Answer:
1. increases
2. increases
3. increases
Explanation:
Part 1:
First of all, since the box remains at rest, the horizontal net force acting on the box must equal zero:
F1 - fs = 0.
And this friction force fs is:
fs = Nμs,
where μs is the static coefficient of friction, and N is the normal force.
Originally, the normal force N is equal to mg, where m is the mass of the box, and g is the constant of gravity. Now, there is an additional force F2 acting downward on the box, which means it increases the normal force, since the normal force by Newton's third law, is the force due to the surface acting on the box upward:
N = mg + F2.
So, F2 is increasing, that means fs is increasing too.
Part 2:
As explained in the part 1, N = mg + F2. F2 is increasing, so the normal force is thus increasing.
Part 3:
In part 1 and part 2, we know that fs = Nμs, and since the normal force N is increasing, the maximum possible static friction force fs, max is also increasing.
Answer:
b. $96,914
Explanation:
360-day borrowing rate = 5%
spot rate = 0.48
360-day deposit rate = 6%
Borrow at the rate of 5% to get
SF200,000/1.05 = $190,476.19
Convert at the spot rate of $0.48 to get
190,476.19*0.48 = $91,428.57
Invest at the interest rate of 6% to get
91,428.57/1.06 = 96,914.28
Therefore, Parker Company will receive $96,914 in 360 days.
Answer: 12Mg/h
Explanation:
Let the spring is compressed by a distance x,before the lift stops,then
Mg(h+x)= 1/2 kx^2 ............... 1
Kx - Mg = M ( 5g ) ............ 2
Make x the subject in equation 2
Kx = 5Mg + Mg
Kx = 6Mg
x = 6Mg/k ............ 3
Put equation 3 into 1
Mg ( h + x ) = 1/2 kx^2
Mgh + Mgx = 1/2kx^2
Mgh + Mg × 6Mg/k = 1/2k × ( 6Mg/k )^2
Mgh + Mg× 6Mg/k = 1/2k 36M^2g^2/ k^2
h =18Mg/k - 6Mg/h
K = 12Mg/h
They are said to be directly related.
a) directly related.
This is Charles' Law.
The force result in stretching the spring 10.0 centimeters is 2.5N.
<h3>
What is Hooke's law?</h3>
If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.
F = kx
where k is the proportionality constant called the spring constant or force constant.
Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.
Let the spring constant be very low 0.04N/m
The force applied is
F = 10 cm / 0.04
F = 0.1 m / 0.04
F = 2.5 N
Thus, the force result in stretching the spring 10cm is 2.5 N.
Learn more about hooke's law.
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