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ludmilkaskok [199]

2 years ago

10

Lint boy leaves the secret Fuzz Fortress at 10:30 in the morning. He uses his super static cling action to propel himself 2700 m

iles by 1:30 in the afternoon. How fast was he going?

1 answer:

Anon25 [30]2 years ago

4 0

I have no clue lol ok

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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

olga nikolaevna [1]

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

The circumference of any circle = <em>2 pi (radius)</em>
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 =

   <em> 73.5133 meter/min =</em>

   <em>4.411 km/hr =</em>

   <em>2.74 miles/hour

</em>(After the last appearance of pi,
all numbers are rounded.)<em>

</em>

8 0

3 years ago

The balance between incoming solar energy and out going energy radiated into space is called...

marin [14]

Based on the physics principle of conservation of energy, this radiation budget represents the accounting of the balance between incoming radiation, which is almost entirely solar radiation, and outgoing radiation, which is partly reflected solar radiation and partly radiation emitted from the Earth system, including the atmosphere.

5 0

3 years ago

radio signals emitted from and received by an airplane have her fruit frequency of 3.00 * 10 to the 12th Hertz and travel at the

Elanso [62]

here we know that the speed of the signal is same as the speed of light

so here we will have

$v = 3 \times 10^8 m/s$

the altitude of the airplane is given as

$h = 1 \times 10^4 m$

now we know that time taken by the signal to reach the control tower is given as

$t = \frac{d}{v}$

now it is given as

$t = \frac{1 \times 10^4 }{3 \times 10^8}$

$t = 3.33 \times 10^{-5} s$

so above is the time taken by the signal to reach the control tower

3 0

3 years ago

Two positive charges are labeled q Subscript 1 baseline and q Subscript 2 baseline are 0.1 m apart. Two positively charged parti

Snowcat [4.5K]

Answer:

Explanation:

Let that point be at a distance x from q1

Then Kq1/x^2= Kq2/ (s-x)^2

Taking square roots and simplifying, x =s /[1+(q2/q1)^0.5]

Assuming an identical distance, the rigidity of Q on 2Q is equivalent in value to the rigidity of 2Q on Q. for that reason, had the area R been stored an identical, the two forces could be equivalent. inspite of the shown fact that, via fact the area is being decreased, we could constantly consult with the equation we use to calculate those forces: F = ok(Q1xQ2)/(R^2) because R is squared and is being halved, the final result's that's it being divided by potential of a million/4. for that reason, the rigidity would be expanded by potential of four, and be 4F.

4 0

3 years ago

A 0.25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target. What is the final veloc

Mademuasel [1]

Answer:20/47 meter per second

Explanation:

Mass of arrow(ma)=0.25kg

Velocity of arrow(va)=12m/s

Mass of target(mt)=6.8kg

Velocity of target(vt)=0 since target is at rest

Conservation of linear momentum says that :

maxva+mtxvt=(ma+mt)V

V=(maxva+mtxvt)/(ma+mt)

V=(0.25x12+6.8x0)/(0.25+6.8)

V=3/(7.05)

V=20/47 meter per second

4 0

3 years ago