Answer:
1.43 kg
Explanation:
We can solve the problem by using Newton's second law:

where
F is the net force on an object
m is its mass
a is its acceleration
For the squirrel in the problem, we have:
is the acceleration
F = 5 N is the net force acting on it
Solving for m, we find its mass:

Answer:
k=694.44
Explanation:
PE=1/2kx^2
5=1/2k*.12^2
5=1/2k(.0144)
divide both sides by .0144
Answer:
(a) The work done is 0.05 J
(b) The force will stretch the spring by 3.8 cm
Explanation:
Given;
work done in stretching the spring from 30 cm to 45 cm, W = 3 J
extension of the spring, x = 45 cm - 30 cm = 15 cm = 0.15 m
The work done is given by;
W = ¹/₂kx²
where;
k is the force constant of the spring
k = 2W / x²
k = (2 x 3) / (0.15)²
k = 266.67 N/m
(a) the extension of the spring, x = 37 cm - 35 cm = 2 cm = 0.02 m
work done is given by;
W = ¹/₂kx²
W = ¹/₂ (266.67)(0.02)²
W = 0.05 J
(b) force = 10 N
natural length L = 30 cm
F = kx
x = F / k
x = 10 / 266.67
x = 0.0375 m
x = 3.75 cm = 3.8 cm
Thus a force of 10 N will stretch the spring by 3.8 cm
No it isn't. (Unless you connect it backwards.)
If the primary has 10 turns and the secondary has 70 turns,
then the voltage that appears across the secondary is
7 times the voltage that you feed to the primary.
If you're 'exciting' the primary with 170 volts, then you need
to be cautious around the secondary terminals, because there's
1,190 volts there !
If you want to use your transformer in a step-down configuration,
you can certainly connect it up the other way around.
Feed the 170 volts to the winding with 70 turns. You've reversed
the labels 'primary' and 'secondary', and that's OK. The voltage
at the terminals of the 10-turn winding will be (170/7) = 24.3 volts.