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3 years ago

How do you state a hypothesis! Branliest

Physics

2 answers:

VLD [36.1K]3 years ago

4 0

Answer:

State your hypothesis as concisely, and to the point, as possible. A hypothesis is usually written in a form where it proposes that, if something is done, then something else will occur. Usually, you don't want to state a hypothesis as a question. You believe in something, and you're seeking to prove it.

Explanation:

blondinia [14]3 years ago

3 0

Use an “If...then...” statement!

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True or false earth orbit is nearly circular

Simora [160]

The answer would be true

4 0

2 years ago

Read 2 more answers

Earth exerts a 100 N gravitational force on a metal box. What is the magnitude of the gravitational force the metal box exerts o

jeka94

We have that the magnitude of the gravitational force is mathematically given as

f=6.377N

Question Parameters:

Earth exerts a 100 N gravitational force on a metal box.

(Mass of the earth is 6e24 kg and radius of the earth is 6.4e6m.)

Generally the equation for the Gravitational mForce is mathematically given as

$F=\frac{GMm}{r^2}\\\\Therefore\\\\F=\frac{100/9.8* 6e116e24}{6.4e6^2}$

f=6.377N

For more information on Force visit

brainly.com/question/26115859

7 0

1 year ago

MATHPHYS HELP

rusak2 [61]

Answer:

16.4287

Explanation:

The force and displacement are related by Hooke's law:

F = kΔx

The period of oscillation of a spring/mass system is:

T = 2π√(m/k)

First, find the value of k:

F = kΔx

78 N = k (98 m)

k = 0.796 N/m

Next, find the mass of the unknown weight.

F = kΔx

m (9.8 m/s²) = (0.796 N/m) (67 m)

m = 5.44 kg

Finally, find the period.

T = 2π√(m/k)

T = 2π√(5.44 kg / 0.796 N/m)

T = 16.4287 s

3 0

2 years ago

Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed,

KATRIN_1 [288]

Answer:

a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

V = i (R₁ + R₂)

with R₁ connected

V = (i + 0.5) R₁

with R₂ connected

V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

V = i ( $\frac{V}{ i+0.5} + \frac{V}{i+0.25}$ )

1 = i ( $\frac{1}{i+0.5} + \frac{1}{i+0.25}$ )

1 = i ( $\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }$ ) = $\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}$

i² + 0.75 i + 0.125 = 2i² + 0.75 i

i² - 0.125 = 0

i = √0.125

i = 0.35355 A

with the second equation we look for R1

R₁ = $\frac{V}{i+0.5}$

R₁ = 12 /( 0.35355 +0.5)

R₁ = 14.1 Ω

with the third equation we look for R2

R₂ = $\frac{V}{i+0.25}$

R₂ = $\frac{12}{0.35355+0.25}$

R₂ = 19.9 Ω

3 0

2 years ago

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

vivado [14]

This question can be solved from the Kepler's law of planetary motion.

As per this law the square of time period of a planet is proportional to the cube of semi major axis.

Mathematically it can be written as $T^{2} \alpha R^{3}$

⇒ $T^{2} = KR^{3}$

Here K is the proportionality constant.

If $T_{1}$ and $T_{2}$ are the orbital periods of the planets and

$R_{1}$ and $R_{2}$ are the distance of the planets from the sun, then Kepler's law can be written as-

$\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{2} }$

⇒ $R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }$

Here we are asked to calculate the the distance of Saturn from sun.It can solved by comparing it with earth.

Let the distance from sun and orbital period of Saturn is denoted as $R_{1}$ and $T_{1}$ respectively.

Let the distance from sun and orbital period of earth is denoted as $R_{2}$ and $T_{2}$ respectively.

we are given that $T_{1} =29.46 years$

we know that $R_{2} =$ 1 AU and $T_{2} =$ 1 year.

1 AU is the mean distance of earth from the sun which is equal to 150 million kilometre.

Hence distance of Saturn from sun is calculated as -

From Kepler's law as mentioned above-

$R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }$

= $[1 ]^{3} *\frac{[29.46]^{2} }{[1]^{2} } AU$

$=867.8916 AU^{3}$

⇒ $R_{1} =\sqrt[3]{867.8916}$

=9.5386 AU [ans]

5 0

2 years ago