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slamgirl [31]
3 years ago
15

PLEASE HELP.You hit a hockey puck across an empty hockey rink. At first the puck moves quickly, but then it slows and comes to a

stop. What caused the hockey puck to stop?
Physics
2 answers:
saveliy_v [14]3 years ago
7 0
Hi there!

Two possible answers are air resistance and friction.

Friction is caused by the rubbing of the surface of the ground and the surface of the object. Although ice doesn't have much friction, it can still cause friction.

Air resistance is caused by friction between the air and the object. As the object moves along a surface, it collides into many air particles; thus, it slows down.

Hope this helps.
Have an awesome day! :)
Greeley [361]3 years ago
6 0
As far as it goes the slower it goes cause it is heavy and when its slowing down the ice makes it stop.
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On what principle does a bottle opener work
Genrish500 [490]

Answer:

Bottle opener works on a fulcrum.

Explanation:

A bottle opener is a second-class lever because the pivot point is at one end of the opener and the load is in the middle.

4 0
3 years ago
Which of the following correctly describes the law of conservation of matter?
Tems11 [23]

Answer:

C

Explanation:

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7 0
2 years ago
Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.
Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge q_{1}= 3.8\times10^{-6}\ C

Second charge q_{2}=3.2\times10^{-6}\ C

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

E_{1}=E_{2}

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}

\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}

\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}

x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}

Put the value into the formula

x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}

x=1.69\ m

Hence, The distance is 1.69 m.

5 0
3 years ago
The energy of an electromagnetic wave changes proportionally to which
ozzi

Answer:

D. Frequency

Explanation:

The energy of an  electromagnetic wave is proportional to frequency, mathematically it is expressed as;

E ∝ f

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ Js

The equation above can also be expanded to;

E = hf = h \frac{c}{\lambda}

where;

c is speed of light = 3 x 10⁸ m/s

λ is the wavelength of the electromagnetic wave

Since the speed of light is constant, we can conclude that the energy of the electromagnetic wave is directly proportional to its frequency and inversely proportional to its wavelength.

Therefore, the correct option for direct proportionality is FREQUENCY

8 0
3 years ago
Consider three identical electric bulbs of power P. Two of bulbs are connected in series and the third one is connected in paral
Tamiku [17]

Answer: 3P/2

Explanation: Let the resistance of the bulbs be R.

now lets consider a Voltage V is supplied to the parallel circuit  such that

P=VI=V^2/R

V=IR

both single bulb( bulb 3) and the two bulbs ( bulb 1 and bulb 2) are provided the same Voltage

( as the voltage remains same in parallel circuit)

we can calculate the Current across both circuits

At Bulb 3

Current 1=V/R

Power1=Voltage * Current1

Power1=V*V/R

Power1=P

At Bulb 1 and Bulb 2

Total Resistance= R+R=2R

Current2=\frac{V}{2R}

Power2=Voltage * Current2

Power2=V*\frac{V}{2R} \\Power2=\frac{V^2}{2R} \\Power2=P/2

TotalPower=Power1+Power2\\TotalPower=P+P/2\\TotalPower=\frac{3P}{2}

6 0
3 years ago
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