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professor190 [17]

4 years ago

14

The charges Q1=Q and Q2=4Q that are a distance d apart, repel each other with a force of 1.60 N. What would be the force between

the charges Q3=2Q and Q4=7Q that are a distance d/3 apart?

1 answer:

gladu [14]4 years ago

7 0

Answer:

50.4 N

Explanation:

Q1 = Q

Q2 = 4 Q

Distance = d

The force is given by

$F = \frac{KQ_{1}Q_{2}}{d^{2}}$

$1.60 = \frac{4KQ^{2}}{d^{2}}$ .... (1)

Now,

Q3 = 2 Q

Q4 = 7 Q

distance = d/3

$F' = \frac{9KQ_{3}Q_{4}}{d^{2}}$

$F' = \frac{126KQ^{2}}{d^{2}}$ .... (2)

Divide equation (2) by equation (1), we get

F' / 1.60 = 126 / 4

F' = 50.4 N

Thus, the force is 50.4 N.

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cluponka [151]

Answer:

25.2563 m/s

Explanation

This is the equation needed

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3 years ago

A constant-volume perfect gas thermometer indicates a pressure of 6.69 kPa at the triple point temperature of water (273.16 K).

mina [271]

Answer:

(a) ΔP=0.0245 kPa

(b) P=9.14 kPa

(c)ΔP=0.0245 kPa

Explanation:

(a) As it is perfect gas we can use

(P₁V₁)/T₁=(P₂V₂)/T₂

Since this constant volume so

P₁/T₁=P₂/T₂

T₂ is change in temperature

T₂=1.00+273.16

T₂=274.16 K

$P_{2}=(\frac{6.69}{273.16} )*274.16\\P_{2}=6.71449 kPa$

ΔP=6.71449-6.69

ΔP=0.0245 kPa

(b) As

$P_{2}=(\frac{6.69}{273.16} )*373.16\\P_{2}=9.14 kPa$

(c) Same steps as in part (a)

$P_{2}=(\frac{9.14}{373.16} )*374.16\\P_{2}=9.164kPa$

ΔP=9.164-9.14

ΔP=0.0245kPa

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3 years ago

Two velocity vectors, one is twice of the other, and separated by 90 Degree angle. If their resultant is calculated 90 m/s, what

Artyom0805 [142]

Answer:

Vy = 80.5 [m/s]

Explanation:

In order to solve this problem we must use the Pythagorean theorem.

V = 90 [m/s]

The components are Vx and Vy:

Therefore:

$v=\sqrt{v_{x}^{2} + v_{y}^{2} }$

where:

Vy = 2*Vx ; because one is twice of the other.

$90 = \sqrt{v_{x}^{2} +(2*v_{x})^{2} }\\ 90 =\sqrt{v_{x}^{2}+4*v_{x}^{2}} \\90 =\sqrt{5v_{x}^{2}} \\90=2.23*v_{x} \\v_{x}=40.25[m/s]$

and the bigger vector is:

Vy = 40.25*2

Vy = 80.5 [m/s]

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3 years ago

A ski jumper travels down a slope and leaves

Serhud [2]

Question seems to be missing. Found it on google:

a) How long is the ski jumper airborne?

b) Where does the ski jumper land on the incline?

a) 4.15 s

We start by noticing that:

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$v_x = 28 m/s$

and the distance covered along the horizontal direction in a time t is

$d_x = v_x t$

- The vertical motion of the skier is a uniformly accelerated motion, with initial velocity $u_y = 0$ and constant acceleration $g=9.8 m/s^2$ (where we take the downward direction as positive direction). Therefore, the vertical distance covered in a time t is

$d_y = \frac{1}{2}gt^2$

The time t at which the skier lands is the time at which the skier reaches the incline, whose slope is

$\theta = 36^{\circ}$ below the horizontal

This happens when:

$tan \theta = \frac{d_y}{d_x}$

Substituting and solving for t, we find:

$tan \theta = \frac{\frac{1}{2}gt^2}{v_x t}= \frac{gt}{2v_x}\\t = \frac{2v_x}{g}tan \theta = \frac{2(28)}{9.8} tan 36^{\circ} =4.15 s$

b) 143.6 m

Here we want to find the distance covered along the slope of the incline, so we need to find the horizontal and vertical components of the displacement first:

$d_x = v_x t = (28)(4.15)=116.2 m$

$d_y = \frac{1}{2}gt^2 = \frac{1}{2}(9.8)(4.15)^2=84.4 m$

The distance covered along the slope is just the magnitude of the resultant displacement, so we can use Pythagorean's theorem:

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