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Lyrx [107]
1 year ago
5

A block of ice is sliding down a ramp of slope 40 to the horizontal. What is the rate of acceleration of the block? Assume the f

orce of friction is not significant.

Physics
1 answer:
uysha [10]1 year ago
8 0

The given problem can be exemplified in the following diagram:

Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:

\Sigma F=ma

Replacing the values:

mg\sin 40=ma

We may cancel out the mass:

g\sin 40=a

Using the gravity constant as 9.8 meters per square second:

9.8\frac{m}{s^2}\sin 40=a

Solving the operations:

6.3\frac{m}{s^2}=a

Therefore, the acceleration is 6.3 meters per square second.

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A convex lens of focal length 35 cm produces a magnified image 2.5 times the size of the object. What is the object distance if
Zanzabum

Answer:

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

Explanation:

u = Object distance

v = Image distance

f = Focal length = 35

m = Magnification = 2.5

m=-\frac{v}{u}\\\Rightarrow 2.5=-\frac{v}{u}\\\Rightarrow v=-2.5 u

Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{35}=\frac{1}{u}+\frac{1}{-2.5u}\\\Rightarrow \frac{1}{35}=\frac{3}{5u}\\\Rightarrow u=21\ cm

v=-2.5\times 21=-52.5\ cm

Image distance is -52.5 cm

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3 years ago
A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal
slamgirl [31]

Answer:

18.89cm

Explanation:

As we know that the person is standing 5m in front of the camera

d_0=5m=500cm

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By Lens formula we have:

\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm

By the formula of magnification

\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm

The height of the image formed is 18.89cm.

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