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2 years ago

Shadow and eclipses result from

Physics

2 answers:

Basile [38]2 years ago

7 0

$\huge\mathsf{\red{\underline{\underline{Answer}}}}$

${\green{\dashrightarrow}}$ When an opaque obstacle is placed between a source of light and a screen, a shadow of the obstacle is formed on the screen. The kind of shadow depends on the size of the source of light. In other words, the earth casts its shadow on the moon. The solar eclipse occurs when the moon comes between the sun and the earth.

poizon [28]2 years ago

5 0

Lunar eclipses occur whenever the full moon passes into the shadow of the earth. Solar eclipses occur when the new moon slips in front of the sun and blocks it from view. The moon's shadow then falls upon the earth.

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Match each branch of science with a related career.

Lapatulllka [165]

A: geologist
b: physical science
c: space science
d: ecologist

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2 years ago

Read 2 more answers

An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Dete

dezoksy [38]

Answer:43.34 m

Explanation:

Given

acceleration(a) $=2 m/s^2$

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

$v=0+2\times 6=12 m/s$

After this object will start moving under gravity

height reached in first 6 s

$s=ut+\frac{at^2}{2}$

$s=0+\frac{2\times 6^2}{2}$

s=36 m

After fuel run out distance traveled in upward direction is

$v^2-u^2=2as_0$

here v=0

u=12 m/s

$a=9.8 m/s^2$

$0-12^2=2(-9.8)(s)$

$s_0=\frac{144}{2\times 9.8}=7.34 m$

$s+s_0=36+7.34=43.34 m$

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2 years ago

A scientist is observing a eukarotic cell and a prokaryotic cell. Which structure could she only observe in the eukaryotic cell?

Ksenya-84 [330]

The answer is nucleus

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2 years ago

Read 2 more answers

You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0.654 mm in radius. the resul

photoshop1234 [79]

1) First of all, let's find the resistance of the wire by using Ohm's law:
$V=IR$
where V is the potential difference applied on the wire, I the current and R the resistance. For the resistor in the problem we have:
$R= \frac{V}{I}= \frac{5.70 V}{17.6 A}=0.32 \Omega$

2) Now that we have the value of the resistance, we can find the resistivity of the wire $\rho$ by using the following relationship:
$\rho = \frac{RA}{L}$
Where A is the cross-sectional area of the wire and L its length.
We already have its length $L=2.90 m$ , while we need to calculate the area A starting from the radius:
$A=\pi r^2 = \pi (0.654\cdot 10^{-3}m)^2=1.34 \cdot 10^{-6}m^2$

And now we can find the resistivity:
$\rho = \frac{RA}{L}= \frac{(0.32 \Omega)(1.34 \cdot 10^{-6}m^2)}{2.90m}= 1.48 \cdot 10^{-7}\Omega \cdot m$

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3 years ago

A space shuttle with a mass of approximately 7.08 E5 kg is sitting on the launch pad. What would be the weight of the space shut

Afina-wow [57]

The weight of the shuttle is $6.94\cdot 10^6 N$

Explanation:

The weight of an object on Earth is the gravitational force exerted by the Earth on the object.

The magnitude of the weight of an object is given by:

$W=mg$

where

m is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity on Earth's surface

And the direction is downward (towards the Earth's centre).

For the shuttle in this problem, its mass is

$m=7.08\cdot 10^5 kg$

So, its weight is

$W=(7.08\cdot 10^5)(9.8)=6.94\cdot 10^6 N$

Note that while the mass of an object (m) does not change, its weight (W) changes according to the location, since the value of g can be different at different location (for example, on the Moon, the value of g is about 1/6 the value of g on the Earth).

Learn more about weight and forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

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2 years ago

Shadow and eclipses result from​

Shadow and eclipses result from