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3 years ago

a 1 gram spiders sits on a platform rotating at 78 rpm. the spider is 15 cm from the centre disk. find the speed of the spider

1 answer:

8 0

The spider is traveling in a circle with radius = 15cm

The circumference of any circle = 2 pi (radius)
The circumference of the spider's path = 2 pi (15 cm) = 30 pi cm

The spider completes a trip around this path 78 times per minute.
Its speed, relative to you, is

 (78) x (30 pi) cm/min =

 2,340 pi cm/min = 7,351.33 cm/min =

  73.5133 meter/min =

 4.411 km/hr =

 2.74 miles/hour

(After the last appearance of pi,
all numbers are rounded.)

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Name two technologies that people in developing nations could benefit from.

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Water filters and internet to get more knowledge

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2 years ago

The gnaphosid spider Drassodes cupreus has evolved a pair of lensless eyes for detecting polarized light. Each eye is sensitive

BartSMP [9]

Answer:

Part A

The intensity is $I = 618 W/m^2$

Part B

The intensity is $I_1= 81.884 W/m^2$

Explanation:

From the question we are told that

The intensity of the light detected by first eye is $I = 700 W/m^2$

Now at initial state according the question the light ray is perpendicular to the eye so it means that it is at 90° the eye

Now the first question is to obtain the intensity the first eye (the first in this case is the one focused on the light )would detect when the head is rotated by 20° its previous orientation

This is mathematically evaluated as

$I = I_i cos^2 ( 20^o)$

$I = 700\ cos^2 (20)$

$I = 618 W/m^2$

Now the second question is to obtain the intensity the first eye (the first eye in this case is the one that is not focused on the light )would detect when the head is rotated by 20° its previous orientation

Now in this case the angle between the eye and the light is 90-20 = 70°

$I_1 = 700 \ cos^2 (70)$

$I_1= 81.884 W/m^2$

5 0

2 years ago

a container of water is knocked off a 10.0 meter high ledge with a horizontal velocity of 1.00 meters/second. calculate the time

Evgen [1.6K]

Answer:

1.43 s

Explanation:

The time it takes for the container to reach the ground is determined only by the vertical motion of the container, which is a free-fall motion, so a uniformly accelerated motion with a constant acceleration of g=9.8 m/s^2 towards the ground.

The vertical distance covered by an object in free fall is given by

$S=ut + \frac{1}{2}at^2$

where

u = 0 is the initial vertical speed

t is the time

a= g = 9.8 m/s^2 is the acceleration

since u=0, it can be rewritten as

$S=\frac{1}{2}gt^2$

And substituting S=10.0 m, we can solve for t, to find the duration of the fall:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(10.0 m)}{9.8 m/s^2}}=1.43 s$

3 0

3 years ago

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.550 m/s. The to

Arisa [49]

Answer: 0.5 m/s

Explanation:

Given

Speed of the sled, v = 0.55 m/s

Total mass, m = 96.5 kg

Mass of the rock, m1 = 0.3 kg

Speed of the rock, v1 = 17.5 m/s

To solve this, we would use the law of conservation of momentum

Momentum before throwing the rock: m*V = 96.5 kg * 0.550 m/s = 53.08 Ns

When the man throws the rock forward

rock:

m1 = 0.300 kg

V1 = 17.5 m/s, in the same direction of the sled with the man

m2 = 96.5 kg - 0.300 kg = 96.2 kg

v2 = ?

Law of conservation of momentum states that the momentum is equal before and after the throw.

momentum before throw = momentum after throw

53.08 = 0.300 * 17.5 + 96.2 * v2

53.08 = 5.25 + 96.2 * v2

v2 = [53.08 - 5.25 ] / 96.2

v2 = 47.83 / 96.2

v2 = 0.497 ~= 0.50 m/s

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3 years ago

Why does a skier wear polarized glasses instead of glasses that are used for reading?

Afina-wow [57]

A skier wears polarized glasses instead of glasses that are used for reading because polarized glasses decrease reflected glare compared to regular glasses. (D)

Hope this answers your question correctly.

8 0

2 years ago

Read 2 more answers