An object accelerates 13.0 m/s2 when a force of 4.4 newtons is applied to it. What is the

Silver has a mass of 10.5 grams and a volume of 19.3 cm3. What is its density?

Neko [114]

Answer:

<h3>The answer is 0.54 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question we have

$density = \frac{10.5}{19.3} \\ = 0.54404145...$

We have the final answer as

Hope this helps you

7 0

3 years ago

A trip is taken that passes through the following points in order

MaRussiya [10]

Answer:

40.0 m

...............

3 0

3 years ago

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

Why is the rotation system used in volleyball? Physics was the closest thing to PE

tankabanditka [31]

Answer:

A rotation occurs after every side out, which is when the receiving team gains the right to serve by winning a rally. ... The new serving team will rotate clockwise one spot. The purpose of this is to rotate all the players through the serving position. If you continue winning points, you stay in position.

3 0

2 years ago

Read 2 more answers

Translation is the process by which the DNA code is reads to create the RNA strand

Leya [2.2K]

True............................................

5 0

3 years ago