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tester [92]
2 years ago
5

ANSWER THIS FOR BRAINY CROWN B) ITS SO EASY!

Physics
2 answers:
Stolb23 [73]2 years ago
5 0
So, if there’s two collide Non-sticky collision. And there’s 600 KG.M/S for 2 cars they we know that the unit rate is 300/KG.M/S for 1 car.
Luda [366]2 years ago
3 0

Answer:

300

Explanation:

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In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point
Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ  X volume of the oil drop

volume of the oil drop (spherical) =  (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

=  (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0
2 years ago
Is electrical conductivity?
expeople1 [14]

Answer:

can't tell if this is question, it is not written correctly

Explanation:

Electrical conductivity is the measure of a material's ability to allow the transport of an electric charge. Its SI is the siemens per meter, (A2s3m−3kg−1) (named after Werner von Siemens) or, more simply, Sm−1. It is the ratio of the current density to the electric field strength.

8 0
2 years ago
Example of items that changed chemically
Ostrovityanka [42]
Burning of gases is one the example of chemical change
3 0
3 years ago
Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0 incline. The coeffi cient of kinetic friction between the bl
UkoKoshka [18]

Answer:

a) W_g =61.25J

b) W_k = -46.25J

c) W_N = 0

d) W_g would be the same.

   W_k would decrease.

   W_N would be the same.

Explanation:

a) On an inclined plane the force of gravity is the sine component of the weight of the block.

F_g = mg\sin(\theta) = 5(9.8)\sin(30^\circ)\\W_g = F_g x = 5(9.8)\sin(30^\circ)2.5 = 61.25J

b) The friction force is equal to the normal force times coefficient of friction.

F_k = -mg\cos(\theta)\mu_k = -5(9.8)cos(30^\circ)0.436 = -18.5 N\\W_k = -F_kx = -46.25J

c) The work done by the normal force is zero, since there is no motion in the direction of the normal force.

d) The relation between the vertical height and the distance on the ramp is

h = x\sin(\theta)

According to this relation, the work done by the gravity wouldn't change, since the force of gravity includes a term of x\sin(\theta).

The work done by the friction force would decrease, because both the cosine term and the distance on the ramp would decline.

The work done by the normal force would still be zero.

5 0
3 years ago
A reaction mixture in a 3.67 L flask at a certain temperature initially contains 0.763 g H2 and 96.9 g I2, At equilibrium, the f
Alexxandr [17]

Answer:

13 530 482

Explanation:

                            H2    +          I2     ------>      2HI

start (mol)             0.3785         0.3818                   0

change (mol)       -0.3534        -0.3534            +0.7067

equilibrium (mol)  0.0251         0.0284             0.7067

concentra (mol/L) 0.0068        0.0077              0.1926

K_{c} = \frac{0.1926^{2}}{0.0068^{2}*0.0077^{2} } = 13530482

7 0
2 years ago
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