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3 years ago

9

a car with a mass of 2,000 kilograms is moving around a circular curve at a uniform velocity of 25 meters per second. the curve

has a radius of 80 meters what is the centripetal force on a car

1 answer:

Ber [7]3 years ago

4 0

Fc=mv^2/r so we get

2000kg*(25m/s)^2/(80m)= 15625N of force

hope this helps! Thank You!!

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A 53.0 kg sled is sliding on snow with μk=0.110. how much friction force does it feel?

Anon25 [30]

Answer:

57N

Explanation:

$F_F = \mu F_N = \mu mg = 0.11 \times 53 kg \times 9,8 \frac{m}{s^{2} }$

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3 years ago

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Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t

n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

$y=\frac{\lambda (m+\frac{1}{2})D}{d}$

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light used

In this problem we have:

$\lambda=683 nm = 683\cdot 10^{-9} m$ is the wavelength of the light

$d=1.1 mm = 0.0011 m$ is the width of the slit

m = 13 is the order of the minimum

$y=8.57 cm = 0.0857 m$ is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

$D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m$

6 0

3 years ago

The speed of a moving bullet can be deter-

Bad White [126]

Answer:

<em>v = 381 m/s</em>

Explanation:

<u>Linear Speed</u>

The linear speed of the bullet is calculated by the formula:

$\displaystyle v=\frac{x}{t}$

Where:

x = Distance traveled

t = Time needed to travel x

We are given the distance the bullet travels x=61 cm = 0.61 m. We need to determine the time the bullet took to make the holes between the two disks.

The formula for the angular speed of a rotating object is:

$\displaystyle \omega=\frac{\theta}{t}$

Where θ is the angular displacement and t is the time. Solving for t:

$\displaystyle t=\frac{\theta}{\omega}$

The angular displacement is θ=14°. Converting to radians:

$\theta=14*\pi/180=0.2443\ rad$

The angular speed is w=1436 rev/min. Converting to rad/s:

$\omega = 1436*2\pi/60=150.3776\ rad/s$

Thus the time is:

$\displaystyle t=\frac{0.2443\ rad}{150.3776\ rad/s}$

t = 0.0016 s

Thus the speed of the bullet is:

$\displaystyle v=\frac{0.61}{0.0016}$

v = 381 m/s

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3 years ago

Does the densoty of an object change if you add more of it because i did an expirement that proved that the density does change

Marysya12 [62]

No the density does not change. Density is a ratio D=m/v no matter how much of a substance you have its mass will be proportional.
When doing density labs sometimes you might get different answers due to errors that are unavoidable.

8 0

3 years ago

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The objective lens and the eyepiece of a telescope are spaced 85 cm apart. If the eyepiece is123 D what is the total magnificati

dedylja [7]

Answer:

The magnification would be "103.55". A further explanation is given below.

Explanation:

The given values are:

Distance between lens and eyepiece,

L = 85 cm

Eyepiece is,

= 123 D

Now,

The refractive power of eye piece will be:

⇒ $\frac{1}{f_e}=123D$

$f_e=\frac{1}{123D}$

$f_e=0.813 \ cm$

The length of the telescope will be:

⇒ $L=f_0+f_e$

⇒ $f_0=L-f_e$

On substituting the values, we get

⇒ $=85-0.813$

⇒ $=84.187 \ cm$

Now,

The magnification of the telescope will be:

⇒ $M=\frac{f_0}{f_e}$

⇒ $=\frac{84.187}{0.813}$

⇒ $=103.55$

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2 years ago