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3 years ago

Which of the following represents an upright image? A. -do B. +m C. -m D. +do

Physics

1 answer:

Tom [10]3 years ago

5 0

Answer:

B. +m

Explanation:

The magnification of an image is defined as the ratio between the size of the image and of the object:

$m = \frac{y'}{y}$

where we have

y' = size of the image

y = size of the object

There are two possible situations:

- When m is positive, y' has same sign as y: this means that the image image is upright

- When m is negative, y' has opposite sign to y: this means that the image is upside down

Therefore, the correct option representing an upright image is

B. +m

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IrinaVladis [17]

Answer:

13.18 m/s

Explanation:

Let the velocity of sports utility car is

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mc = 1200 kg, uc = 31.1 m/s

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Using conservation of momentum

mc × uc + ms × us = 0

1200 × 31.1 - 2830 × u = 0

u = 13.18 m/s

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3 years ago

The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid

faust18 [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

$w=2250*2\pi *\frac{1}{60}\\ w=235.61$

$I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}$

Now using Inertia an w

$E=\frac{1}{2}*I*(w)^{2} \\ E=\frac{1}{2}*0.416x10^{-3}*(235.61)^{2} \\E=11.54J$

average power= $\frac{11.4J}{0.230s}=50.2 W$

b).

power=t*w

P= $11.5465*0.25*235.61$

P=2.957 kW

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crimeas [40]

Answer:

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Explanation:

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zubka84 [21]

Answer:

vp = 0.94 m/s

Explanation

Formula

Vp = position/ time

position: Initial position - Final position

Position = 25 m - (-7 m) = 25 m + 7 m = 32 m

Then

Vp = 32 m / 34 seconds

Vp = 0.94 m/s

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