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Degger [83]
2 years ago
8

An object is 70 micrometer long and 47.66 micrometer wide. How long and wide is the object in km

Physics
1 answer:
MatroZZZ [7]2 years ago
7 0

Answer:

The length of the object in kilometer (km) is 70 x 10⁻⁹ km

The width of the object in kilometers (km) is 47.66 x 10⁻⁹ km

Explanation:

Given;

length of the object = 70 micrometer = 70 μm

the width of the object = 47.66 micrometer =   47.66  μm

The length of the object in meter:

70 micrometer = 70 μm = 70 x 10⁻⁶ m

The length of the object in kilometer (km):

70 x 10⁻⁶ m = 70 x 10⁻⁹ km

The width of the object in meters:

47.66 micrometer = 47.66 μm = 47.66 x 10⁻⁶ m

The width of the object in kilometers (km):

47.66 x 10⁻⁶ m = 47.66 x 10⁻⁹ km

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The answer is a.12.5kg because i just did the test and it was correct.

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3 years ago
A ball is thrown into the air with a vertical velocity of 50 m/s and a horizontal
daser333 [38]

\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Projectile Motion.

Since, here given that, vertical velocity= 50m/s

we know that u*sin(theta) = vertical velocity

so the time taken to reach the maximum height or the time of Ascent is equal to

T = Usin(theta) ÷ g, here g = 9.8 m/s^2

so we get as,

T = 50/9.8

T = 5.10 seconds

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5 0
2 years ago
Explain why incremental development is the most effective approach for developing business software systems. Why is this model l
Ksenya-84 [330]

Explanation:

Typically, business software technologies are complex, and software strenuous. Business software applications are also often upgraded for changes in business goals or procedures. Real-time systems usually require a lot of hardware components that are quite difficult to change and cannot be upgraded Usually, actual-time safety critical systems that required to be built based on well-planned processes.

3 0
2 years ago
What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a
Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

M=-\frac{f_o}{f_e}

Here, M is the magnification of an astronomical telescope, f_e is the focal length of the eyepiece lens and f_o is the focal length of the objective lens.

It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.

Put f_o=74 cm and f_e=2.4 cm in the above expression.

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Therefore, the magnification of an astronomical telescope is -30.83.

5 0
3 years ago
When you jump from an elevated position you usually bend your knees upon reaching the ground. By doing this, you make the time o
dsp73

Answer:

c. about 1/10 as great.

Explanation:

While jumping form a certain height when we bend our knees upon reaching  the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.

This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.

Mathematically:

F\propto\frac{d}{dt} (p)

\Rightarrow F=\frac{d}{dt} (m.v)

since mass is constant

F=m\frac{d}{dt}v

when dt=10t

then,

F'=m.\frac{v}{10\times t}

F'=\frac{1}{10} \times \frac{m.v}{t}

F'=\frac{F}{10} the body will experience the tenth part of the maximum force.

where:

\frac{d}{dt} = represents the rate of change in dependent quantity with respect to time

p= momentum

m= mass of the person jumping

v= velocity of the body while hitting the ground.

7 0
2 years ago
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