What is the focal length of the mirror? follow the sign conventions?

1 answer:

4 0

Well, the focal length is a distance to the point in which all of the rays of a parallel beam cross. Since a parallel beam reflected by the mirror is still parallel and does not converge, no matter how far from the mirror it travels, the focal distance of a mirror is

f = -∞

minus sign denotes that the image in the mirror is imaginary

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Which of the following is a true statement?

pishuonlain [190]

I think the answer is A.........

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2 years ago

A ball is rolled of a 2.14m table at 8m/s. Find the time of flight for the ball and the horizontal range

Mama L [17]

Answer:

0.661 s, 5.29 m

Explanation:

In the y direction:

Δy = 2.14 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(2.14 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.661 s

In the x direction:

v₀ = 8 m/s

a = 0 m/s²

t = 0.661 s

Find: Δx

Δx = v₀ t + ½ at²

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Δx = 5.29 m

Round as needed.

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2 years ago

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist

e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > $38 \frac{ft}{s^2} x \frac{1mi}{5280ft} x \frac{(3600s)^2}{(1h)^2} = 93272.27 \frac{mi}{h^2}$

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > $v_2^2=v_1^2 + 2as$

The initial velocity is 50mi/h $(v_1=50)$

When it stops the final velocity is $(v_2=0)$

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

$0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s = 2500/(185644.54)

s=0.0134
$

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > $0.0134 mi * \frac{5280ft}{1mi} = 70.8 ft$
So this means that the car traveled in feet 70.8 ft before it came to a stop.