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dangina [55]
3 years ago
10

an optician uses a plane mirror to help him. suppse a patient sits in a chair 2.5m away from him. He views the image of a chart

which faces the mirror and is located 30m behindd him. how far is the chart as viewed by his eyes​
Physics
1 answer:
viva [34]3 years ago
5 0

Answer:

I think 75 m

Explanation:

tell if it was correct

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the diagram below shows the situation described in the problem. the focal length of the lens is labeled f; the scale on the opti
Artist 52 [7]

Answer:

I have no clue

Explanation:

8 0
2 years ago
A driver averaged 64 mph and took 3½ hours to travel from st. Louis to chicago. Based on this, what is the distance between st.
sammy [17]

Average speed of the driver is given as

v = 64 mph

if he moved for total time t = 3.5 hours

so the distance between the tow is given as

d = v* t

d = 64 * 3.5

d = 224 miles

so the distance between St. Louis and Chicago is 224 miles

6 0
3 years ago
Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water
Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

v_2^2 = v_1^2 + 2gh

Where,

V_i = Velocity in each state

g= Gravity

h = Height

Our values are given as,

A_1 = 2.4*10^{-4} m^2

v_1 = 0.8 m/s

h = 0.11m

Replacing at the kinetic equation to find V_2 we have,

v_2 = \sqrt{v_1^2 + 2gh}

v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}

v_2= 1.67 m/s

Applying the concepts of continuity,

A_1v_1 = A_2v_2

We need to find A_2 then,

A_2= \frac{A_1v_1 }{v_2}

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

A_2= \frac{A_1v_1 }{v_2}

A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}

A_2= 1.14*10^{-4} m2

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 1.14*10^{-4} m2

8 0
3 years ago
Hooke's law describes a certain light spring of unstretched length 33.6 cm. when one end is attached to the top of a doorframe a
masha68 [24]
Missing question: "What is the spring's constant?"

Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
F=mg=(6.89 kg)(9.81 m/s^2)=67.6 N
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
\Delta x=43.2 cm-33.6 cm=9.6 cm=0.096 m
And by using Hook's law, we can find the constant of the spring:
k= \frac{F}{\Delta x}= \frac{67.6 N}{0.096 m}=704.2 N/m
4 0
3 years ago
Which of your groups should you NOT change anything for? (in other
ella [17]

Answer:

B

Explanation:

The control is something that is meant to not be changed, the control is a comparison of the experimental.

7 0
3 years ago
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