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3 years ago

A 31.0 kg child on a swing reaches a maximum height of 1.92 m above their rest position.

Physics

1 answer:

ValentinkaMS [17]3 years ago

6 0

<span>c) Assuming this maximum height was the result of one push from her parent, what was the </span>

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What describes how the spring constant affects the potential energy of an object for a given displacement from an equilibrium po

Anit [1.1K]

Answer:

The higher the spring constant, the greater the elastic potential energy.

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3 years ago

A charge +3q is placed at x = 0 and a charge +11q is placed at x = 5 units. Where, along the x-axis is the net force on a charge

RideAnS [48]

At x = 1.72 units the net force on a charge Q equal to zero when a charge +3q is placed at x = 0 and a charge +11q is placed at x = 5 units.

A net force is defined as the sum of all the forces acting on an object. The equation below is the sum of N forces acting on an object. There may be several forces acting on an object, and when you add up all of those forces, the result is what we call the net force acting on the object.

This equation is the sum of n forces acting on an object. The magnitude of the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object, as shown in this formula.

+3q ⇔ +11q

x = ? ⇔ x = 5

$\frac{1}{4\pi E0} (\frac{3q}{x^{2} } -\frac{11q}{(5-x)^{2} } ) = 0$

$\frac{3}{x^{2} } = \frac{11q}{(5-x)^{2} }$

$\frac{5-x}{x} = \sqrt{\frac{11}{3} } = 1.91\\$

$5-x = 1.91x\\x = 1.715 = 1.72 units$

∴ At x = 1.72 units the net force on a charge Q equal to zero

Learn more about net force here

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5 0

1 year ago

What are the characteristics of a gas ?

LenKa [72]

Answer:

c no definite shape and volume

4 0

3 years ago

Read 2 more answers

a golfer tees off and hits a golf ball at a speed of 31 m/s and an angle of 35 degrees. how far did the ball travel before hitti

poizon [28]

Ans: R = Ball Travelled = 92.15 meters.

Explanation:
First we need to derive that formula for the "range" in order to know how far the ball traveled before hitting the ground.

Along x-axis, equation would be:
$x = x_o + v_o_xt + \frac{at^2}{2}$

Since there is no acceleration along x-direction; therefore,
$x = x_o + v_o_xt$

Since $v_o_x = v_ocos \alpha$ and $x_o$ =0; therefore above equation becomes,

$x = v_ocos \alpha t$ --- (A)

Now we need to find "t", and the time is not given. In order to do so, we shall use the y-direction motion equation. Before hitting the ground y ≈ 0 and a = -g; therefore,

=> $y = y_o + v_o_yt - \frac{gt^2}{2}$
=> $t = \frac{2v_o_y}{g}$

Since $v_o_y = sin \alpha$ ; therefore above equation becomes,
$t = \frac{2v_osin \alpha }{g}$

Put the value of t in equation (A):

(A) => $x = v_ocos \alpha \frac{2v_osin \alpha }{g}$

Where x = Range = R, and $2sin \alpha cos \alpha = sin(2 \alpha )$ ; therefore above equation becomes:

=> $R = (v_o)^2 *\frac{sin(2 \alpha )}{g}$

Now, as:
$v_o = 31 m/s$

and $\alpha = 35$ °
and g = 9.8 m/(s^2)

Hence,
$R = (31)^2 *\frac{sin(2 *35 )}{9.8}$

Ans: R = 92.15 meters.

-i

7 0

3 years ago

A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 65.5-gram mass is attached at the 21.0-cm

Gnesinka [82]

Answer:

113.53 g

Explanation:

Please see attached photo for explanation.

In the attached photo, M is the mass of the meter stick.

The value of M can be obtained as shown below:

Clockwise moment = M × 10.5

Anticlockwise moment = 65.5 × 18.2

Anticlockwise moment = Clockwise moment

65.5 × 18.2 = M × 10.5

1192.1 = M × 10.5

Divide both side by 10.5

M = 1192.1 / 10.5

M = 113.53 g

Thus, the mass of the meter stick is 113.53 g

7 0

2 years ago