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2 years ago

How did Galileo increase public support for Copernicus’s model?

Physics

2 answers:

miskamm [114]2 years ago

8 0

I think by using data collected by Tycho Brahe

just olya [345]2 years ago

3 0

Answer:

Explanation:

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Two rocks, a and b, are thrown horizontally from the top of a cliff. rock a has an initial speed of 10 meters per second and roc

Arte-miy333 [17]

Both rocks hit the ground at the same time and at the same distance from the base of the cliff. Option A is correct.

<h3>What is speed ?</h3>

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while u for the final speed. its SI unit is m/sec.

If the resistance is ignored, things falling near the Earth's surface have the same estimated acceleration due to gravity of 9.8 meters per second.

As a result, the objects' acceleration is the same, and their velocity is growing at a consistent pace.

So that both rocks hit the ground at the same time and at the same distance from the base of the cliff.

Hence, option A is correct.

To learn more about the sped, refer to the link;

brainly.com/question/7359669

#SPJ1

5 0

1 year ago

How to solve these two questions?

hammer [34]

1) See attached figure

The relationship between charge and current is:

$i = \frac{Q}{t}$

where

i is the current

Q is the charge

t is the time

Therefore, the current is the rate of change of the charge passing through a given point over time.

This means that for a graph of charge over time, the current is just equal to the slope of the graph.

For the graph in this problem:

- Between t = 0 and t = 2 s, the slope is

$\frac{50-0}{2-0}=25 C/s$

therefore the current is

i = 25 A

- Between t = 2 s and t = 6 s, the slope is

$\frac{-50-(50)}{6-2}=-25 C/s$

therefore the current is

i = -25 A

- Between t = 6 s and t = 8 s, the slope is

$\frac{0-(-50)}{8-6}=25 C/s$

therefore the current is

i = 25 A

The figure attached show these values plotted on a graph.

2) $15 \mu C$

The previous equation can be rewritten as

$Q = i t$

This equation is valid if the current is constant: if the current is not constant, then the total charge is simply equal to the area under a current vs time graph.

Here we have the current vs time graph, so we gave to find the area under it.

The area of the first triangle is:

$A_1 = \frac{1}{2}(0.001 s)(0.010 A)=5\cdot 10^{-6} C$