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3 years ago

compare 51.5 hectograms to 51500 decigrams. How do each of these values compare to a gram, and which represents a larger mass?

1 answer:

7 0

These values when compared seem to be the same. They are equal. If we convert them to the same units, they results to the value which is:

51.5 Hectograms = 5100 Grams

51,500 Decigrams = 5100 Grams

Hope this answers the question. Have a nice day.

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A frog hops 5m east and 2m north. What is the magnitude of the frogs total displacement in m?

TiliK225 [7]

As per the question a frog jumps 5 m towards east.

Frog again jumps 2 m north.

Let the displacement along east is denoted by vector A and the displacement towards north is denoted as vector B.

Hence magnitude of A = 5 m

Magnitude of B = 2 m

We are asked to calculate the total displacement.

Here the angle between them is 90 degree as A is towards east and B is towards north.

As per parallelogram law of vector addition,the magnitude of total displacement [R] will be-

$R=\sqrt{ A^{2} +B^{2}+2AB\cos\theta}$

$=\sqrt{ 5^{2}+ 2^{2}+2*5*2 cos 90}$

$=\sqrt{25+4+0} m$ [cos90= 0]

$=\sqrt{29} m$

$= 5.38516 m$ [ans]

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3 years ago

two idential train engines are parked on parallel track. one engine has ten cars attached and the other has no cars attached. ea

nikklg [1K]

Answer:

The second engine with no cars will accelerate more.

Explanation:

Since the engines are identical so they hacve the same mass. The first engine has 10 cars attached which adds more mass to it whereas second engine has no cars attached.

Therefore second engine is lighter in mass than the first one.

We know from the Newton's second law of motion:

$F=m.a$

$\therefore a=\frac{F}{m}$

Now, according to question both the engines exert same force for moving.

We see that :

$a\propto\frac{1}{m}$

So more is the mass, lesser is the acceleration, hence the second engine with no cars will accelerate more.

6 0

3 years ago

Calculate the frequency of each of the following wave lengths of electromagnetic radiation. Part A 488.0 nm (wavelength of argon

Drupady [299]

Answer:

A. $f=6.1475*10^{14}Hz$

B. $f=5.9642*10^{14}Hz$

Explanation:

The frequency has an inversely proportional relationship with the concept of wavelength, the greater the wavelength, the lower the frequency. For electromagnetic waves, the frequency is equal to the speed of light, divided by the wavelength.

$f=\frac{c}{\lambda}$

$f=\frac{3*10^8\frac{m}{s}}{488*10^{-9}m}\\\\f=6.1475*10^{14}Hz$

$f=\frac{3*10^8\frac{m}{s}}{503*10^{-9}m}\\\\f=5.9642*10^{14}Hz$

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3 years ago

How could you keep an objects acceleration the same if the force acting on the object were double?

Bogdan [553]

You can keep an object's acceleration constant if the force acting on it is double by adding another force going in the opposite direction of the first force. The second force will have to be as strong as the first force was originally in order for the acceleration to be the same. This is the same concept as if you have 2 of something, double it to get 4, and subtract 2 again. You'll end up with the same value you started with.

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