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Rudiy27
2 years ago
12

A toy car that 0.12 m long is used to model the actions of an actual car that is 6m long. Which ratio shows the relationship bet

ween the sizes of the model and the actual car?
Physics
1 answer:
ra1l [238]2 years ago
3 0

Answer:

We know that a 0.12m long toy car is used as a model of a 6m long car.

We want to find the ratio that shows the relationship between both cars (the toy one and the real one)

The ratio that shows the relationship between both measures is just the quotient between both lengths, for example, if we take:

0.12m/6m = 0.02

This means that the length of the toy car is 0.02 times the length of the long car.

We also could have:

6m/0.12m = 50

This means that the length of the real car is 50 times the length of the toy car.

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The fact I can get is22
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3 years ago
Explain why it is esaier to climb a mountain on a zigzag path rather than one straight up the side. is your increase in gravitat
pogonyaev

Answer:

More work done with less power

The increase in gravitational energy is the same as the height which is a function of gravitational energy is the same in both cases

Explanation:

Climbing the mountain in zigzag pattern is easier because

1. The time it takes to climb increases so that the required power or rate of doing work decreases

2. Climbing in zigzag pattern affords the use of leverages by the sides

3. Similar mechanical power gain and efficiency from using a drive screw instead of a nail to fasten items together can be achieved

The increase in gravitational energy is the same gravitational energy ~ mass × gravity ×height

6 0
3 years ago
What is the highest pHoney bees beat their wings, making a buzzing sound at a frequency of 2.3 × 102 hertz. What is the period o
VMariaS [17]

The period of the wave is 4.35 ms. The sound waves are called longitudinal waves

Explanation:

The period of a wave is related to its frequency by the equation:

T=\frac{1}{f}

where

T is the period

f is the frequency

For the bee in this problem, the frequency of the sound wave emitted by it is

f=2.3 \cdot 10^2 Hz = 230 Hz

Therefore, the period of the sound wave is

T=\frac{1}{230}=4.35\cdot 10^{-3}s = 4.35 ms

The sound wave is a type of wave called longitudinal wave. In longitudinal waves, the oscillation of the medium occurs in a direction parallel to the direction of motion of the wave: therefore in a sound wave, the particle of the medium (air, in this case) oscillate back and forth along the direction of propagation of the wave, forming alternating areas of higher density of particles (called compressions) and of lower density of particle (called rarefactions).

The other type of wave, instead, is called transverse wave. In a transverse wave, the oscillation of the wave occurs in a direction perpendicular to the direction of motion of the wave. An example of transverse waves are the electromagnetic waves, which consists of electric field and magnetic fields that vibrate in a plane perpendicular to the direction of motion of the wave itself.

Learn more about waves:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

5 0
3 years ago
A coin is placed on a large disk which rotates uniformly at a rate of 1 rot/s. The coefficient of friction between the coin and
s2008m [1.1K]

Answer:

r = 0.02 m

Explanation:

from the question we have :

speed = 1 rps = 1x 60 = 60 rpm

coefficient of friction (μ) = 0.1

acceleration due to gravity (g) = 9.8 m/s^{2}

maximum distance without falling off (r) = ?

to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force

mv^2 / r =  m x g x μ

v^2 / r =  g x μ  .......equation 1

where

velocity (v) = angular speed (rads/seconds) x radius

angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm

angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds

now

velocity = 6.28 x r = 6.28 r

now substituting the value of velocity into equation 1

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(6.28r)^2 / r = 9.8 x 0.1

39.5 x r = 0.98

r = 0.02 m

6 0
2 years ago
suppose that you look into a photometer's eyepiece and the fluorescent disks appear to be equal in intensity. If the distance be
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L1/L2=(D2*D2)/(D1*D1)

L1/15=(200*200)/(400*400)
L1=15*0.25
L1=3.75 <span>candela</span>

7 0
3 years ago
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