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evablogger [386]

3 years ago

15

When you take your 1900-kg car out for a spin, you go around a corner of radius 53m with a speed of 13m/s. The coefficient of st

atic friction between the car and the road is 0.88. Assuming your car doesnt skid, what is the force exerted on it by static friction?

1 answer:

anastassius [24]3 years ago

3 0

Answer:

Ff = 6058.5N

Explanation:

The sum of forces is:

$Ff = m*a_c$

$Ff = m*V^2/R$

$Ff = 1900*13^2/53$

$Ff = 6058.5N$

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How to intercept a slope

Harlamova29_29 [7]

Answer:

The slope intercept form is probably the most frequently used way to express equation of a line. To be able to use slope intercept form, all that you need to be able to do is 1) find the slope of a line and 2) find the y-intercept of a line.

Explanation:

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3 years ago

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An object is placed to the left of a convex mirror, such that the object-to-image distance is 140 cm.

labwork [276]

Answer:

The focal length of the mirror is 52.5 cm.

Explanation:

Given that,

Object to Image distance d = 140 cm

Image distance v= 35 cm

We need to calculate the object distance

$u = d-v$

$u = 140-35=105\ cm$

We need to calculate the focal length

Using formula of mirror

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

Put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-105}+\dfrac{1}{35}$

$\dfrac{1}{f}=\dfrac{2}{105}$

$f=52.5\ cm$

Hence, The focal length of the mirror is 52.5 cm.

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3 years ago

A body builder exerts 15 N of force over 3 m. How much work did she do?

tangare [24]

Work done = force x distance

work done = 15 x 3

work done = 45J

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3 years ago

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How does the force the Earth exerts on you compare with the force you exert on it?

Pie

Answer:

C.Earth and you exert equal and opposite forces on each other.

Explanation:

Hello, I can help you with this

The force that the earth exerts on any object is due to the force of gravity and is known as weight, F = mg, this force is directed towards the center of the earth, in addition, according to the laws of physics for any action must have a reaction, in this case it is a force that goes out from the ground and it is directed upwards, this force is called normal and it has the same magnitude of your weight but the opposite direction.

Have a great day.

6 0

3 years ago

A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t

Arisa [49]

A) $2.4\cdot 10^{-16}kg$

The radius of the oil droplet is half of its diameter:

$r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m$

Assuming the droplet is spherical, its volume is given by

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3$

The density of the droplet is

$\rho=885 kg/m^3$

Therefore, the mass of the droplet is equal to the product between volume and density:

$m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg$

B) $1.5\cdot 10^{-18}C$

The potential difference across the electrodes is

$V=17.8 V$

and the distance between the plates is

$d=11 mm=0.011 m$

So the electric field between the electrodes is

$E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m$

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

$qE=mg$

So, from this equation, we can find the charge of the droplet:

$q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C$

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

$q=-1.5\cdot 10^{-18}C$

we can think this charge has made of N excess electrons, so the net charge is given by

$q=Ne$

where

$e=-1.6\cdot 10^{-19}C$ is the charge of each electron

Re-arranging the equation for N, we find:

$N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9$

so, a surplus of 9 electrons.

3 0

3 years ago