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Anna71 [15]
3 years ago
7

Negatively charged particle that orbits the nucleus

Physics
1 answer:
Ede4ka [16]3 years ago
6 0

Answer: Electrons.

Explanation:

Electrons are the negatively charged particles that orbit the nucleus of an atom. Protons, on the other hand, are the positively charged particles that orbit the nucleus of an atom.

in the picture, the protons and neutrons are “in” the nucleus and you’ll also see the the electrons are orbiting all around the nucleus.

If this helped please mark me brainliest.

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I dont know how to do question 3,4
dlinn [17]

<span>Prior to the Cambrian, organisms with hard parts (eg. shells and bones) weren't present for most the time, and it's generally such hard parts that get to fossilize. There's also been more time for Precambrian deposits to get destroyed by natural processes, </span>

<span>The supply of Precambrian fossils has actually greatly improved over the last few decades. The so called Ediacaran faunas have been found at many parts of the world, and would be worth looking into.hope it helps</span>
7 0
3 years ago
If the mass of the block is 5 kg and the speed 7 m/s, what is the work done <br> on the block?
Kamila [148]

Answer:

A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40N is applied, the acceleration of the block will be then (g=10ms

2 ).

Mass of the block=5kg

Coeffecient of friction=0.2

external applied force, F=40N

The angle at which the force is applied=30degree

So the horizontal component of force=Fcos30=40×

23 =20 3 N

While the uertical component of the force acting in upward direction=Fsin30=40× 21

​ =20N

The normal reaction from the surface (N)=mg−Fsin30=50−20=30N

So the ualue of limiting friction=μN=0.2×30=6N

Hence the net horizontal force on the block=Fcos30=μN=20

3

​

N−6N=28.64N

The horizontal acceleration of the block=

m

Fcos30−μN = 528.64

​  =5.73m/s 2

8 0
3 years ago
Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light
motikmotik

Answer:

n_cladding = 1.4764

Explanation:

We are told that θ_max = 5 °

Thus;

θ_max + θ_c = 90°

θ_c = 90° - θ_max

θ_c = 90° - 5°

θ_c = 85°

Now, critical angle is given by;

θ_c = sin^(-1) (n_cladding/n_core)

sin θ_c = (n_cladding/n_core)

n_cladding = (n_core) × sin θ_c

Plugging in the relevant values, we have;

n_cladding = 1.482 × sin 85

n_cladding = 1.4764

7 0
3 years ago
A circular loop of wire with a radius of 20 cm lies in the xy plane and carries a current of 2 A, counterclockwise when viewed f
Zina [86]

To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be

A=\pi r^2

A =\pi (0.2)^2

A =0.1256 m^2

Once the area is obtained, it is possible to calculate the magnetic dipole moment considering the previously given definition:

\mu=IA

\mu=2(0.1256)

\mu=0.25 A\cdot m^2

Therefore the magnetic dipole moment is 0.25A\cdot m^2

6 0
3 years ago
A nonconducting container filled with 25 kg of water at 23°C is fitted with a stirrer, which is made to turn by gravity acting o
Paul [167]

Explanation:

Given that,

Weight of water = 25 kg

Temperature = 23°C

Weight of mass = 32 kg

Distance = 5 m

(a). We need to calculate the amount of work done on the water

Using formula of work done

W=mgh

W=32\times9.8\times5

W=1568\ J

The amount of work done on the water is 1568 J.

(b). We need to calculate the internal-energy change of the water

Using formula of internal energy

The change in internal energy of the water equal to the amount of the  work done on the water.

\Delta U=W

\Delta U=1568\ J

The  change in internal energy is 1568 J.

(c). We need to calculate the final temperature of the water

Using formula of the change internal energy

\Delta U=mc_{p}\Delta T

\Delta U=mc_{p}(T_{2}-T_{1})

T_{2}=T_{1}+\dfrac{\Delta U}{mc_{p}}

T_{2}=23+\dfrac{1568}{25\times4.18\times10^{3}}

T_{2}=23.01^{\circ}\ C

The final temperature of the water is 23.01°C.

(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.

The amount of heat is 1568 J.

Hence, This is the required solution.

6 0
3 years ago
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