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3 years ago

10

How fast is an object going if it travels from San Diego to Anaheim in 1.25 hours (hr)? The distance from San Diego to Anaheim i

s 93 miles (mi).
a) 74.4 mi/hr

b) 116.25 mi/hr

c) 0.013 mi/hr

d) 84.25 mi/hr

e) None of the answers

1 answer:

Y_Kistochka [10]3 years ago

4 0

Answer:

74.4 mph

Explanation:

Since we have the distance in miles and the time it took in hours, we can divide the two to get miles per hour.

speed = mi/hr

speed = 93/1.25

speed = 74.4 mph

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A car travels at a constant velocity of 70 mph for one hour. At the second hour, the car’s velocity was 60 mph. At the third hou

katovenus [111]

When the velocity increases, then the acceleration will be positive and when the velocity decrease then the acceleration will be negative.

During the first hour, the velocity was 70 mph and during the seconds hour the velocity was 60 mph. Hence, the velocity decrease in the seconds hour. So, the acceleration will be negative during the second hour.

Now, during the third hour the velocity increases as it is 80 mph. Hence, the acceleration will be positive during the third hour.

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The current theory of the structure of the

IRISSAK [1]

1) The mass of the continent is $3.3\cdot 10^{21} kg$

2) The kinetic energy of the continent is 624 J

3) The speed of the jogger must be 4 m/s

Explanation:

1)

We start by finding the volume of the continent. We have:

$L = 5850 km = 5.85\cdot 10^6 m$ is the side

$t = 35 km = 3.5\cdot 10^4 m$ is the depth

So the volume is

$V=L^2 t = (5.85\cdot 10^6)^2 (3.5\cdot 10^4)=1.20\cdot 10^{18} m^3$

We also know that its density is

$d=2750 kg/m^3$

Therefore, we can find the mass by multiplying volume by density:

$m=dV=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg$

2)

The kinetic energy of the continent is given by:

$K=\frac{1}{2}mv^2$

where

$m=3.3\cdot 10^{21} kg$ is its mass

v = 3.2 cm/year is the speed

We have to convert the speed into m/s. We have:

3.2 cm = 0.032 m

$1 year = 1(365)(24)(60)(60)=3.15\cdot 10^7 s$

So, the speed is:

$v=\frac{0.032 m}{3.15 \cdot 10^7 s}=1.02\cdot 10^{-9} m/s$

So, we can now find the kinetic energy:

$K=\frac{1}{2}(1.20\cdot 10^{21})(1.02\cdot 10^{-9})^2=624 J$

3)

Here we have a jogger of mass

m = 78 kg

And the jogger has the same kinetic energy of the continent, so

K = 624 J

The kinetic energy of the jogger is given by

$K=\frac{1}{2}mv^2$

where v is the speed of the jogger.

Solving for v, we find the speed that the jogger must have:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(624)}{78}}=4 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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3 years ago

Lots of points and brainiest for first correct answer

sineoko [7]

That's the answer: I've attached the pic

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Consider a steel guitar string of initial length L=1.00L=1.00 meter and cross-sectional area A=0.500A=0.500 square millimeters.

erma4kov [3.2K]

Answer:

The extent to which it would stretch is $\Delta L = 0.015 \ m$

Explanation:

From the question we are told that

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The area is $A = 0.500 mm^2 = \frac{0.500}{1 *10^6} = 0.500*10^6 \ m^2$

The Young modulus of the steel is $Y = 2.0*10^{11} Pa$

The tension is $T =1500 N$

The Young modulus is mathematically represented as

$Y = \frac{\sigma}{e}$

Where $\sigma$ is the stress which is mathematically represented as

$\sigma = \frac{F}{A}$

Substituting values

$\sigma = \frac{1500}{0.500*10^{-6}}$

$\sigma = 3.0*10^9 N/m^2$

And e is the strain which is mathematically represented as

$e = \frac{\Delta L}{L }$

Where $\Delta L$ The extension of the steel string

Substituting these into the equation above

$Y = \frac{3.0*10^9}{\frac{\Delta L}{L} }$

Substituting values

$2.0 *10^{11} = \frac{3.0*10^9}{\frac{\Delta L}{L} }$

$\Delta L = \frac{3.0*10^9 * 1}{2.0 *10^{11}}$

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Wave An oscillation that transfers energy and momentum.

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In a transverse wave, perpendicular to the direction the wave travels, the particles are displaced. Examples of transverse waves include on a string vibrations and on the water surface ripples. By moving the slinky up and down vertically, we can create a horizontal transverse wave.

In a longitudinal wave, parallel to the direction the wave travels, the particles are displaced. Compressions that move along a slinky are an example of longitudinal waves. By pushing and pulling the slinky horizontally, we can make a horizontal longitudinal wave.

Common mistakes and misconceptions

Sometimes people forget that wave velocity is not the same as the velocity of the medium particles. How fast the disturbance travels through a medium is the wave speed. The velocity of the particle is how fast a particle moves about its position of equilibrium.

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