Answer:
P = 75 W
Explanation:
given,
Distance, L = 8 m
Force,F = 150 N
Time, t = 16 s
Work by the climber
Work done = Force x displacement
W = F. L
W = 150 x 8
W = 1200 J
We know,


P = 75 W
Hence, Power climber is using to climb is equal to 75 W.
Answer:
Thermal expansion ,thermal contraction
Explanation:
They are different because it has different cause and chemical reactions
thermal contraction is a chemical reaction
Answer:
The potential between the plates will decrease.
Explanation:
An insulator is usually placed between the parallel plates and is also called a dielectric because it makes the amount of charge a capacitor can accommodate to increase at a particular potential difference.
Furthermore, the dielectric effect will make the electric field of the charged capacitor which is not connected to a source of supply to decrease.
Now, when the battery is removed, the charge Q remains constant and Capacity C will increase.
Formula for the potential difference is here;
V = Q/C
Since the numerator Q is constant and the denominator C increases, it means the potential difference V will decrease
Answer:
1. Largest force: C; smallest force: B; 2. ratio = 9:1
Explanation:
The formula for the force exerted between two charges is

where K is the Coulomb constant.
q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.
For simplicity, let's combine Kq₁q₂ into a single constant, k.
Then, we can write

1. Net force on each particle
Let's
- Call the distance between adjacent charges d.
- Remember that like charges repel and unlike charges attract.
Define forces exerted to the right as positive and those to the left as negative.
(a) Force on A

(b) Force on B

(C) Force on C

(d) Force on D

(e) Relative net forces
In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

2. Ratio of largest force to smallest

Correct temperature is 80°F
Answer:
T_f = 38.83°F
Explanation:
We are given;
Volume; V = 8 ft³
Initial Pressure; P_i = 100 lbf/in² = 100 × 12² lbf/ft²
Initial temperature; T_i = 80°F = 539.67 °R
Time for outlet flow; t_o = 90 s
Mass flow rate at outlet; m'_o = 0.03 lb/s
Final pressure; P_f = 30 lbf/in² = 30 × 12² lbf/ft²
Now, from ideal gas equation,
Pv = RT
Where v is initial specific volume
R is ideal gas constant = 53.33 ft.lbf/°R
Thus;
v = RT/P
v_i = 53.33 × 539.67/(100 × 12²)
v_i = 2 ft³/lb
Formula for initial mass is;
m_i = V/v_i
m_i = 8/2
m_i = 4 lb
Now change in mass is given as;
Δm = m'_o × t_o
Δm = 0.03 × 90
Δm = 2.7 lb
Now,
m_f = m_i - Δm
Thus; m_f = 4 - 2.7
m_f = 1.3 lb
Similarly in above;
v_f = V/m_f
v_f = 8/1.3
v_f = 6.154 ft³/lb
Again;
Pv = RT
Thus;
T_f = P_f•v_f/R
T_f = (30 × 12² × 6.154)/53.33
T_f = 498.5°R
Converting to °F gives;
T_f = 38.83°F