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Oxana [17]
3 years ago
6

What objects did Archimedes use to test the density of the gold crown? How Can a modern investigator identify the elements in pa

int
Physics
1 answer:
alekssr [168]3 years ago
6 0
:/........?????????,,,??????
You might be interested in
Which method do acoustic use to limit reverberations in music halls
dalvyx [7]

Answer:

reverberation time appropriate to the use and size of the room, adequate balance between direct and reverberant sound, intimacy and good sound diffusion in the room to obtain a uniform sound.

Explanation:The process of ... second method measured the speed of sound propagation by the phase shift.

8 0
3 years ago
If a cars velocity is slowing down is it considered a positive or negative acceleration?
Natali [406]

Answer:

Negative

Explanation:

Observe that the object below moves in the positive direction with a changing velocity. An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration).

4 0
3 years ago
We are running late for school and we want to make our 0.5 kg tea (it’s at 90 C) colder. Let’s assume we can drink tea when it’s
PSYCHO15rus [73]

Answer:

x=0.154kg

Explanation:

(x*L)+(0.5kg*4200*50)+(x*4200*(-50)=0

(x*333 000J/kg*c)+(0.5kg*4200J/kg*C*(-40C))+(x*4200J/kg*C*50C)=0

6 0
2 years ago
A 90-kilogram physics student would weigh 2970 Newtons on the surface of planet X. What is the magnitude of the acceleration due
KiRa [710]
<u>Weight = (mass) x (acceleration of gravity)</u>

Divide each side by (mass),and we have

                     Acceleration of gravity = (weight) / (mass)

Acceleration of gravity = 2,970/90 = 33 newtons per kilogram = <em>33 m/s²</em>
5 0
3 years ago
An astronaut on the moon drops a feather from rest (v = 0). The acceleration due to gravity is 1.67 m/s​ 2​ . If the feather beg
olga_2 [115]

Given :

Initial velocity , u = 0 m/s .

Acceleration due to gravity on moon , g_m=1.67\ m/s^2 .

Height , h = 2 m .

To Find :

Final position after falling for 1.5 seconds .

Solution :

We know , by equation of motion :

s=ut+\dfrac{at^2}{2}

Here , a = g_m .

So , equation will transform by :

s=ut+\dfrac{g_mt^2}{2}\\\\s=0+\dfrac{1.67\times 1.5^2}{2}\ m\\\\s=1.88\ m

Therefore , the height form moon's surface is 1.88 m .

Hence , this is the required solution .

6 0
3 years ago
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