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algol [13]
3 years ago
6

A 50.0 N box sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at on

e instant the box is sliding to the right at 1.75 m/s and that is stops in 2.25 s with uniform acceleration. The force that friction exerts on this box is closest to:
Physics
1 answer:
Vikki [24]3 years ago
6 0

Answer:

-4.0 N

Explanation:

Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):

F_f = ma (1)

We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

m=\frac{W}{g}=\frac{50.0 N}{9.8 m/s^2}=5.1 kg

And we can fidn the acceleration by using the formula:

a=\frac{v-u}{t}

where

v = 0 is the final velocity

u = 1.75 m/s is the initial velocity

t = 2.25 s is the time the box needs to stop

Substituting, we find

a=\frac{0-1.75 m/s}{2.25 s}=-0.78 m/s^2

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)

Therefore, substituting into eq.(1) we find the force of friction:

F_f = (5.1 kg)(-0.78 m/s^2)=-4.0 N

Where the negative sign means the direction of the force is opposite to the motion of the box.

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Answer:

65.625 m/s

Explanation:

F=20\times 10^9 Hz

\Delta F=F_r-F_t=4750 Hz

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Relative speed

v=\frac{\Delta F}{2F}\times c\\\Rightarrow v=\frac{4750}{2\times 20\times 10^9}\times 3\times 10^8\\\Rightarrow v=35.625\ m/s

Relative speed = 35.625 m/s

Velocity of speeder = Velocity of police car + Relative speed

⇒Velocity of speeder = 30 + 35.625 = 65.625 m/s

∴ Velocity of speeder is 65.625 m/s

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3 years ago
When using the lens equation, a negative value as the solution for di indicates that the image is
nirvana33 [79]

Answer:

The Anatomy of a Lens

Refraction by Lenses

Image Formation Revisited

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Converging Lenses - Object-Image Relations

Diverging Lenses - Ray Diagrams

Diverging Lenses - Object-Image Relations

The Mathematics of Lenses

Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a lens. The use of these diagrams was demonstrated earlier in Lesson 5 for both converging and diverging lenses. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and image size. To obtain this type of numerical information, it is necessary to use the Lens Equation and the Magnification Equation. The lens equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f)

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The annoying sound from a mosquito is produced when it beats its wings at the average rate of 600. wingbeats per second. What is
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The frequency is exactly the rate at which the bug beats its wings.
If that "600" that you mentioned is 600 beats per second, then
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3 years ago
A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou
Sergeeva-Olga [200]

Answer:

The wavelength of the radiation absorbed by ozone is 319.83 nm

Explanation:

Given;

frequency of absorbed ultraviolet (UV) radiation, f = 9.38×10¹⁴ Hz

speed of the absorbed ultraviolet (UV) radiation, equals speed of light, v = 3 x 10⁸ m/s

wavelength of the absorbed ultraviolet (UV) radiation, λ = ?

Apply wave equation for speed, frequency and wavelength;

v = fλ

λ = v / f

λ = (3 x 10⁸) / (9.38×10¹⁴)

λ = 3.1983 x 10⁻⁷ m

λ = 319.83 x  10⁻⁹ m

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Therefore, the wavelength of the radiation absorbed by ozone is 319.83 nm

5 0
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What is light intensity? ​
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Hope this helps.

7 0
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