Answer:
Weight on Jupiter will be equal to 2940 N
Explanation:
We have given given acceleration due to gravity on Jupiter is 3 times of acceleration due to gravity on earth
Acceleration due to gravity on earth 
So acceleration due to gravity on Jupiter = 
Mass is given m = 100 kg
We have to find the weight
Weight is equal to W = mg, here m is mass and a is acceleration
So weight 
... the angular tilt of the Earth's position on its axis relative to the sun
Answer:
17.4 cm
Explanation:
Power of lens = +1.75 diopters
Focal length of lens
This is a convex lens as focal the diopter given is positive which makes the focal length positive. Image distance will be negative.
v = -25
∴ The new near point is 17.4 cm
Answer:
<u><em>0.03 m/s</em></u>
Explanation:
<em>Applying law of conservation of momentum, </em>
- <em>m₁v₁ + m₂v₂ = (m₁ + m₂)v</em>
- <em>0.105(24) + 75(0) = (0.105 + 75)v</em>
- <em>75.105v = 2.52</em>
- <em>v = 2.52/75.105</em>
- <em>v = </em><u><em>0.03 m/s</em></u>
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows:
