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beks73 [17]
2 years ago
13

Light rays from an object after passing through a convex lens form an image at the focal point behind the lens (opposite side of

the object). What is the actual position of the object?. . . A. at the focal point in front of the lens B. at twice the focal length . C. between the focal point and twice the focal length . D. at a great distance, effectively infinite . E. between the focal point and the lens
Physics
1 answer:
Delvig [45]2 years ago
7 0
The actual position of the object is <span>at a great distance, effectively infinite. The other options given in the question are not at all correct. The correct option among all the options that are given in the question is the last option or option "D". I hope that this answer has actually come to your great help.</span>
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A 65 Kg skier starts at rest at the top of a 150 m long hill that has an incline of 28 degrees. How fast will she be going at th
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Answer:

34 m/s

Explanation:

Potential energy at top = kinetic energy at bottom + work done by friction

PE = KE + W

mgh = ½ mv² + Fd

mg (d sin θ) = ½ mv² + Fd

Solving for v:

½ mv² = mg (d sin θ) − Fd

mv² = 2mg (d sin θ) − 2Fd

v² = 2g (d sin θ) − 2Fd/m

v = √(2g (d sin θ) − 2Fd/m)

Given g = 9.8 m/s², d = 150 m, θ = 28°, F = 50 N, and m = 65 kg:

v = √(2 (9.8 m/s²) (150 m sin 28°) − 2 (50 N) (150 m) / (65 kg))

v = 33.9 m/s

Rounded to two significant figures, her velocity at the bottom of the hill is 34 m/s.

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How does your body’s flexibility naturally change as you age?
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A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw
8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

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2 years ago
Find the pressure exerted by a 3000 N crate that has an area of 2m squared
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Pressure = Force/ Area = 3000/2 = 1500 pascal.
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