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marusya05 [52]
3 years ago
9

A playground slide is 8.80 ft long and makes an angle of 25.0° with the horizontal. A 63.0-kg child, initially at the top, slide

s all the way down to the bottom of the slide. (a) Choosing the bottom of the slide as the reference configuration, what is the system's potential energy when the child is at the top and at the bottom of the slide? What is the change in potential energy as the child slides from the top to the bottom of the slide? (Include the sign of the value in your answer.)
Physics
1 answer:
mario62 [17]3 years ago
7 0

Answer:

initial: 1654.6 J, final: 0 J, change: -1654.6 J

Explanation:

The length of the slide is

L = 8.80 ft = 2.68 m

So the height of the child when he is at the top of the slide is (with respect to the ground)

h = L sin \theta = (2.68 m)sin 25.0^{\circ}=1.13 m

The potential energy of the child at the top is given by:

U = mgh

where

m = 63.0 kg is the mass of the child

g = 9.8 m/s^2 is the acceleration due to gravity

h = 1.13 m

Substituting,

U=(63.0 kg)(9.8 m/s^2)(2.68 m)=1654.6 J

At the bottom instead, the height is zero:

h = 0

So the potential energy is also zero: U = 0 J.

This means that the change in potential energy as the child slides down is

\Delta U = 0 J - (1654.6 J) = -1654.6 J

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Natasha_Volkova [10]

Answer:

weight = mg \\  = 70 \times 10 \\  = 700 \: newtons

force = 700 \: newtons

force = mass \times acceleration \\ 700 = 70 \times a \\ a = 10 \:  {ms}^{ - 2}

4 0
3 years ago
Describe the electron transfers that occur in the formation of calcium fluoride from elemental calcium and elemental fluorine.
boyakko [2]

Answer:

Check explanation

Explanation:

In the formation of calcium fluoride we take calcium and fluorine.

in elemental form calcium exist in solid form and fluorine in gaseous form.

formation of compound takes place to complete their octet, in case of calcium  need to remove two electron and need to add one elecron in fluorine to complete their octet so two electron will ransferred from calcium to two fluorine atom.  

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3 years ago
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2 years ago
When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6
Stella [2.4K]

Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 = +8.4\mu C

q_2 = +5.6 \mu C

force between two charges is given as

F = 0.66 N

now we have

F = \frac{kq_1q_2}{r^2}

0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}

r = 0.8 m

so it is separated by 80 cm distance

3 0
3 years ago
Read 2 more answers
calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end
sdas [7]

Answer:0.8\ mm

Explanation:

Given

length of wire l=75\ cm

change in length \Delta l=1.85\ mm

mass of wire m=10\ kg

Young's modulus for silver E=7.9\times 10^{10}\ N/m^2

load on wire F=mg

F=10\times 9.8=98\ kg

change in length is given by

\Delta l=\dfrac{Pl}{AE}

Where A=area of cross-section

A=\dfrac{Pl}{\Delta lE}

A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}

A=\dfrac{73.5}{14.615\times 10^{7}}

A=5.029\times 10^{-7}\ m^2

also wire is the shape of cylinder so cross-section is given by

A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2

\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }

\Rightarrow d^2=64.02\times 10^{-8}

\Rightarrow d=8\times 10^{-4}\ m

\Rightarrow d=0.8\ mm

4 0
3 years ago
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