negative charge and protons carry a positive charge. Matter is positively charged if it contains more protons than electrons, and negatively charged if it contains more electrons than protons.
Learn more about charge here:
brainly.com/question/19886264
#SPJ4
Answer:
ФE = 9.403W
Explanation:
In order to calculate the magnitude of the electric flux trough the sheet, you use the following formula:
(1)
A: area of the rectangular sheet = (0.400m)(0.600m) = 0.24m^2
E: magnitude of the electric field = 95.0N/C
θ: angle between the direction of the electric field and the normal to the surface of the sheet
You replace the values of the parameters in the equation (1):
The magnitude of the electric flux is trough the sheet is 9.403W
