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oksian1 [2.3K]
2 years ago
7

Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally

100m north What distance did he cover? What was his displacement
Also every Kilometer is = to .6 cm
Find Displacement and Distance
Physics
1 answer:
Bess [88]2 years ago
3 0

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0


east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero


distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m  = 3600m


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A counter attendant in a diner shoves a ketchup bottle with a mass 0.30 kg along a smooth, level lunch counter. The bottle leave
balu736 [363]

Answer:

1.176 N

Explanation:

m = mass of the bottle = 0.30 kg

v_{o} = initial speed of the bottle = 2.8 m/s

v = final speed of the bottle = 0 m/s

d = stopping distance traveled = 1.0 m

f = magnitude of frictional force acting on bottle

Using work-change in kinetic energy theorem

- f d = (0.5) m (v^{2} - v_{o}^{2} )\\- f (1) = (0.5) (0.30) (0^{2} - 2.8^{2} )\\-f = - 1.176 \\f = 1.176 N

direction :

frictional force acts in opposite direction of motion.

8 0
2 years ago
If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

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8 0
2 years ago
A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision
BigorU [14]

Complete Question

The  complete question is  shown on the first uploaded image  

Answer:

a

  f=  -75 \ cm =  - 0.75 \ m

b

  P  =  -1.33 \ diopters

Explanation:

From the question we are told that

    The  image distance is  d_i =  -75 cm

The value of the image is negative because it is on the same side with the corrective glasses

    The  object distance is  d_o =  \infty

The  reason object distance  is because the object father than it being picture by the eye

General focal length is mathematically represented as

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substituting values

             \frac{1}{f}  =  \frac{1}{-75}  -   \frac{1}{\infty}

=>         f=  -75 \ cm =  - 0.75 \ m

Generally the power of the corrective lens is  mathematically represented as

        P  =  \frac{1}{f}

substituting values

       P  =  \frac{1}{-0.75}

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erica [24]

Answer:

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Throughout the S-turns, on either side of the road, a progressively smaller half-circle is formed, and this turn does not stop until the road or reference line is crossed during the early part of the turn, the twists increase too quickly.

5 0
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Ierofanga [76]

1-125/500)x100=efficiency

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