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3 years ago

10

Which image is observe by seeing a distant thing using magnifying glass?

2 answers:

Snezhnost [94]3 years ago

8 0

Answer: D) Big - straight

<h2>Brainliest pls if correct!</h2>

lesya692 [45]3 years ago

7 0

Answer:

1/f = 1/o + 1 /i thin lens equation

1 / i = 1/f - 1/o = (o - f) / (o f) rearranging

or i = o f / (o - f) for the image distance

This image will be positive (real) with values given

m = -i / o magnification

m = - f / (o - f) shows m is negative and the image inverted

Also, since o here is large m will be small and the image is small

I think they mean A as the right answer

Different texts use different terms but here i is image distance and o the object distance

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An airplane travels at 300 mi/h south for 2.00 h and then at 250 mi/h north for 750 miles. what is the average speed for the tri

iren2701 [21]

<span>Th find the average speed of a trip we need to dived the total distance by the total time. Let's find the total distance d. d = (300 mi/h)(2.00 h) + 750 miles d = 600 miles + 750 miles d = 1350 miles The total distance is 1350 miles Let's find the total time t. t = 2.00 hours + (750 mi / 250 mi/h) t = 2.00 hours + 3.00 hours t = 5.00 hours The total time of the trip is 5.00 hours. We can find the average speed. d / t = 1350 miles / 5.00 hours d / t = 270 miles/ hour The average speed of the trip is 270 mi/h (Note that the direction does not matter when we find the average speed.)</span>

6 0

4 years ago

To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circus performer wh

sp2606 [1]

Answer:

$v = 15.45 m/s$

Explanation:

As per mechanical energy conservation we can say that here since friction is present in the barrel so we will have

Work done by friction force = Loss in mechanical energy

so we will have

$W_f = (U_i + K_i) - (U_f + K_f)$

here we know that

$W_f = F_f . d$

$W_f = 40 \times 4$

$W_f = 160 J$

Initial compression in the spring is given as

$F = kx$

$4400 = 1100 x$

$x = 4 m$

now from above equation

$W_f = (\frac{1}{2}kx^2 + 0) - (mgh + \frac{1}{2}mv^2)$

$160 = (\frac{1}{2}1100(4^2) + 0) - (60 \times 9.8\times 2.50 + \frac{1}{2}(60)v^2)$

$160 = 8800 - 1470 - 30 v^2$

$v = 15.45 m/s$

3 0

3 years ago

During a hurricane in 2008, the Westin Hotel in downtownNew Orleans suffered damage. Suppose a piece of glass dropped near the t

tangare [24]

Answer:

<u>77.8 m/s, downward</u>

Explanation:

For uniform acceleration motion, the average speed is equal to half the soum of the initial velocity, Vi, and the final velocity, Vf

Average speed = (Vf + Vi)/2

Also, by definition, the average speed is the distance divided by the time:

Average speed = distance / time

Then:

(Vf + Vi)/2 = 300m/6.62s

Other kinematic equation for uniform acceleration is:

Vf = Vi + a×t

Since the window is falling and the air resistance is ignored, a = g (gravitational acceleration ≈ 9.8m/s²)

Replacing the known values we can set a system of two equations:

From (Vf + Vi)/2 = 300m/6.62s

(Vf + Vi) = 2 × 300m/6.62s

Vf + Vi = 90.634 equation 1

From Vf = Vi + a×t

Vf - Vi = 9.8 (6.62)

Vf - Vi = 64.876 equation 2

Adding the two equations:

2Vf = 155.510

Vf = 77.8 m/s downward (velocities must be reported with their directions)

8 0

3 years ago

A good description of magnets would be, "Magnets are

madreJ [45]

B-things that can attract iron.

5 0

3 years ago

A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b

jonny [76]

Answer:

Part a)

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

$P_o = 0.162 Pa$

Explanation:

Part a)

Level of sound = 75 dB

now we know that

$L = 10 Log\frac{I}{I_0}$

here we know that

$I_0 = 10^{-12} W/m^2$

now we have

$75 = 10 Log(\frac{I}{10^{-12}})$

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

Intensity of sound wave is given as

$I = \frac{1}{2}\rho A^2\omega^2 c$

here we know that

$A = \frac{P_o}{Bk}$

so we have

$I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c$

$I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}$

now we know

$\rho = 1.2 kg/m^3$

$c = 340 m/s$

$B = 1.4 \times 10^5 Pa$

now we have

$3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}$

$P_o = 0.162 Pa$

5 0

3 years ago