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3 years ago

15. What is a thermograph?

Physics

2 answers:

Genrish500 [490]3 years ago

8 0

Answer: A device that uses infrared sensors.

Explanation:

Anni [7]3 years ago

8 0

Answer:

a device that uses infrared sensors to create pictures that show variations in temperature

Explanation:

Gradpoint

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Stability is a term that means...

Aleksandr-060686 [28]

The correct answer is A

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3 years ago

Any fracture or system of fractures along which Earth moves is known as a

bogdanovich [222]

Any fracture or system of fractures along which Earth moves is known as a D.fault.

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3 years ago

You may make the following measurements of an object 42kg and 22m3. What would the objects density be?

ki77a [65]

Density=mass/volume
to find the density
mass=42kg
volume=22m3
so density=42/22
density=1.9Kgm3

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3 years ago

A student wants to determine the impulse delivered to the lab cart when it runs into the wall. The student measures the mass of

WINSTONCH [101]

The correct answer to the question is : A) The velocity of the cart after it hits the wall.

EXPLANATION:

Before answering this question, first we have to understand impulse.

Impulse of a body is defined as the change in momentum or the product of force with time.

Mathematically impulse = m ( v- u ).

Here, v is the final momentum and u is the initial momentum.

Hence, we need the velocity of the cart after it hits the wall in order to calculate the impulse of the lab cart.

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3 years ago

Read 2 more answers

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

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3 years ago