Answer:
Explanation:
Using
4.01 × 10^3 * 4.186 = 1.72×10^4j
In KJ 17.2kj
We are aware that weight is the product of applied gravitational force and mass. W = MG thus, where W represents the weight, M the mass, and G the gravitational force. As a result, it might also mean that "an object's weight is directly proportionate to its mass."
<h3>What is mass?</h3>
- Mass is a physical body's total amount of matter.
- It also serves as a gauge for the body's inertia, or resistance to acceleration (change in velocity) in the presence of a net force.
- The strength of an object's gravitational pull to other bodies is also influenced by its mass.
- The kilogram is the primary mass unit in the SI (kg).
- Even though weight is frequently measured using a spring scale rather than a balancing scale and directly compared with known masses, mass is not the same as weight in physics.
<h3>What is weight?</h3>
- The force exerted on an object by gravity is known as the weight of the object in science and engineering.
- Weight is sometimes described as a vector quantity, or the gravitational force exerted on the object, in some common textbooks.
- Others define weight as a scalar quantity, the gravitational force's strength.
- Others define it as the strength of the force applied to a body as a result of systems designed to resist the effects of gravity; the weight is the amount that is determined, for instance, by a spring scale.
Learn more about mass here:
brainly.com/question/19694949
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Answer:
The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)
Explanation:
Fundamental frequency = wave velocity/2L
where;
L is the length of the stretched rubber
Wave velocity = 
Frequency (F₁) = 
To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.
Given:
L₂ =2L₁ = 2L
T₂ = 2T₁ = 2T
(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)
F₂ = ![\frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B%5Cfrac%7B2T%7D%7B0.5%28%5Cfrac%7BM%7D%7BL%7D%29%7D%7D%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B4%28%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%29%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B2%7D%7B2%7D%20%5B%5Cfrac%7B%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%7D%7B2%2AL%7D%5D%20%3D%20F_1)
Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).
Answer:
The answer are given above in attachment.
Answer:
Change in electric potential energy ∆E = 365.72 kJ
Explanation:
Electric potential energy can be defined mathematically as:
E = kq1q2/r ....1
k = coulomb's constant = 9.0×10^9 N m^2/C^2
q1 = charge 1 = -2.1C
q2 = charge 2 = -5.0C
∆r = change in distance between the charges
r1 = 420km = 420000m
r2 = 160km = 160000m
From equation 1
∆E = kq1q2 (1/r2 -1/r1) ......2
Substituting the given values
∆E = 9.0×10^9 × -2.1 ×-5.0(1/160000 - 1/420000)
∆E = 94.5 × 10^9 (3.87 × 10^-6) J
∆E = 365.72 × 10^3 J
∆E = 365.72 kJ
