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pantera1 [17]
3 years ago
6

A water rocket has a mass of 0.8kg and is launched in a school playground with an inital upwards force of 12newtons. what is the

weight of the rocket in the playground?
What is the inital acceleration of the rocket in the playground?
If this water rocket could were launched from the moon what would be its initial acceleration (rember to find the new weight first)
Physics
1 answer:
Darina [25.2K]3 years ago
7 0

Weight = (mass) x (gravity).
It always acts downward.

On Earth, the acceleration of gravity is 9.807 m/s².
On the Moon, the acceleration of gravity is 1.623 m/s².

On Earth, the rocket's weight is  (0.8kg) x (9.8 m/s²) = 7.84 newtons

On the Moon, the rocket's weight is  (0.8kg) x (1.62 m/s²) = 1.3 newtons

The force of the rocket engine acts upward.
Its magnitude is 12 newtons. (From the burning chemicals.
Doesn't depend on local gravity. Same force everywhere.)

Now we have all the data we need to mash together and calculate the
answers to the question.  You might choose a different method, but the
machine that I have selected to do the mashing with is Newton's 2nd law
of motion:

                           Net Force = (mass) x (acceleration).
 
Since the question is asking for acceleration, let's first solve Newton's law
for it.  Divide each side by (mass) and we have

                           Acceleration = (net force) / (mass) .

On Earth, the forces on the rocket are

        (weight of 7.84 N down) + (blast of 12 N up) =  4.16 newtons UP (net)

         Acceleration = (4.16 newtons UP) / (0.8 kg) = 5.2 m/s² UP .

On the moon, the forces on the rocket are

         (weight of 1.3 N down) + (blast of 12 N up) = 10.7 newtons UP (net)

         Acceleration = (10.7 newtons UP) / (0.8 kg) = 13.375 m/s² UP   
          
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There are 1,660 megawatts of wind-generated electricity produced globally every year. This amount is equivalent to A. 1,660,000
disa [49]
Let us first write down the known things .

1660 megawatts = 1660 X 10^6 watt
                           = 166000 kilowatt

From the above deduction we can conclude that the correct option among all the options that are given in the question is the second option or option "B". I hope that this is the answer that you were looking for and the answer has actually come to your desired help.
5 0
3 years ago
Read 2 more answers
Solve using correct significant figures and indicating maximum absolute uncertainty.
Vera_Pavlovna [14]

We use the criterion of significant figures to find the result with reliable figures

          X = 9.2 10-5

now with the propagation of errors we obtain the result with its uncertainty

         X ± ΔX = (9.2 ± 0.5) 10⁻⁵

given Parameter

     * expression values ​​with their absolute errors

to find

     * the result with the correct significant figures

     * the absolute error of the expression

Significant figures are defined with the number of decimals that give information, the number of figures in a quantity gives information about the uncertainty of this quantity.

There are two criteria for applying significant figures:

     * Add and subtract the result of going with the number of decimal places of the figure that has the least

    * Product and division as a result of going with the least number of significant figures than the value that has the least.

Remember that the zero to the left do not form a pair of the significant figures

Let's apply this belief to the case presented, let's write the precaution

 

              x = \frac{a-b}{c}

where in this case they are worth

         a = 0.0336 ± 0.0002

         b = 0.010 ± 0.001

         c = 255.4 ± 0.4

We see that the significant figures of each parameterize (a, b, c) and their absolute errors are correct.

Let's apply the criteria to the operation

          a-b = 0.0336 - 0.010

          a- b = 0.0236

we apply the criterion of significant figures for the subtraction, the result must be with 3 decimal places

        a - b = 0.024

let's do the other operation

         X = \frac{a-b}{c}

         X = 0.024 / 255.4

         X = 9.24 10⁻⁵

We apply the criterion of significant figures for the division, in this case the result is left with two significant figures

         X = 9.2 10⁻⁵

The uncertainty or error of the measurements is of most importance as it determines how many significative figures are reliable at a given magnitude.

If the magnitudes are measured with some type of instrument, the absolute error is given by the appreciation of the instrument, if the magnitude is calculated using some equation, the errors must be propagated using the variations of each parameter in the worst case.

           

the uncertainty of the calculated quantity (X) is

        \Delta X = | \frac{dX}{da}| \Delta a + | \frac{dX}{db} | \Delta b + | \frac{dX}{dc}| \Delta c

let's perform the derivatives

        \frac{dX}{da} = \frac{1}{c}

        \frac{dX}{db} = - \frac{1}{c}

        \frac{dX}{dc} = - \frac{a-b}{c^2}

we substitute

remember that the bulk value guarantees that we tune the worst case. So all the mistakes add up

          ΔX = \frac{1}{c}  Δa + \frac{1}{c} Δb + \frac{a-b}{c^2}  Δc

          ΔX = \frac{1}{c} (Δa + Δb) + \frac{a-b}{c^2} Δc

we substitute

         ΔX = \frac{1}{255.4}  (0.0002 + 0.001) + \frac{0.0336-0.010}{255.4^2}  0.4

         ΔX = 4.698 10⁻⁶ + 1.45 10⁻⁷

         DX = 4.8 10-6

Absolute errors must be given with a single significant figure

         ΔX = 5 10⁻⁶

The result of the requested quantity using the criterion of significant figures and propagation of errors is

          X ± ΔX = (9.2 ± 0.5) 10⁻⁵

learn more about   significative figure here:

brainly.com/question/18955573

8 0
3 years ago
A sodium ion, Na+ , has a positive charge because it
goldenfox [79]

D. lost an electron

because they have free electron in the outermost shell which they will like to denote, in order to attain stable configuration

3 0
3 years ago
an object is sliding down in clean plane the velocity change at a constant rate from 10 cm to 15 CM in 2 second what is it accel
AlekseyPX

Initial velocity=10m/s=u

Final velocity=v=15m/s

Time=t=2s

\boxed{\sf Acceleration=\dfrac{v-u}{t}}

\\ \sf\longmapsto Acceleration=\dfrac{15-10}{2}

\\ \sf\longmapsto Acceleration=\dfrac{5}{2}

\\ \sf\longmapsto Acceleration=2.5m/s^2

6 0
3 years ago
Read 2 more answers
The pressure exerted by a column of liquid is equal to the product of the height of the column times the gravitational constant
almond37 [142]

Answer:

The column of methanol is = 12.27 m high

Explanation:

The pressure of mercury  (p) = ghd,................... equation 1

Where g = acceleration due to gravity, h= height, d = density.

d = 13.6 g/ml convert to kg/m³ = 13.6 g/ml(1000 (kg/m³)/(g/ml))

d =13600 kg/m³., h= 713mm = 713/1000 = 0.713 m., g= 9.80 m/s²

Substituting these values into equation 1,

P = 13600 × 0.713 × 9.8 = 95028.64 N/m².

If the pressure of the mercury supports the pressure exerted by the methanol

∴ pressure that supports the mercury = pressure of the methanol

       p = ghd

making h the subject of formula,

h = p/gd ........................... equation 2

Where p = 95028.64 N/m², g = 9.8 m/s², d = 0.79 g/ml = (0.79 × 1000) kg/m³

d = 790 kg/m³.

Substituting these values into equation 2

h = 95028.64/(9.8×790)

h = 95028.64/7742

h = 12.27 m.

The column of methanol is = 12.27 m high

7 0
3 years ago
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