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posledela
3 years ago
13

Two children sit on different sides of a seesaw. The first child of mass 27 kg sits 1.5 m from the center. How far must the seco

nd child of mass 23 kg sit from the center on the opposite side in order to balance the seesaw
Physics
1 answer:
erma4kov [3.2K]3 years ago
5 0

Answer:

dhdhhdshdhdhsjsnsjdnnsjsjshsjsndnsnsnsnzsnsnsnsjzjsjsjsjj

Explanation:

jsjszhxhbssbzbbzbdfnndsndnndndnsndnfndndnsmsnsxnbdbsndndndnxbcbwnnbfndbdxbbdbvh go do some gehehhsdhehndnsbxhdbdj

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The ability to reproduce is essential for a species, but not for an individual. There are some people that can't reproduce, but they're still alive!
6 0
2 years ago
A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

3 0
3 years ago
In a pressure cooker, a lid is screwed tightly onto a pot so that the air pressure in the pot is increased. What effect, if any,
Andreyy89
It will increase the temp
8 0
3 years ago
Read 2 more answers
You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experimen
Nikolay [14]

Answer:

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center

Explanation:

Let  n₁ and n₂ be no of lines per unit length  of grating A and B respectively.

λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,

Distance of first maxima for green light

= λ₁ D/ d₁

Distance of first maxima for red light

= λ₂ D/ d₂

Given that

λ₁ D/ d₁ = λ₂ D/ d₂

λ₁ / d₁ = λ₂ / d₂

λ₁ / λ₂  = d₁ / d₂

But

λ₁  <  λ₂

d₁ < d₂

Therefore no of lines per unit length of grating A will be more because

no of lines per unit length  ∝ 1 / d

If grating B is illuminated with green light first maxima will be at distance

λ₁ D/ d₂

As λ₁ < λ₂

λ₁ D/ d₂ < λ₂ D/ d₂

λ₁ D/ d₂ < 1 m

In this case position of first maxima will be less than 1 meter.

Option a is correct .

5 0
3 years ago
If C is the vector sum of A and B, C = A + B, What must be true if C = A + B?. What must be true if C = 0?.
dybincka [34]
A and B are equal vectors
7 0
3 years ago
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