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3 years ago

Okay i'm totally stuck and nobody I know really gets it either, so i've turned to Yahoo for help :)

2 answers:

5 0

Here is the rule for see-saws here on Earth, and there is no reason
to expect that it doesn't work exactly the same anywhere else:

 (weight) x (distance from the pivot) on one side
is equal to
 (weight) x (distance from the pivot) on the other side.

That's why, when Dad and Tiny Tommy get on the see-saw, Dad sits
closer to the pivot and Tiny Tommy sits farther away from it.

 (Dad's weight) x (short length) = (Tiny Tommy's weight) x (longer length).

So now we come to the strange beings on the alien planet.
There are three choices right away that both work:

#1).
(400 N) in the middle-seat, facing (200 N) in the end-seat.

 (400) x (1) = (200) x (2)

#2).
(200 N) in the middle-seat, facing (100 N) in the end-seat.

 (200) x (1) = (100) x (2)

#3).

On one side: (300 N) in the end-seat (300) x (2) = 600

On the other side:
 (400 N) in the middle-seat (400) x (1) = 400
 and (100 N) in the end-seat (100) x (2) = 200
Total . . . . . . . . . . . . 600

These are the only ones to be identified at Harvard . . . . . . .
There may be many others but they haven't been discarvard.

ss7ja [257]3 years ago

5 0

400N and 100N on one side and 300N and 200N on the other side

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Suppose that two point charges, each with a charge of +3.00 Coulomb are separated by a distance of 3.00 meters. Determine the ma

Sveta_85 [38]

Answer:

$9\cdot 10^9 N$

Explanation:

The magnitude of the electrical force between the two point charges is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

$q_1 =q_2 = +3.0 C$ is the magnitude of each charge

r = 3.00 m is the separation between the two charges

Substituting the numbers into the formula, we find

$F=(9\cdot 10^9 Nm^2C^{-2})\frac{(+3 C)^2}{(3.00 m)^2}=9\cdot 10^9 N$

3 0

3 years ago

Match the vocabulary terms to their definitions.

Vilka [71]

See the attached picture:

8 0

3 years ago

Read 2 more answers

1. On this graph, what is the change in velocity between 5 seconds and 8 seconds?

alukav5142 [94]

Can I see the graph so I can help you

8 0

2 years ago

A laser of wavelength 720 nm illuminates a double slit where the separation between the slits is 0.22 mm. Fringes are seen on a

kumpel [21]

Answer:

The appropriate solution is "2.78 mm".

Explanation:

Given:

$\lambda = 720 \ nm$

or,

$= 720\times 10^{-9} \ m$

$D=0.85 \ m$

$d = 0.22 \ mm$

or,

$=0.22 \times 10^{-3} \ m$

As we know,

Fringe width is:

⇒ $\beta=\frac{\lambda D}{d}$

hence,

Separation between second and third bright fringes will be:

⇒ $\theta=\beta=\frac{\lambda D}{d}$

$=\frac{720\times 10^{-9}\times 0.85}{0.22\times 10^{-3}}$

$=2.78\times 10^{-3} \ m$

or,

$=2.78 \ mm$

8 0

2 years ago

A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p

Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

u = 0.2 m

u = 20 cm

So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

8 0

2 years ago