Answer:
8.9
Explanation:
We can start by calculating the initial elastic potential energy of the spring. This is given by:
(1)
where
k = 35.0 N/m is the initial spring constant
x = 0.375 m is the compression of the spring
Solving the equation,
Later, the professor told the student that he needs an elastic potential energy of
U' = 22.0 J
to achieve his goal. Assuming that the compression of the spring will remain the same, this means that we can calculate the new spring constant that is needed to achieve this energy, by solving eq.(1) for k:
Therefore, Tom needs to increase the spring constant by a factor:
On a planet with different gravity, the molarity of water won't be different as water produces regular natural gravity.
<h3>How to explain the gravity?</h3>
Although sunlight is the energy source, gravity is the main factor driving the water cycle. The Earth's gravity pulls matter downward and toward its center. Gravity is the force that attracts two objects. It pushes water downhill and precipitation from the clouds. Air and ocean water are also moved by gravity.
We understand that even if the gravitational pull varies throughout the universe, the molarity of water would be constant everywhere. This is thus because a substance's mass is unaffected by gravity; only its weight is. The quantity (or mass, indirectly) of a solute is used to calculate the molarity.
Here, on a planet with different gravity, the molarity of water won't be different as water produces regular natural gravity.
Learn more about gravity on:
brainly.com/question/557206
#SPJ1
Answer:
The carrier lengthen is 0.08436 m.
Explanation:
Given that,
Length = 370 m
Initial temperature = 2.0°C
Final temperature = 21°C
We need to calculate the change temperature
Using formula of change of temperature
We need to calculate the carrier lengthen
Using formula of length
Put the value into the formula
Hence, The carrier lengthen is 0.08436 m.
Answer:
K = .3941 × 10³ W/m.K
Explanation:
Qcond = K A ΔT÷ L
∴K = Qcond ×L ÷ A ΔT
J ÷ S = P
P = I × V =Qcond
∴Qcond = I × V
= 0.6 A × 110 V
=66 W
L = 0.12 m
ΔT = 8 °C
Qcond =33 V
Area = (πD²) ÷ 4
= [π (4 × 10⁻² )²] ÷ 4
= 1.256 × 10⁻³ m²
∴A = 1.256 × 10 ⁻³³ m²
So K = ( Qcond × L ) ÷ A ΔT
= (33) (0.12 ) ÷ (1.256 ×10⁻³ ) × 8
= 0.3941 × 10³ W/m .K
