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Elanso [62]
2 years ago
5

A person invests money, or ___, in a business.

Physics
1 answer:
love history [14]2 years ago
4 0

Explanation:

<em>A</em><em> </em><em>person</em><em> </em><em>invests</em><em> </em><em>money</em><em>,</em><em> </em><em>or</em><em> </em><em>resource</em><em>s</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>business </em><em>.</em><em> </em>

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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
2 years ago
What is Aristotle famous for?
Mila [183]

Answer:

History

Explanation:

Im. Sorry. Im. Not. Much. Help

8 0
2 years ago
A pendulum makes 60 vibrations in 15 secs what’s its frequency
o-na [289]

Answer:

4 hertz

Explanation:

The defination of freqyency = the total no of cycle made by a wave in one second .

so,

cycle or vibrations=60

tame taken = 15

now,

frequency = no. of cycle / time taken

= 60/15

=4 hertz

hence, the its frequency = 4hertz

8 0
3 years ago
An instrument used to detect a static electric charge is called an
Eddi Din [679]
It is B. false that an instrument used to detect a static electric charge is called an ammeter. It is actually called an electroscope. Ammeter measures current. 
7 0
3 years ago
Read 2 more answers
(a) Calculate the acceleration due to gravity on the surface of the Sun.
Ira Lisetskai [31]
<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

a) Acceleration due to gravity

                      g=\frac{GM}{r^2}

 Here we need to find acceleration due to gravity of Sun,

                G = 6.67259 x 10⁻¹¹ N m²/kg²

    Mass of sun, M = 1.989 × 10³⁰ kg

    Radius of sun, r = 6.957 x 10⁸ m

Substituting,

                g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²

b) Acceleration due to gravity in earth = 9.81 m/s²

   Ratio of gravity = 274.21/9.81 = 27.95

   Weight = mg

  Factor of increase in weight = 27.95

8 0
3 years ago
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