You are given the mass of a sphere that is 26 kg sphere and it is released from rest when θ = 0°. You are also given the force of the spring that is F = 100 N. You are asked to find the tension of the spring. Imagine that the sphere is connected to a spring. The spring exerts a tension and the spring exerts gravitational pull. This will follow the second law of newton.
T - F = ma
T = ma + F
T = 26kg (9.81m/s²) + 100 N
T = 355.06 N
Answer:
External force on him will be 112 N
Explanation:
We have given the mass of the sprinter m =70 kg
Acceleration of the sprinter 
We have to find the net external force
According to second law of motion force = mass ×acceleration
Force is dependent on the mass and acceleration
So 
So external force will be 112 N
We know, acceleration = final velocity - initial velocity / time
Here, if velocity is increasing, then,
Final velocity > initial velocity, in that case, acceleration is also increasing, as it is directly proportional to velocity
In short, Your Answer would be "Yes"
Hope this helps!
Answer:
A. nuclear fusion reactions
C. it's still hot from the big bang
Explanation:
The inside of the earth is hot due to some reasons. This heat provides the internal energy the drives processes within the earth interior. Here are some of the ways in which the heat has accumulated:
- Nuclear reactions within the earth interior by fusion and other radioactive processes releases a large amount of heat.
- Some heat accreted during the early formation of the earth and has not been lost till this day.
- Heating due to friction
These are some of the sources of the earth's internal heat.
Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force. We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.
If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as 588 newtons or as
132.3 pounds. That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.
If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is
y(t) = y₀ + M sin(2π t/15) .
The vertical speed of the deck is y'(t) = M (2π/15) cos(2π t/15)
and its vertical acceleration is y''(t) = - (2πM/15) (2π/15) sin(2π t/15)
= - (4 π² M / 15²) sin(2π t/15)
= - 0.1755 M sin(2π t/15) .
There's the important number ... the 0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.
The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of 0.1755 x amplitude).
At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of 65kg, when in reality it's only 60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.
Now I'm going to wave my hands in the air a bit:
Apparent weight = (apparent mass) x (real acceleration of gravity)
(Apparent mass) = (65/60) = 1.08333 x real mass.
Apparent 'gravity' = 1.08333 x real acceleration of gravity.
The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.
0.08333 G = 0.1755 M
The 'M' is what we need to find.
Divide each side by 0.1755 : M = (0.08333 / 0.1755) G
'G' = 9.0 m/s²
M = (0.08333 / 0.1755) (9.8) = 4.65 meters .
That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .
