All categories

3 years ago

Will mark brainlist

Physics

1 answer:

Ksenya-84 [330]3 years ago

4 0

Answer:

5.00 x 10^9 j

Explanation:

mass = 100 kg

speed / velocity = 10 km/s x 1000 = 10,000m

K.E =

$\frac{m{v}^{2} }{2}$

10,000x 100/ 2

= 5000000000

= 5.0 x 10^9j

You might be interested in

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block

borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

$T = 2\pi \sqrt{\frac{m}{k} }$

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

$T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\$

Now, determine the amplitude of oscillation, A;

$E = \frac{1}{2} kA^2$

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

$E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm$

Therefore, the amplitude of the oscillation is 2.82 cm

8 0

3 years ago

The noise from a power mower was measured at 104 dB. The noise level at a rock concert was measured at 121 dB. Find the ratio of

UkoKoshka [18]

Based on the calculations, the ratio of the intensity of Ir to Ip is equal to 50.12.

<h3>How to find the ratio of the intensity?</h3>

Let the intensity of the power mower be Ip.

Let the intensity of the rock music be Ir.

Mathematically, sound intensity level can be calculated by using this formula:

$\beta = 10 log(\frac{I}{I_{ref}} )$

Where:

I is the intensity of the sound.

Making I the subject of formula, we have:

I = Io × $10^{\frac{\beta }{10} }$

For Ip, we have:

Ip = Io × $10^{\frac{104 }{10} }$

Ip = Io × $10^{10.4}$

For Ir, we have:

Ir = Io × $10^{\frac{121 }{10} }$

Ir = Io × $10^{12.1}$

Now, we can find the ratio of the intensity:

Ir/Ip = Io × $10^{12.1}$ /Io × $10^{10.4}$

Ir/Ip = $10^{12.1}$ / $10^{10.4}$

Ir/Ip = $10^{12.1 -10.4}$

Ir/Ip = $10^{1.7}$

Ir/Ip = 50.12.

Read more on sound intensity here: brainly.com/question/17062836

#SPJ1

8 0

2 years ago

No matter what values of m and k you used for the spring, what is the ratio of the period to k m.

Leto [7]

The relationship between the period of an oscillating spring and the attached mass determines the ratio of the period to $\sqrt{\dfrac{m}{k} }$ .