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3 years ago

13

2. The Himalayas in central Asia are the tallest mountains in the world. But fossils of seashells can be found high in these mou

ntains, far from any ocean.
How do you think they got there?

1 answer:

IRINA_888 [86]3 years ago

3 0

Answer:

The sea level could have been much higher in the past.

Explanation:

The tectonic plates shifted and the mountain rose from there.

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Draw a neat labelled diagram for a particle moving in a circular path with a constant speed. In your diagram show the direction

Naya [18.7K]

Explanation:

Particle moving in a circular path with a constant speed.

3 0

2 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

8 0

3 years ago

A punter wants to kick a football so that the football has a total flight time of 4.70s and lands 56.0m away (measured along the

Sindrei [870]

Answer:

Explanation:

How long does the football need to rise?

4.70/3 = 2.35 s

What height will the football reach?

h = ½(9.81)2.35² = 27.1 m

With what speed does the punter need to kick the football?

vy = g•t = 9.81(2.35) = 23.1 m/s

vx = d/t = 56.0/4.70 = 11.9 m/s

v = √(vx²+vy²) = 26.0 m/s

At what angle (θ), with the horizontal, does the punter need to kick the football?

θ = arctan(vy/vx) = 62.7°

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2 years ago

Our galaxy, the ______________, contains more than 200 billion stars.

erica [24]

Answer: Milky Way

Explanation:

4 0

3 years ago

Why doesn’t the ball roll on forever after being kicked at a soccer game?

gavmur [86]

My guess would be because the gravity from the Earth's core is constantly pulling the ball towards the ground. It's like the moon. Why doesn't the moon just float away in space? Because Earth's gravitational pull keeps it rotating around it. Therefore, the ball will always be pulled towards the core which keeps it from from rolling forever due to friction. But i may be wrong, even though this a quite a good answer, hope it is right!

7 0

3 years ago