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2 years ago

An oil droplet is sprayed into a uniform electric field of adjustable magnitude. The 0.11 g droplet hovers

Physics

1 answer:

ohaa [14]2 years ago

7 0

Answer:

The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.

F = N e E where E is the value of the field and N e the charge Q

M g = N e E and M g is the weight of the drop

N = M g / (e E)

N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13

.00011 kg is a very large drop

Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs

Check: N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13 electrons

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Hey how's your day going

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3 years ago

. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

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Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

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3 years ago

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Answer:

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3 years ago

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Answer:

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3 years ago

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Pavel [41]

The distance of the canoeist from the dock is equal to length of the canoe, L.

<h3>Conservation of linear momentum</h3>

The principle of conservation of linear momentum states that the total momentum of an isolated system is always conserved.

v(m₁ + m₂) = m₁v₁ + m₂v₂

where;

v is the velocity of the canoeist and the canoe when they are together

u₁ is the velocity of the canoe
u₂ velocity of the canoeist
m₁ mass of the canoe
m₂ mass of the canoeist

<h3>Distance traveled by the canoeist</h3>

The distance traveled by the canoeist from the back of the canoe to the front of the canoe is equal to the length of the canoe.

Thus, the distance of the canoeist from the dock is equal to length of the canoe, L.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

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