If a net force of 25 N is exerted over a distance of 4 m to the right on a 2 kg mass initially at rest and moves it, what is

How much time would Simpson save by increasing his average velocity to 26m/s East?

igor_vitrenko [27]

Answer:

8 minutes

Explanation:

Part of the question is missing. Complete question is:

<em>Simpson drives his car 144 km with an average velocity of 24 m/s toward the east. How much time would Simpson save by increasing his average velocity to 26m/s East?</em>

Solution:

First of all, we have to calculate the time it takes for Simpson to complete the drive moving at the initial velocity, 24 m/s. We can use the formula:

$t=\frac{d}{v}$

where

$d=144 km = 1.44\cdot 10^5 m$ is the displacement

$v=24 m/s$ is the velocity

Substituting,

$t=\frac{1.44\cdot 10^5}{24}=6000 s = 100 min$

While driving at 26 m/s, the time taken will be

$t=\frac{1.44\cdot 10^5}{26}=5538 s=92 min$

So, Simpson will save 8 minutes.

7 0

2 years ago

An atomic nucleus is composed of A) protons. B) protons and neutrons. C) protons and electrons. D) protons, neutrons, and electr

8090 [49]

Answer:

B

Explanation:

The nucleus is made up of protons and neutrons, and electrons float around the neucleus in the electron cloud.

3 0

2 years ago

Read 2 more answers

Question 11 of 11 | Page 11 of 11

KiRa [710]

Answer:

Decreases the time period of revolution

Explanation:

The time period of Cygnus X-1 orbiting a massive star is 5.6 days.

The orbital velocity of a planet is given by the formula,

v = √[GM/(R + h)]

In the case of rotational motion, v = (R +h)ω

ω = √[GM/(R + h)] /(R +h)

Where 'ω' is the angular velocity of the planet

The time period of rotational motion is,

T = 2π/ω

By substitution,

Hence, from the above equation, if the mass of the star is greater, the gravitational force between them is greater. This would reduce the time period of revolution of the planet.

3 0

2 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:

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

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