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Sedbober [7]
4 years ago
14

I need a creative science fair title about how music affects your run

Physics
2 answers:
Ganezh [65]4 years ago
6 0

how about

1) music on the run

2) music on the road

3) run to the music

WITCHER [35]4 years ago
5 0
How about
Music on the go
You might be interested in
0.3-L glass of water at 20C is to be cooled with ice to 5C. Determine how much ice needs to be added to the water, in grams, i
abruzzese [7]

Answer:

a. m_i_c_e=54.6g\\b. m_i_c_e=48.7g\\m_c_o_l_d_w_a_t_e_r=900g

Explanation:

First we need to state our assumptions:

Thermal properties of ice and water are constant, heat transfer to the glass is negligible, Heat of ice h_i_f=333.7KJkg

Mass of water,m_w=\rho V =1\times0.3=0.3Kg.

Energy balance for the ice-water system is defined as

E_i_n-E_o_u_t=\bigtriangleup E_s_y_s\\0=\bigtriangleup U=\bigtriangleup U_i_c_e+\bigtriangleup U_w

a.The mass of ice at 0\textdegree C is defined as:

[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[0+333.7+418\times(5-0)]+0.3\times4.18\times(5-20)=0\\m_i_c_e=0.0546Kg=54.6g

b.Mass of ice at 20\textdegree C is defined as:

[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[2.11\times(0-(20))+333.7+4.18\times(5-0)]+0.3\times4.18\times(5-20)=0\\\\m_i_c_e=0.0487Kg=48.7g

c.Mass of cooled water at T_c_w=0\textdegree C

\bigtriangleup U_c_w+\bigtriangleup U_w=0

[mc(T_2-T_1)]_c_w+[mc(T_2-T_1)]_w=0\\m_c_w\times4.18\times(5-0)+0.3\times4.18\times(5-20)\\m_c_w=0.9kg=900g

8 0
3 years ago
At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is
Katyanochek1 [597]

Answer:

4.47\ \text{km/h}

Explanation:

\dfrac{da}{dt} = Rate at which the distance between A and starting point of B is changing = -20 km/h

\dfrac{db}{dt} = Rate at which the distance of B is changing = 15 km/h

\dfrac{dc}{dt} = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = 20\times 4=80\ \text{km}

a = Distance remaining to the start point of B = 110-80=30\ \text{km}

b = Distance covered by B in 4 hours = 15\times 4=60\ \text{km}

Distance between A and B after 4 hours

c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}

c^2=a^2+b^2

Differentiating with respect to time we get

c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}

The rate at which the distance between the ships is changing at 4 PM is 4.47\ \text{km/h}.

7 0
3 years ago
the resistivity of gold is 2.44×10−8Ω⋅m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a cur
cricket20 [7]

Answer:

0.03605 V/m is the electric field in the gold wire.

Explanation:

Resistivity of the gold = \rho = 2.44\times 10^{-8} \Omega.m

Length of the gold wire = L = 14 cm = 0.14 m ( 1 cm = 0.01 m)

Diameter of the wire = d = 0.9 mm

Radius of the wire = r = 0.5 d = 0.5 × 0.9 mm = 0.45 mm = 0.45\times 0.001 m

( 1mm = 0.001 m)

Area of the cross-section = A=\pi r^2=\pi r^(0.45\times 0.001 m)^2

Resistance of the wire = R

Current in the gold wire = 940 mA = 0.940 A ( 1 mA = 0.001 A)

R=\rho\times \frac{L}{A}

V(voltage)=I(current)\times R(Resistance) ( Ohm's law)

\frac{V}{I}=\rho\times \frac{L}{A}

We know, Electric field is given by :

E=\frac{dV}{dr}

E=\frac{V}{L}

E=\frac{V}{L}=\rho\times \frac{I}{A}

E=2.44\times 10^{-8} \Omega.m\times \frac{0.940 A}{\pi r^(0.45\times 0.001 m)^2}=0.03605 V/m

0.03605 V/m is the electric field in the gold wire.

3 0
4 years ago
A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with
Ray Of Light [21]

Answer:

See below

Explanation:

Vertical position = 45 +  20 sin (30) t  - 4.9 t^2

 when it hits ground this = 0

               0 = -4.9t^2 + 20 sin (30 ) t + 45

                0 = -4.9t^2 + 10 t +45 = 0     solve for t =4.22 sec

  max height is at  t= - b/2a = 10/9.8 =1.02

     use this value of 't' in the equation to calculate max height = 50.1 m

      it has  4.22 - 1.02 to free fall = 3.2 seconds free fall

           v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL

      it will <u>also</u> still have horizontal velocity =  20 cos 30 = 17.32 m/s

        total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s

Horizontal range = 20 cos 30  * t  =  20 *  cos 30  * 4.22 = 73.1 m

8 0
2 years ago
Energy is transmitted only in indivisible discrete quantities. True or False
alex41 [277]

Answer:

False

Explanation:

From the fact that light is emitted in portion, it does not follow that light ray is discrete in structure.

The way Einstein put it was that if a beer is always sold in pint bottles, that does not mean that beer contain of indivisible pint bottles. Yet experiments in which light  knocked electrons out of materials insistently made the point that light is absorbed only in the form of discrete portions. An emitted portion or quantum of light energy retains it's individuality in subsequent events as well.

This concept was first stated by Einstein in 1905. From the heuristic point of view that he developed light is always made up of discrete portions possessing energy and momentum.

5 0
3 years ago
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