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butalik [34]
4 years ago
5

What is the difference between virtual images produced by concave, plane and convex mirrors? What does the negative sign in the

value of magnification produced by a mirror indicates about an image?
Physics
1 answer:
vampirchik [111]4 years ago
4 0
Hey.. the virtual image formed by a plane mirror would be the same size. If formed by a convex mirror it would be diminished and if formed by a concave mirror it would be magnified.
the negative sign in the value of magnification indicates the type of mirror, that the mirror being used is convex, and if it is positive then the mirror being used is concave mirror. The negative sign denotes that the image projected is in the inverted direction.
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The speed of car is 106.67 miles per hour

To solve problems involving speed and distance we are required to first find speed or distance travelled in an hourly timeframe

It is given that car takes 4 hours to cover the distance at a speed of 40 miles per hour

Therefore, total distance travelled by the car= speed x time

                                                                          = 40 miles per hour x 4 hour

                                                                          =160 miles

We are required to find the speed of the car if the distance is to be covered in 1.5 hours

We are required to find the speed by the formula speed= Distance/time

distance= 160 miles     time = 1.5 hour

speed= 160/1.5=106.666=106.67 miles per hour

Hence the speed is 106.67 miles per hour

For further reference:

brainly.com/question/10930186?referrer=searchResults

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7 0
2 years ago
An astronaut floating alone in outer space throws a baseball. If the ball moves away at 20 m/s, the astronaut will:.
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Move the opposite direction ( at a lesser magnitude of velocity)

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7 0
2 years ago
An 8 Ω resistor is connected to a battery. The current that flows in the circuit is 2 A. Calculate the voltage of the battery.
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Here is the highly detailed, arcane, complex, technical form of Ohm's Law that is needed in order to answer this question  ===> V = I · R .

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4 years ago
The system below uses massless pulleys and ropes. The coefficient of friction is μ. Assume that M1 and M2 are sliding. Gravity i
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T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1

Block 2

T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2

Block 3

- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3

Solving Eq1,2,3 simultaneously

Divide 1 and 2

\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)}  \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} =  \frac{M_{1} *T_{2} }{M_{2} } .... Eq4

Put Eq 4 into Eq3

T_{2} = \frac{M_{3}*(g-a) }{1+\frac{M_{1} }{M_{2} } }  ...Eq5

Put Eq 5 into Eq2 and solve for a

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Substitute back in Eq2 and use Eq4 and solve for T2 & T1

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5 0
4 years ago
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