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Juliette [100K]
3 years ago
11

Dimensional formula of current density

Physics
2 answers:
kozerog [31]3 years ago
6 0
Ampere per meter squared A/mx^{2}
avanturin [10]3 years ago
6 0
<em>The Dimensional Formula is [M ^{-2} A]</em>

When we are talking about current density, we are referring to "Ampere per unit area", so the unit we would use is A/m^{2}

Thank you for your question! I hope this helped! Have an amazing rest of your week and feel free to let me know if you need any further help with anything! :D
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(Fill the blank)<br> the ________ g____ makes saliva that make your food wet and easy to swallow
blagie [28]

Answer:

What make saliva that make your food wet and easy to swallow?

enzyme amylase

Explanation:

The digestive functions of saliva include moistening food, and helping to create a food bolus, so it can be swallowed easily. Saliva contains the enzyme amylase that breaks some starches down into maltose and dextrin. Thus, digestion of food occurs within the mouth, even before food reaches the stomach.

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6 0
2 years ago
A 0.18 m radius pulley is free to rotate about a horizontal axis. A mass and a mass are attached by a massless string, which is
Musya8 [376]

Answer:

T = 1.766(M-m) Nm where M and m are the 2 masses of the objects

Explanation:

Let m and M be the masses of the 2 objects and M > m so the system would produce torque and rotational motion on the pulley. Force of gravity that exert on each of the mass are mg and Mg. Since Mg > mg, the net force on the system is Mg - mg or g(M - m) toward the heavier mass.

Ignore friction and string mass, and let g = 9.81 m/s2, the net torque on the pulley is the product of net force and arm distance to the pivot point, which is pulley radius r = 0.18 m

T = Fr = g(M - m)0.18 = 0.18*9.81(M - m) = 1.766(M-m) Nm

8 0
3 years ago
The stone, which weighs 400 g, is thrown upwards at a speed of 20 m / s. Climbed to a height of 12 m. Determine: what is equal t
maxonik [38]

Given that,

Mass of the stone, m = 400 g = 0.4 kg

Initial speed, u = 20 m/s

It is climbed to a height of 12 m.

To find,

The work done by the resistance force.

Solution,

Let v is the final speed. It can be calculated by using the conservation of energy.

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 12} \\\\v=15.33\ m/s

Work done is equal to the change in kinetic energy. It can be given as follows :

W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 0.4\times (15.33^2-20^2)\\\\=-32.99\ J

So, the required work done is 32.99 J.

3 0
2 years ago
From the top of the engineering building, you throw a ball vertically upward. the ball strikes the ground 4.00 s later. the engi
anastassius [24]
Equations of the vertical launch:

Vf = Vo - gt

y = yo + Vo*t - gt^2 / 2

Here yo = 35.0m
Vo is unknown
y final = 0
t = 4.00 s
and I will approximate g to 10m/s^2

=> 0 = 35.0 + Vo * 4 - 5 * (4.00)^2 => Vo = [-35 + 5*16] / 4 = - 45 / 4 = -11.25 m/s

The negative sign is due to the fact that the initial velocity is upwards and we assumed that the direction downwards was positive when used g = 10m/s^2.

Answer: 11.25 m/s


5 0
3 years ago
Read 2 more answers
Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at
PSYCHO15rus [73]

Answer:

|\Delta v |=\sqrt{\frac{4Gm}{d} }

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

\mu= \frac{m_1m_2}{m_1+m_2}

now, since both the masses have mass m

therefore,

\mu= \frac{m^2}{2m}

= m/2

The final K.E. of the particles is

KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2

Distance between two particles is d and the gravitational potential energy between them is given by

PE_{Gravitational}= \frac{Gmm}{d}

By law of conservation of energy we have

KE_{initial}+KE_{final}= PE_{gravitaional}

Now plugging the values we get

0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}

|\Delta v |=\sqrt{\frac{4Gm}{d} }

=\sqrt{\frac{Gm}{d} }

This the required relation between G,m and d

5 0
3 years ago
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