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3 years ago

What is the best way to support your hypothesis

Physics

1 answer:

Diano4ka-milaya [45]3 years ago

4 0

Answer:

Making a Hypothesis

Explanation:

-Research the subject of your question. Review the literature and find out as much as you can about previous information and discoveries surrounding your question.

-Develop an educated guess that answers your initial question. This is your hypothesis. Make a prediction based on your hypothesis and state it as a cause-effect relationship.

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Explanation:

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3 years ago

A truck going 15 kmôŹ°€h has a head-on collision with a small car going 30 kmôŹ°€h. Which statement best describes the situation

zlopas [31]

1. e) None of the above is necessarily true.

2.d) Without knowing the mass of the boat and the sack, we cannot tell.

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3 years ago

Please help Math Phys

Lorico [155]

Answer:

269 m

45 m/s

-58.6 m/s

Explanation:

Part 1

First, find the time it takes for the package to land. Take the upward direction to be positive.

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 5.98 s

Next, find the horizontal distance traveled in that time:

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²

Δx = 269 m

Part 2

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: v

v = at + v₀

v = (0 m/s²) (5.98 s) + (45 m/s

v = 45 m/s

Part 3

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)

v = -58.6 m/s

6 0

3 years ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

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3 years ago

What must two objects have for gravity to exist between them

Musya8 [376]

The two factors are mass and distance between them.

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3 years ago

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