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4 years ago

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

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Would this one be correct?(C.)

BaLLatris [955]

No. The correct one would be D .

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3 years ago

Molecules move from areas of low concentration to high concentration through a process of

weeeeeb [17]

Active transport move molecules from low concentrations to high concentrations.

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3 years ago

Question 2: Start-Up

yuradex [85]

Answer:

The car starts moving in the positive direction at x = 0.2 seconds. Initially it moves very little, but it covers a greater distance with each time increment.

Explanation:

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3 years ago

At a certain location, Earth has a magnetic field of 0.60 ✕ 10−4 T, pointing 75° below the horizontal in a north-south plane. A

saveliy_v [14]

Answer with Explanation:

We are given that

Magnetic field,B= $0.6\times 10^{-4} T$

$\theta=75^{\circ}$

Length of wire,l=15 m

Current,I=19 A

a.We have to find the magnitude of magnetic force and direction of magnetic force.

Magnetic force,F= $IBlsin\theta$

Using the formula

$F=0.6\times 10^{-4}\times 15\times 19sin75$

$F=16.5\times 10^{-3} N$

Direction= $tan\theta=cot(90-75)=tan15^{\circ}$

$\theta=15^{\circ}$

15 degree above the horizontal in the northward direction.

5 0

3 years ago

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

3 years ago