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vladimir1956 [14]

2 years ago

6

For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 m and sto

pped in a 1.2-ms-long encounter with a hard floor?

1 answer:

arsen [322]2 years ago

6 0

Answer:

Acceleration will be equal to $3689.16m/sec^2$

Explanation:

We have given initial height h = 1 m

Time instant t = 1.2 m-sec = 0.0012 sec

Initial velocity u = 0 m/sec

From third equation of motion we know that $v^2=u^2+2gh$

So $v^2=0^2+2\times 9.8\times 1$

v = 4.427 m/sec

We have to find the acceleration

Acceleration is time rate of change of velocity

So acceleration $a=\frac{v}{\Delta t}=\frac{4.427}{0.0012}=3689.16m/sec^2$

So acceleration will be equal to $3689.16m/sec^2$

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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

How does increasing the distance between charged objects affect the electric force between them? the electric force increases be

konstantin123 [22]

the electric force decreases because the distance has an indirect relationship to the force

Explanation:

The electric force between two objects is given by

$F=k \frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the distance between the two objects

As we can see from the formula, the magnitude of the force is inversely proportional to the square of the distance: so, when the distance between the object increases, the magnitude of the force decreases.

3 0

2 years ago

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The wattage of a bulb is 24 W when it is connected to a 12 V battery. Calculate its effective wattage if it operates on a 6 V ba

tresset_1 [31]

It would be 12W because: 6v is half of 12v so half of 24w would be 12w

3 0

2 years ago

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The question is in the picture

Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

$v_0-gt = 0$ .

where $v_0$ is the initial velocity.

Solving for $t$ we get

$t = \dfrac{v_0}{g}$

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time $t_0$ since it had reached the highest point, the ball has traveled downwards and the velocity $v_f$ it has gained is

$v_f = gt_0$ ,

and we are told that this is twice the initial velocity $v_0$ ; therefore,

$v_f = 2v_0 = gt_0$

which gives

$t_0 = \dfrac{2v_0}{g}.$

Thus, the total time taken to reach velocity $2v_0$ is

$t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}$

$t_{tot} = \dfrac{3v_0}{g}.$

This $t_{tot}$ , we are told, is 36 seconds; therefore,

$36= \dfrac{3v_0}{g},$

and solving for $v_0$ we get:

$v_0 = \dfrac{36g}{3}$

$v_0 = \dfrac{36s(10m/s^2)}{3}$

$\boxed{v_0 = 120m/s}$

which from the options given is choice e.

7 0

3 years ago

types of plate boundaries where two plates separate or move apart?? does anyone know please help ??????

Shtirlitz [24]

Erosion i belive it is called

7 0

3 years ago

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