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AfilCa [17]
2 years ago
7

A chief finds a glittering stone which he shows to the family. The family assures him that the stone is pure gold but he doubts.

Prepare a message of what you can do with the stone to give the chief and his family the advice
Physics
1 answer:
Nina [5.8K]2 years ago
5 0

I will tell the Chief about the high worth of the Gold and the various

substances it can be used to produce.

<h3>What is Gold?</h3>

Gold is a naturally occurring element with a bright, slightly orange yellow

color. It has a high value and can be used in the production of various

substances such as:

  • Jewelries
  • Medals and awards
  • Arts etc.

The Chief should however be told about the high worth value of Gold and

the various things it can be used for so he can make a decision.

Read more about Gold here brainly.com/question/1159978

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Nicolaus Copernicus was a Polish astronomer who is best known for the theory that the sun is near the center of the universe and
abruzzese [7]

Answer:B) It was not understood by scientific world.

Explanation:

4 0
2 years ago
Particles at the very outer edge of Saturn’s A Ring are in a 7:6 orbital resonance with the moon Janus. If the orbital period of
vivado [14]

Answer:

14 hours 18 minutes.

Explanation:

ratio of number of orbits, so it completes 7 orbits in the time Janus does 6.

(16*60+41)*6/7=858 minutes or 14 hours 18 minutes

3 0
3 years ago
The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr
MA_775_DIABLO [31]

Answer:

q=49Q/64

and

x =16R/15

Explanation:

See  attached figure.

E_{Q}= E due to sphere

E_{q}= E due to particule

E_{total}=E_{Q}-E_{q}=0  (1)

according to the law of gauss and superposition Law:

E_{Q}=E_{1}+E_{2}=E_{2} ; electric field due to the small sphere with r1=R/4

E_{Q}=kq_{2}/(r_{1}^{2})=

q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}

then: E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})  (2)

on the other hand, for the particule:

E_{q}=kq/(r_{p}^{2})

r_{p}=2R-R/4=7R/4   ⇒    E_{q}=16kq/(49R^{2})   (3)

We replace (2) y (3) in (1):

E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})

q=49Q/64

--------------------

if R<x<2R   AND E_{total}=E_{Q}-E_{q}=0

E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})

remember that  q=49Q/64

then:

Q(2R-x^{2})=49/64*x^{2}

solving:

x_{1} =16R/15

x_{2} =16R

but: R<x<2R  

so : x =16R/15

7 0
3 years ago
A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just
bagirrra123 [75]

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

5 0
3 years ago
An ice cream maker has a refrigeration unit which can remove heat at 120 Js'. Liquid ice
Rom4ik [11]

Answer:

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C is 45,360 J

Explanation:

The given parameters for the refrigeration unit and the ice cream are;

The power of the refrigeration unit = 120 J/s

The mass of the liquid ice cream, m = 0.6 kg

The initial temperature of the liquid ice cream, T₁ = 20°C

The freezing point temperature of the ice cream, T₂ = -16°C

The specific heat capacity of the ice cream, c = 2,100 J/kg⁻¹·°C⁻¹

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, ΔQ, is given as follows;

ΔQ = m × c × ΔT

Where;

ΔT = T₁ - T₂

∴ ΔQ = m × c × (T₁ - T₂)

Therefore, by substituting the known values, we have;

ΔQ = 0.6 × 2,100 × (20 - (-16)) = 45,360

The amount of heat energy that must be removed from the mixture to cool it to its freezing point, of -16°C = ΔQ = 45,360 J.

8 0
3 years ago
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