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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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A toy rocket moving vertically upward passes by a 2.0-m-high window whose sill is 8.0 m above the ground. The rocket takes 0.15

sveta [45]

Answer:

$v_i = 18.86 m/s$

Explanation:

As we know that the speed of the rocket is v1 and v2 at the bottom and top of the window

then we will have

$d = (\frac{v_1 + v_2}{2}) t$

$2 = (\frac{v_1 + v_2}{2})(0.15)$

$26.67 m/s = v_1 + v_2$

also we know that

$v_2 - v_1 = (-9.81)(t)$

$v_2 - v_1 = (-9.81)(0.15) = -1.47$

now we have

$v_2 = 12.6$

also we have

$v_1 = 14.1 m/s$

now if the sill of the window is at height 8 m from the ground then we have

$v_1^2 - v_i^2 = 2 a h$

$(14.1^2) - v_i^2 = 2(-9.81)(8)$

$v_i = 18.86 m/s$

8 0

3 years ago

A sound wave has a frequency of 295 Hz and travels the length of a football field, 91.4 m in 0.506 s. What is the period of the

dem82 [27]

Answer:1/295 seconds

Explanation:

Period=1/frequency

Period=1/295 seconds

7 0

2 years ago

How is it that even light precipitation can still cause the collection of large amounts of water.

Firdavs [7]

Answer:

the heat of the light.

Explanation:

no matter the light, there's always heat being produced from it. and heat makes liuqid rise

4 0

2 years ago

How is one second time defined in the si system

Sphinxa [80]

Answer:The place to go for the answer to such an easy question is the SI Brochure, the document which defines the SI and all its units.

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3 years ago

The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an

Readme [11.4K]

1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
$d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
substituting the coordinates of the two charges, we get
$d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m$

2) Then, we can calculate the electrostatic force between the two charges $q_1$ and $q_2$ , which is given by
$F=k_e \frac{q_1 q_2}{d^2}$
where $k_e=8.99\cdot10^{9} Nm^2C^{-2}$ is the Coulomb's constant.
Substituting numbers, we get
$F=8.99\cdot10^{9} Nm^2C^{-2} \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N$
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.

7 0

3 years ago