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3 years ago

A car moves from a position 50 m south of the library to a new position 250

Physics

1 answer:

Charra [1.4K]3 years ago

6 0

Answer:

A. 20 m/s north

Explanation:

The velocity of the car can be calculated as:

$v=\frac{d}{t}$

where

d is the displacement

t is the time interval

Taking north as positive direction, the initial position of the car is

$r_i = -50 m$ (south)

while the final position is

$r_f = +250 m$ (north)

So, the displacement is

$d=r_f -r_i = +250-(-50)=+300 m$ (north)

The time interval for the motion is

t = 15 s

So, the velocity of the car is

$v=\frac{+300}{15}=+20 m/s$ (north)

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Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11

tiny-mole [99]

Answer:

The line charge density is $1.59\times10^{-4}\ C/m$

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

$V=EA$

$V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2$

$\lambda=\dfrac{V\times2\epsilon_{0}}{r}$

Where, r = radius

V = potential difference

Put the value into the formula

$\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}$

$\lambda=1.59\times10^{-4}\ C/m$

Hence, The line charge density is $1.59\times10^{-4}\ C/m$

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3 years ago

Of all the planets, which is the most different from the others? Explain.

Alenkinab [10]

Answer: VENUS

Explanation:

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3 years ago

Read 2 more answers

HELP ME PLEASE!!!!!!!!!

Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

$F = \frac{kq_1q_3}{d_{13}^2}$

now from the above equation

$F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}$

$F = 0.288 N$

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

$F_{net} = 2Fcos\theta$

here we know that angle is

$\theta = 37 degree$

now we have

$F_{net} = 2\times 0.288 cos37$

$F_{net} = 0.46 N$

so net force on q3 is 0.46 N vertically upwards along +Y axis

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3 years ago

if you have 34 pounds of meat at 200F and you immerse it in 15 liters of water at 25C what is the final temp of the water? ( spe

hram777 [196]

Answer:

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3 years ago

So u see what had happened was i need help again..

liubo4ka [24]

first off lemme just say this is really easy man, just look at the directions

Blank #1: -23

Blank #2: 23

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2 years ago