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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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An elevator is moving is an upwards
Minchanka [31]

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The elevator's free-body diagram has three forces, the force of gravity, a downward normal force from you, and an upward force from the tension in the cable holding the elevator. The combined system of you + elevator has two forces, a combined force of gravity and the tension in the cable.

Explanation:

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A force of 20N changes the position of a body. If mass of the body is 2kg, find the acceleration produced in the body.2. A ball
shepuryov [24]

Explanation:

<em>Hello</em><em> </em><em>there</em><em>!</em><em>!</em><em>!</em>

<em>You</em><em> </em><em>just</em><em> </em><em>need</em><em> </em><em>to</em><em> </em><em>use</em><em> </em><em>simple</em><em> </em><em>formula</em><em> </em><em>for</em><em> </em><em>force</em><em> </em><em>and</em><em> </em><em>momentum</em><em>, </em>

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<em>and</em><em> </em><em>momentum</em><em> </em><em>(</em><em>p</em><em>)</em><em>=</em><em> </em><em>m.v</em>

<em>where</em><em> </em><em>m</em><em>=</em><em> </em><em>mass</em>

<em>v</em><em>=</em><em> </em><em>velocity</em><em>.</em>

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5 0
3 years ago
In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor
leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}.....(I)

Where, F(x) = net force

We know, the net force is the sum of forces.

So, \sum{F}=ma

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

F(x)+(-F(f))=ma

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

We need to find the value of \int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

Using newton's second law

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}...(II)

We know that,

Acceleration is rate of change of velocity.

a=\dfrac{dv}{dt}

Put the value of a in equation (II)

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

Now, the work done by the net force on the block is,

W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

7 0
3 years ago
A ball is shot at an angle of 45 degrees into the air with initial velocity of 46 ft/sec. Assuming no air resistance, how high d
mestny [16]

Answer:

5.02 m

Explanation:

Applying the formula of maximum height of a projectile,

H = U²sin²Ф/2g...................... Equation 1

Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.

Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°

Constant: g = 9.8 m/s²

Substitute these values into equation 1

H = (14.021)²sin²45/(2×9.8)

H = 196.5884×0.5/19.6

H = 5.02 m.

Hence the ball goes 5.02 m high

8 0
3 years ago
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