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Mamont248 [21]
4 years ago
10

Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate

, produces y widgets every 5 minutes. If machines M and N work simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N at that time?
Physics
1 answer:
Westkost [7]4 years ago
3 0

Answer:

Yes

Explanation:

In 4 minutes, machine M produces x widgets

In 20 minutes, machine M produces 5x widgets

Assuming x = 1, in 20 minutes machine M produces 5 widgets

In 5 minutes, machine N produces y widgets

In 20 minutes, machine N produces 4y widgets

Assuming y = 1, in 20 minutes machine N produces 4 widgets

Machine M produces 1 more widget than machine N in 20 minutes assuming x=1 and y=1

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<span>Both objects receive the same impulse.</span>
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what do radio waves transfer between a cell phone and a cell phone tower? A.Sound. B.Energy. C. the medium D. air particles
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The answer would be energy, because all wave emitters give off at least one type of energy....
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4 years ago
A canon is tilled back 30.0 degrees and shoots a cannon ball at 155 m/s. What is the
Umnica [9.8K]

Answer:

= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m

Therefore, highest point that the cannon ball reaches is 168.7m

Explanation:

the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s

highest point that the cannon ball reaches?

H_{max}=\frac{V^2\sin ^2 \theta}{2g}

g = 9.8m/s2

= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m

Therefore, highest point that the cannon ball reaches is 168.7m

6 0
4 years ago
Convert 1.5 radian into milliradians.
Alex

Answer:

1500 milliradians

Explanation:

Data provided in the question:

1.5 radians

Now,

1 radians consists of  1000 milliradians

1 milli = 1000

thus for the 1.5 radians, we have

1.5 radians = 1.5 multiplied by 1000 milliradians

or

1.5 radians = 1500 milliradians

Hence, after the conversion

1.5 radians equals to the value 1500 milliradians

4 0
3 years ago
Two identical 0.200kg mass are pressed against opposite ends of a light spring of force constant 1.75N/cm compressing the spring
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This type of a problem can be solved by considering energy transformations. Initially, the spring is compressed, thus having stored something called an elastic potential energy. This energy is proportional to the square of the spring displacement d from its normal (neutral position) and the spring constant k:

E_p=\frac{1}{2}kd^2= \frac{1}{2}175\frac{N}{m}\cdot 0.37^2m^2=11.98J

So, this spring is storing almost 12 Joules of potential energy. This energy is ready to be transformed into the kinetic energy when the masses are released. There are two 0.2kg masses that will be moving away from each other, their total kinetic energy after the release equaling the elastic energy prior to the release (no losses, since there is no friction to be reckoned with).

The kinetic energy of a mass m moving with a velocity v is given by:

E_k = \frac{1}{2}mv^2

And we know that the energies are conserved, so the two kinetic energies will equal the elastic potential one:

E_p = 2E_k=mv^2

From this we can determine the speed of the mass:

E_p =mv^2\implies v=\pm \sqrt{\frac{E_p}{m}}=\pm\sqrt{\frac{11.98J}{0.2kg}}=\pm 7.74\frac{m}{s}

The speed will be 7.74m/s in in one direction (+), and same magnitude in the opposite direction (-).

4 0
3 years ago
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