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3 years ago

. A paper airplane of .025kg is falling at a rate 2m/s2. What is the amount of net force?

Physics

1 answer:

Pie3 years ago

6 0

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 0.025 × 2

We have the final answer as

Hope this helps you

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Increasing the two objects will cause the gravitational force between the objects to decrease.

ELEN [110]

This statement is false. Increasing the two objects' mass (I'm guessing) will actually increase their gravitational force. This is because of the equation:

$F_g = \frac{Gm_1m_2}{d^2}$

If the distance was increased, then the statement would be true, but since you are increasing mass, which is proportional to the Force of Gravity, you are in fact, increasing the gravitational force between the two objects.

4 0

3 years ago

Read 2 more answers

For this discussion, you will work in groups to answer the questions. In a video game, airplanes move from left to right along t

Mariulka [41]

Answer:

When fired from (1,3) the rocket will hit the target at (4,0)

When fired from (2, 2.5) the rocket will hit the target at (12,0)

When fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

When fired from (4,2.25) the rocket will hit the target at (40,0)

Explanation:

All of the parts of the problem are solved in the same way, so let's start with the first point (1,3).

Let's assume that the rocket's trajectory will be a straight line, so what we need to do here is to find the equation of the line tangent to the trajectory of the airplane and then find the x-intercept of such a line.

In order to find the line tangent to the graph of the trajectory of the airplane, we need to start by finding the derivative of such a function:

$y=2+\frac{1}{x}$

$y=2+x^{-1}$

$y'=-x^{-2}$

$y'=-\frac{1}{x^{2}}$

so, we can substitute the x-value of the given point into the derivative, in this case x=1, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(1)^{2}}$

m=y'=-1

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-3=-1(x-1})$

$y-3=-1x+1$

$y=-x+1+3$

$y=-x+4$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-x+4=0$

and solve for x

x=4

so, when fired from (1,3) the rocket will hit the target at (4,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2, 2.5)

so, we can substitute the x-value of the given point into the derivative, in this case x=2, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2)^{2}}$

$m=y'=-\frac{1}{4}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.5=-\frac{1}{4}(x-2})$

$y-2.5=-\frac{1}{4}x+\frac{1}{2}$

$y=-\frac{1}{4}x+\frac{1}{2}+\frac{5}{2}$

$y=-\frac{1}{4}x+3$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{1}{4}x+3=0$

and solve for x

x=12

so, when fired from (2, 2.5) the rocket will hit the target at (12,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2.5, 2.4)

so, we can substitute the x-value of the given point into the derivative, in this case x=2.5, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2.5)^{2}}$

$m=y'=-\frac{4}{25}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.4=-\frac{4}{25}(x-2.5})$

$y-2.4=-\frac{4}{25}x+\frac{2}{5}$

$y=-\frac{4}{25}x+\frac{2}{5}+2.4$

$y=-\frac{4}{25}x+\frac{14}{5}$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{4}{25}x+\frac{14}{5}=0$

and solve for x

$x=\frac{35}{20}$

so, when fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

Now, let's calculate the coordinates where the rocket will hit the target if fired from (4, 2.25)

so, we can substitute the x-value of the given point into the derivative, in this case x=4, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(4)^{2}}$

$m=y'=-\frac{1}{16}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.25=-\frac{1}{16}(x-4})$

$y-2.25=-\frac{1}{16}x+\frac{1}{4}$

$y=-\frac{1}{16}x+\frac{1}{4}+2.25$

$y=-\frac{1}{16}x+\frac{5}{2}$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{1}{16}x+\frac{5}{2}=0$

and solve for x

$x=40$

so, when fired from (4,2.25) the rocket will hit the target at (40,0)

I uploaded a graph that represents each case.

8 0

3 years ago

With a bit of algebraic reasoning find your gravitational acceleration toward any planet of mass M a distance d from its center.

grandymaker [24]

The acceleration due to gravity is given as:

g = GM/r²

<h3>Derivation of gravitational acceleration:</h3>

According to Newton's second law of motion,

F = ma

where,

F = force

m = mass

a = acceleration

According to Newton's law of gravity,

F<em>g </em>= GMm/(r + h)²

F<em>g = </em>gravitational force

From Newton's second law of motion,

F<em>g </em>= ma

a = F<em>g</em>/m

We can refer to "a" as "g"

a = g = GMm/(m)(r + h)²

g = GM/(r + h)²

When the object is on or close to the surface, the value of g is constant and height has no considerable impact. Hence, it can be written as,

g = GM/r²

Learn more about gravitational acceleration here:

brainly.com/question/2142879

#SPJ4

5 0

2 years ago

Scientist who first proposed the theory that the continents drifted is __________

Aloiza [94]

Answer:

Alfred Wegener

Explanation:

Alfred Wegener is a german meteorologist who proposed the theory that the continents drifted, and he presented it to the German Geological Society on January 1912.

4 0

3 years ago

(a) Calculate the buoyant force on a 2.00-L helium balloon.

iragen [17]

Answer:

0.0239364 N

0.0057879 N

Explanation:

$\rho$ = Density of the gas

g = Acceleration due to gravity = 9.81 m/s²

V = Volume

Mass of rubber = 1.5 g

Buoyant force is given by

$F_b=\rho gV\\\Rightarrow F_b=1.22\times 9.81\times 2\times 10^{-3}\\\Rightarrow F_b=0.0239364\ N$

The buoyant force is 0.0239364 N

Net vertical force is given by

$F_n=F_b-W_{He}-W_{r}\\\Rightarrow F_n=0.0239364-0.175\times 2\times 10^{-3}\times 9.81-1.5\times 10^{-3}\times 9.81\\\Rightarrow F_n=0.0057879\ N$

The net vertical force is 0.0057879 N

6 0

3 years ago