All categories

3 years ago

Which of the following has no energy?

Physics

2 answers:

tankabanditka [31]3 years ago

7 0

D. is correct because a wrecking ball has potential energy.

Readme [11.4K]3 years ago

4 0

The correct answer is D: which is none of the above.
Hint: a wreckling ball contains pontential energy that acts like a pendulum

a pot of water contains pontential thermal energy

You might be interested in

Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiment

Drupady [299]

Answer:

$u_K=0.862$

Explanation:

The force of friction between the quails feet and the ground is:

$F=m*a$

$F_K=m*a$

$F_K=u_k*m*g$

$u_K*m*g=m*a_c$

$u_K*g=a$

$u_K=\frac{a_c}{g}$

$a_c=\frac{v^2}{r}$

So the coefficient of static is solve

$u_K=\frac{\frac{v^2}{r}}{g}$

$u_K=\frac{v^2}{r*g}=\frac{(2.6m/s)^2}{0.80m*9.8m/s^2}$

$u_K=0.862$

4 0

3 years ago

Please help me whit this question

mariarad [96]

Answer:

Gypsum

Explanation:

Gypsum is a soft sulfate mineral composed of Calcium Sulfate di-hydrate. It is widely used in composition of fertilizer. It is also known as land plaster. On adding gypsum to soil, the quality of soil is improved. It conditions the soil and adds nutrients. The properties of soil is improved. Hence, the correct answer is Gypsum.

Hope this helped:)

pls mark brainlist

6 0

3 years ago

What could be the possible answer to the question ?<br><br>thankyou ~

Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$