Answer:
Work done, W = 1786.17J
Explanation:
The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "
Mass of a painter, m = 75 kg
He climbs 2.75-m ladder that is leaning against a vertical wall.
The ladder makes an angle of 30 degrees with the wall.
We need to find the work done by the gravity on the painter.
The angle between the weight of the painter and the displacement is :
θ = 180 - 30
= 150°
The work done by the gravity is given by :
Hence, the required work done is 1786.17 J.
Answer:
The correct answer is
a) 1, 2, 3
Explanation:
In rolling down an inclined plane, the potential energy is Transferred to both linear and rotational kinetic energy thus
PE = KE or mgh = 1/2×m×v² + 1/2×I×ω²
The transformation equation fom potential to kinetic energy is =
m×g×h =
=
=
=
Therefore the order is with increasing rotational kinetic energy hence
the first is the sphere 1 followed by the disc 2 then the hoop 3
the correct order is a, 1, 2, 3
It’s both a solid and a liquid. It can thicken and soften depending on how it’s handled. It can be used to cover wounds to stop bleed, and used to drown enemies. Bungee Gum has the properties of both rubber and gum.
Answer: D. 5cm
Explanation:
Given the following :
Focal length (f) = - 6.0 cm
Height of object = 15.0cm
Distance of object from mirror (u) = 12.0cm
Height of image produced by the mirror =?
Firstly, we calculate the distance of the image from the mirror.
Using the mirror formula
1/f = 1/u + 1/v
1/v = 1/f - 1/u
1/v = 1/-6 - 1/12
1/v = - 1/6 - 1/12
1/v = (- 2 - 1) / 12
1/v = - 3 / 12
v = 12 / - 3
v = - 4
Using the relation :
(Image height / object height) = (- image distance / object distance)
Image height / 15 = - (-4) / 12
Image height / 15 = 4 / 12
Image height = (15 × 4) / 12
Image height = 60 / 12
Image height = 5cm
Answer:
q = 8.61 10⁻¹¹ m
charge does not depend on the distance between the two ships.
it is a very small charge value so it should be easy to create in each one
Explanation:
In this exercise we have two forces in balance: the electric force and the gravitational force
F_e -F_g = 0
F_e = F_g
Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.
Let's write Coulomb's law and gravitational attraction
In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.
k q² = G m²
q = m
we substitute
q = m
q = m
q = 0.861 10⁻¹⁰ m
q = 8.61 10⁻¹¹ m
This amount of charge does not depend on the distance between the two ships.
It is also proportional to the mass of the ships with the proportionality factor found.
Suppose the ships have a mass of m = 1000 kg, let's find the cargo
q = 8.61 10⁻¹¹ 10³
q = 8.61 10⁻⁸ C
this is a very small charge value so it should be easy to create in each one