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3 years ago

Newton’s second law of motion states that when a force acts on an object, the object’s acceleration is equal to _____.

1 answer:

3 0

Newton’s second law of motion states that when a force acts on an object, the object’s acceleration is equal to the net force divided by the object’s mass, and is in the same direction as the net force.

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The longest banana split ever made was 7.32 km long (obviously they used more than one banana). If an archer were to shoot an ar

elixir [45]

Answer:

The horizontal displacement of the arrow is not larger than the banana split.

Explanation:

Using y - y₀ = ut - 1/2gt², we find the time it takes the arrow to drop to the ground from the top of mount Everest.

So, y₀ = elevation of Mount Everest = 29029 ft = 29029 × 1ft = 29029 × 0.3048 m = 8848.04 m, y = final position of arrow = 0 m, u = initial vertical speed of arrow = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for arrow to fall to the ground.

y - y₀ = ut - 1/2gt²

0 - y₀ = 0 × t - 1/2gt²

-y₀ = -1/2gt²

t² = 2y₀/g

t = √(2y₀/g)

Substituting the values of the variables, we have

t = √(2y₀/g)

= √(2 × 8848.04 m/9.8 m/s²)

= √(17696.08 m/9.8 m/s²)

= √(1805.72 s²)

= 42.5 s

The horizontal distance the arrow moves is thus d = vt where v = maximum firing speed of arrow = 100 m/s and t = 42.5 s

So, d = vt

= 100 m/s × 42.5 s

= 4250 m

= 4.25 km

Since d = 4.25 km < 7.32 km, the horizontal displacement of the arrow is not larger than the banana split.

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2 years ago

What happens when a main-sequence star exhausts its core hydrogen fuel supply?

Ivanshal [37]

Explanation:

when a main-sequence star exhausts its core hydrogen fuel supply the core starts to shrink ( lack of fusion reactions) and the rest of the star starts to expands. The fusion reaction leaves the main sequence and begin to fuse helium in a shell outside the core. This mass stars become red supergiant and then evolve to become blue super giant.

8 0

3 years ago

Which image shows an example of the electromagnetic force in action?

yarga [219]

Answer:

Where are the images?

Explanation:

3 0

3 years ago

Read 2 more answers

What is the speed of a wave with a wavelength of 3.0 m and a period of 0.40 s?

Zolol [24]

The formula you need for this is
v = f λ
(velocity = frequency • wavelength)
A built-in reminder for this relationship is the units:
meters / second = meters • 1/seconds (aka hz, frequency)
therefore v = 3m / 0.4 seconds = 7.5 m/s

8 0

3 years ago

Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is F = GmM r2

Vladimir [108]

<h2>Answers:</h2>

<h2><u>Answer 1 (a): </u></h2><h2> </h2>

According to Newton's Law of Gravitation, the Gravity Force is:

$F=\frac{GMm}{{r}^{2}}$ (1)

This expression can also be written as:

$F=GMm{r}^{-2}$ (2)

If we derive this force $F$ respect to the distance $r$ between the two masses:

$\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr$ (3)

Taking into account $GMm$ <u>are constants:</u>

$\frac{dF}{dr}dFdr=-2GMm{r}^{-3}$ (4)

$\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}$ (5)

<h2><u>Answer 2 (a):</u> dF/dr represents the rate of change of the force with respect to the distance between the bodies. </h2>

In other words, this means how much does the Gravity Force changes with the distance between the two bodies.

More precisely this change is inversely proportional to the distance elevated to the cubic exponent.

<u>As the distance increases, the Force decreases.</u>

<h2><u>Answer 3 (a):</u> The minus sign indicates that the bodies are being forced in the negative direction. </h2>

This is because Gravity is an attractive force, as well as, a central conservative force.

This means it does not depend on time, and both bodies are mutually attracted to each other.

<h2><u>Answer 4 (b):</u> $X=-32N/km$ </h2>

In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit $N/km$ :

$\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}$ (5)

We have a force that decreases with a rate 1 $\frac{dF_{1}}{dr}dFdr=4N/km$ when $r=20000km$ :

$4N/km=-2\frac{GMm}{{(20000km)}^{3}}$ (6)

Isolating $-2GMm$ :

$-2GMm=(4N/km)({(20000km)}^{3})$ (7)

In addition, we have another force that decreases with a rate 2 $\frac{dF_{2}}{dr}dFdr=X$ when $r=10000km$ :

$XN/km=-2\frac{GMm}{{(10000km)}^{3}}$ (8)

Isolating $-2GMm$ :

$-2GMm=X({(10000km)}^{3})$ (9)

Making (7)=(9):

$(4N/km)({(20000km)}^{3})=X({(10000km)}^{3}$ (10)

Then isolating $X$ :

$X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}$

Solving and taking into account the units, we finally have:

$X=-32N/km$ >>>>This is how fast this force changes when $r=10000 km$

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3 years ago