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bogdanovich [222]
3 years ago
12

Rank the ten objects from loudest to softest.

Physics
1 answer:
frez [133]3 years ago
5 0
Um what are the ten objects..?
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A piece of solar system debris that passes through Earth's atmosphere and strikes the ground is a/an _______.
Flauer [41]
The answer is meteorite. can i get brainliest?

5 0
3 years ago
A single mass (m1 = 3.5 kg) hangs from a spring in a motionless elevator. The spring constant is k = 278 N/m. 1)What is the dist
artcher [175]
<h2>Answer:</h2>

0.126m

<h2>Explanation:</h2>

According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;

F = k x e            -------------------(i)

Where;

k = the spring's constant.

From the question, the force acting on the spring is the weight(W) of the mass. i.e

F = W               -----------------------(ii)

<em>But;</em>

W = m x g;

where;

m = mass of the object

g = acceleration due to gravity [usually taken as 10m/s²]

<em>From equation (ii), it implies that;</em>

F = W = m x g

<em>Now substitute F = m x g into equation(i) as follows;</em>

F = k x e

m x g = k x e      ------------------(iii)

<em>From the question;</em>

m = m1 = 3.5kg

k = 278N/m

<em>Substitute these values into equation (iii) as follows;</em>

3.5 x 10 = 278 x e

35 = 278e

<em>Now solve for e;</em>

e = 35/278

e = 0.126m

Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m

3 0
3 years ago
Newton’s second law states that the acceleration of an object is dependent on 1 variable which is the net force
Darya [45]
I think it’s false. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and then mass of the object.
7 0
2 years ago
When do net force is applied to a moving object it still comes to rest because of its inertia
Paraphin [41]
That is false for that.
6 0
3 years ago
A parallel-plate capacitor has square plates that have edge length equal to 1.20×102cm and are separated by 1.00 mm. It is conne
lora16 [44]

Answer:

Energy stored in the capacitor will be 1.835\times 10^{-6}J

Explanation:

We have given edge length of capacitor = 1.2\times 10^2cm=1.2m

So area A = 1.2×1.2 = 1.44 m^2

Separation is given as d  1 mm = 10^{-3}m

We know that capacitance is given by C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 1.44}{10^{-3}}=12.744\times 10^{-9}F

Voltage is given as V = 12 volt

We know that energy stored is given by E=\frac{1}{2}CV^2=\frac{1}{2}\times 12.744\times 10^{-9}\times 12^2=1.835\times 10^{-6}J

3 0
3 years ago
Read 2 more answers
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