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Olenka [21]
3 years ago
9

You are on a new planet, and are trying to use a simple pendulum to find the gravity experienced on this new planet. You find th

at a pendulum having a length of 35 cm has a period of 2 seconds. What is the acceleration due to gravity on this planet?
A. 13.82 m/s^2
B. 345.4 m/s^2
C. 3.45 m/s^2
D. 1.1 m/s^2
Physics
1 answer:
levacccp [35]3 years ago
4 0

To solve this problem it is necessary to apply the concepts related to the Period based on the length of its rope and gravity, mathematically it can be expressed as

T= 2\pi \sqrt{\frac{L}{g}}

g = Gravity

L = Length

T = Period

Re-arrange to find the gravity we have

g = \frac{4\pi^2 L}{T^2}

Our values are given as

L = 0.35m\\T = 2s\\

Replacing we have

g = \frac{4\pi^2 L}{T^2}

g = \frac{4\pi^2 0.35}{2^2}

g = 3.45 m/s^2

Therefore the correct answer is C.

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What is the weight of the body under the influence of a force of 10N reaches a speed of 1.5 m / s in 3s?
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Answer:

Explanation:

F=ma

F=10N

A=v-u upon t

=1.5-0 upon 3

=0.5m per second squared

Then do force divided by. Accelaration and multiply by gravity.

8 0
3 years ago
How is wave movement affected by the various mediums it may travel through?
Juliette [100K]

Answer:

speed is affected by the medium in wich the wave travels.

Explanation:

The speed of a wave is affected when traveling in different medium, depending on the medium in which the wave travels it can go faster or slower, or not even be able to travel in that medium.

For light waves, the denser the medium through which the wave travels, the slower it will go through it, as there is more resistance to the light particles.

And in waves like sound, which instead of particles is a vibration or disturbance in the medium, they travel faster in denser mediums such as metal, because the disturbance is transmitted more efficiently thanks to how close together that the particles are in the metal, This is why sound does not travel in a vacuum; there are not even particles to transmit the vibration.

5 0
3 years ago
5- A 2500g object is pushed with 55N for 12m in 11s, there was a force of friction of 30N.
Assoli18 [71]

Answer:

1kg =1000g

2.5kg

D=12m

t=11s

F=2.5KG

Explanation:

work done =f.d

=2.5×12

=30Nm

55-30

average speed

final - initial

divide by time t(s)

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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})

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
Which condition must be met in order for an equation to be balanced?
vitfil [10]

Answer:

There must be an equal amount of each element on both sides of the equation. Hope this helps and please marks as the brainliest.

Explanation:

4 0
3 years ago
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