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Novay_Z [31]
3 years ago
11

G (where g=9.8 m/s2). If an object’s mass is m=10. kg, what is its weight?

Physics
1 answer:
Karolina [17]3 years ago
8 0

An object's weight is

           (its mass) x (the acceleration of gravity wherever the object is) .

The acceleration of gravity on or near the Earth's surface is about 9.8m/s².

So an object with 10kg of mass, as long as it's located on or near the
Earth's surface, weighs
                                       (10 kg) x (9.8 m/s²) = 98 kg-m/s² = <em>98 Newtons</em>.

                                                                      That's about 22.05 pounds.

If you take the object somewhere else, it's still 10 kg of mass, but
it weighs something different, depending on the gravity of the place
where you take it.


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Answer:

d= 794.4 cmExplanation:

Given that

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=286\times \dfrac{1000}{3600}\ m/s

V=79.44 m/s

Given that time ,t= 100 ms

t= 0.1 s

We know that ( if acceleration is zero)

Distance = Speed x time

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Now by putting the values in the above equation

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3 years ago
A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre
AleksandrR [38]

Answer:

A) 1568.60 Hz

B) 1437.15 Hz

Explanation:

This change is frequency happens due to doppler effect

The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

f_(observed)=\frac{(c+-V_r)}{(C+-V_s)} *f_(emmited)\\

where

C = the propagation speed of waves in the medium;

Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;

Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.

A) Here the Source is moving towards the receiver(C-Vs)

and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

f_(observed)=\frac{C}{C-V_s} *f_(emmited)\\=\frac{343}{343-15}*1500\\ =1568.60 Hz

B)Here the Source is moving away the receiver(C+Vs)

and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser

f_(observed)=\frac{C}{C+V_s} *f_(emmited)\\=\frac{343}{343+15}*1500\\ =1437.15 Hz

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The product or products of a reaction are determined by the type of reaction taking place. true or false
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Answer:

True

Explanation:

The products of a reaction are determined by the type of chemical reaction that are taking place. This is very true.

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