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Darya [45]
4 years ago
6

An ice cream machine produced 44 ice creams per minute. After reconditioning, its speed increase to 55 ice creams per minute. By

what percent did the speed of the machine increase
Physics
2 answers:
dolphi86 [110]4 years ago
7 0

Answer:

25%

Explanation:

The way to find this answer is to:

Subtract 44 from 55 (11)

Then, you can put 11 over 44

Simplified, this is equal to 1/4.

1/4 as a percentage is 25%.

Good luck with your RSM work!

Slav-nsk [51]4 years ago
3 0

Answer:

an ice cream machine produced 44 ice cream per minute.

Now, after reconditioning it  speeds up to 55 ice cream per minute.

=> 55 - 44 = 11

=> 11 / 44 = 0.25

=> 0.25 * 100% = 25%.

thus it increased it production to 25%.

Explanation:

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Explanation:

Yes, translations are rigid transformations. You can verify that the distances between the points are preserved.

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Which of the following statements is correct about the force of gravity between two objects
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bagirrra123 [75]

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Cocoa mix is the: Solute

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3 years ago
A level of
Wittaler [7]

Answer:

x = 2 meters.

Explanation:

Let the position (distance) of fulcrum to the load be x.

Given the following data;

Load = 40 kg

Effort (force) = 40 Newton

Effort arm = 4 - x

To find the position of the fulcrum, we would use the expression;

Effort * effort arm = load * load arm

40 * (4 - x) = 40 * x

160 - 40x = 40x

160 = 40x + 40x

160 = 80x

x = 160/80

x = 2 meters

Therefore, the position (distance) of fulcrum to the load is 2 meters.

7 0
3 years ago
The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ
LenaWriter [7]

The orbital period of the satellite[T] is given as 2*10^{6} S.

The radius of the satellite is given [R] 2.5*10^{12} m.

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way  the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V]   written as -

                                V=\frac{2\pi R}{T}

where T is the time period of the satellite and R is the radius of the satellite.

                        V=\frac{2*3.14*10^{12} }{2*10^{6} }

                               = 7.85*10^{6} m/s

There is also another way through which we can get  the solution as explained below-

We know that the tangential speed of a satellite V=\sqrt{\frac{GM}{R^{2} } }

where G is the gravitational constant and M is the mas of central object.

But we know that g=\frac{GM}{R^{2} }

                               ⇒GM=gR^{2}  where g is the acceleration due to gravity of that central object.


Hence    V=\sqrt{\frac{gR^{2} }{R} }

               ⇒   V=\sqrt{gR}

By knowing the value of g due to that central object we can also calculate its tangential speed.

                           

 




7 0
3 years ago
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