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White raven [17]
3 years ago
7

Runner A had the highest velocity. Runner C had the lowest velocity. Runners A and B started at the same position, but runner C

started ahead of runners A and B. What did you include in your response? Check all that apply.
Runner A had the highest velocity.
Runner C had the lowest velocity.
Runners A and B started at the same position.
Runner C started ahead of runners A and B.
Physics
2 answers:
galina1969 [7]3 years ago
6 0

Answer:

A. C. and D.

Explanation:

Elan Coil [88]3 years ago
5 0

Answer:

A, C, and D

Just confirming

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In which part of the ear is the sound wave converted into electrical impulse
Kryger [21]
The part of the ear where the sound wave converted into electrical impulse would be the cochlea. This part is the  auditory portion of the inner ear which produces nerve impulses in response to sound vibrations. Hope this answers the question. Have a nice day.
3 0
3 years ago
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A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider
shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

d=v_x t

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

y(t) = h + v_y t - \frac{1}{2}gt^2

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s

So now we can find the magnitude of the initial velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s

4 0
3 years ago
The series in the He spectrum that corresponds to the set of transitions where the electron falls from a higher level to the nf
Galina-37 [17]
Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:  
<span>E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| </span>
<span>E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J </span>

<span>After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:</span>

<span>E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m </span>

so 6.56×10^-7 m or better written 656 nm is in the visible spectrum
7 0
3 years ago
A penudulum has a period of 6.5s what is its frequency
Sergio039 [100]

Answer:0.153 Hz

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5 0
3 years ago
A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons
shepuryov [24]

Answer:

v=2.556m/s

Explanation:

From the conservation of mechanical energy

K_{E1}+U_1=K_{E2}+U_2

\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2

x_2=0.08m

v_1=0 m/s

Solve to velocity v2

m*v_2^2=k*x_1^2-k*x_2^2

v^2=\frac{k}{m}*(x_1^2-x_2^2)

v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)

v=\sqrt{6.54m^2/s^2}=2.556m/s

4 0
3 years ago
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