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3 years ago

The even distribution of weight in a compostion is known as

Physics

1 answer:

mojhsa [17]3 years ago

3 0

The answer is Balance. hopei helped? ahahahaha

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An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t

elixir [45]

To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

$\mu = \frac{f_0}{f_e}$

Here,

$\mu$ = Magnification

$f_e$ = Focal length eyepieces

$f_0$ = Focal length of the Objective

Rearranging to find the focal length of the objective

$f_0 = \mu f_e$

Replacing with our values

$f_0 = 7.5* 3.7cm$

$f_0 = 27.75cm$

Therefore the focal length of th eobjective lenses is 27.75cm

5 0

3 years ago

In the following diagram, the voltage is 1.5 volts and the resistance is 7.5 ohms. Use Ohm's law to determine the current in the

Basile [38]

I believe it is 0.2 amps Hope this helps sorry if wrong

5 0

4 years ago

Read 2 more answers

A string has an abrupt change in linear density at its midpoint so that the speed of a pulse on the left side is 2/3 of that on

goldenfox [79]

Answer:

Explanation:

a) Linear density is greater on the left side because the velocity of wave in a string is inversely proportional to the linear mass density of the string

b) We should start pulse from the left hand side so the reflected wave does not get inverted because the wave traveling from the denser to lighter medium gets reflected in the same phase.

3 0

3 years ago

Jerome is learning how the model of the atom has changed over time as new evidence was gathered. He has images of four models of

Stolb23 [73]

Answer:

Y, X, Z, W

Explanation:

You know W is the most recent because it features the nucleus in the middle and the electron cloud which was shown in models after the others.

5 0

4 years ago

Read 2 more answers

A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is

mezya [45]

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate = 150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

$h= \frac{1}{2}at^2$

$h= \frac{1}{2}4.71\times4.96^2$

h= 58.012 m

h≅ 58 m

4 0

3 years ago