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3 years ago

The velocity of 200kg object changes from 10m/s to 22m/s in 12 sec.what is the force on the accelerated object?

Physics

1 answer:

Vikentia [17]3 years ago

4 0

Initial velocity (u) = 10 m/s
Final velocity (v) = 22 m/s
Time (t) = 12 s
Mass (m) = 200 Kg
Let the acceleration be a.
By using the equation of motion,

v = u + at, we have

22 m/s = 10 m/s + 12 s × a
or, 22m/s - 10 m/s = 12 s × a
or, 12 m/s = 12 s × a
or, a = 1 m/s^2
Let the force be F.
We know, F = ma
Therefore, the force on the accelerated object (F)
= ma
= (200 × 1) N
= 200 N

Answer:

b) 200 N

Hope you could understand.

If you have any query, feel free to ask.

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The two forces acting on a skydiver are __________ and air _____________.

Stolb23 [73]

Answer:

Gravity and

Air resistance

Explanation:

The two forces acting on a skydiver are gravitational force and air resistance.

Gravitational force is a force that tends to pull all massive bodies towards the center of the earth. It works on all bodies that has mass. The larger or bigger the mass, the more the pull of gravity on the body.

Air resistance is the drag of air on a body as it passes to it. It is resisting force.

When a sky diver jumps out of a plane, he/she encounters both gravity and air resistance.
It soon balances both force and attain terminal velocity.
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3 0

4 years ago

Read 2 more answers

86 Km/h convertir a (m/min)

gogolik [260]

Answer: 1433.3 m/min

Explanation:

For 86 Km/h converted to a (m/min), convert kilometers to meters, and hour to minutes

So, 86 Km/h means 86 kilometers per 1 hour

- If 1 kilometer = 1000 metres

86 kilometers = 86 x 1000 = 86,000m

- If 1 hour = 60 minute

1 hour = 60 minutes

In m/min: (86,000m / 60 minute)

= 1433.3 m/min

Thus, 86 Km/h convert to 1433.3 m/min

5 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:

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

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