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aleksklad [387]
3 years ago
7

The pedals on a bicycle give a mechanical advantage by allowing you to turn the pedals a __________ distance to turn the _______

___ circumstance of the wheels.
Physics
2 answers:
denpristay [2]3 years ago
6 0
Short distance to turn the circumference of the wheels. Mechanical advantage is load/effort I think
Goshia [24]3 years ago
3 0

Answer:

C. Short, larger

Explanation:

It's the correct answer on USA test prep.

You might be interested in
PART ONE
stira [4]

Answer:

3.64×10⁸ m

3.34×10⁻³ m/s²

Explanation:

Let's define some variables:

M₁ = mass of the Earth

r₁ = r = distance from the Earth's center

M₂ = mass of the moon

r₂ = d − r = distance from the moon's center

d = distance between the Earth and the moon

When the gravitational fields become equal:

GM₁m / r₁² = GM₂m / r₂²

M₁ / r₁² = M₂ / r₂²

M₁ / r² = M₂ / (d − r)²

M₁ / r² = M₂ / (d² − 2dr + r²)

M₁ (d² − 2dr + r²) = M₂ r²

M₁d² − 2dM₁ r + M₁ r² = M₂ r²

M₁d² − 2dM₁ r + (M₁ − M₂) r² = 0

d² − 2d r + (1 − M₂/M₁) r² = 0

Solving with quadratic formula:

r = [ 2d ± √(4d² − 4 (1 − M₂/M₁) d²) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(1 − (1 − M₂/M₁)) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(1 − 1 + M₂/M₁) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(M₂/M₁) ] / 2 (1 − M₂/M₁)

When we plug in the values, we get:

r = 3.64×10⁸ m

If the moon wasn't there, the acceleration due to Earth's gravity would be:

g = GM / r²

g = (6.672×10⁻¹¹ N m²/kg²) (5.98×10²⁴ kg) / (3.64×10⁸ m)²

g = 3.34×10⁻³ m/s²

4 0
3 years ago
Naomi is taking a train from DC to New York which is 360km apart. If the train is traveling at an average speed of 32 meters per
rodikova [14]

Answer:

11250 seconds or 187.5 mins

Explanation:

The formula to be used here is that of speed.

Speed (in m/s) = distance (meter) ÷ time (secs)

The speed provided is 32 meters per second

The distance provided will have to be converted to meters; 360 km = 360 × 1000 = 360000 meters

Thus,

32 = 360000 ÷ time

time = 360000 ÷ 32

time = 11250 seconds or 187.5 mins

It will take Naomi 11250 seconds to get to New York

4 0
2 years ago
An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec
lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

  • x = initial distance of the electron from the proton = 6 cm = 0.06 m
  • y = initial distance of the electron from the proton = 3 cm = 0.03 m
  • u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

  • m = mass of an electron = 9.1\times 10^{-31}\ kg
  • v = final velocity of the electron
  • e = magnitude of charge on an electron = 1.6\times 10^{-19}\ C
  • p = magnitude of charge on a proton = 1.6\times 10^{-19}\ C

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\

\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0
3 years ago
When Jennifer is out for a
Nady [450]
The acceleration is the principal subordinate of the speed if the speed is steady the subsidiary is invalid if the speed is diminishing the subsidiary is negative. When discussing so much stuff we consider the momentary esteem.

<span>Note that when you back off, you back off by and large yet can locally in time quicken a tiny bit, suppose amid 1/tenth of a sec since you achieved a segment of the street which was slanting. In any case, this does not change the way that when the speed diminishes, the quickening is negative.</span>
4 0
3 years ago
The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu
Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

E=\phi + K

where

E=\frac{hc}{\lambda} is the energy of the incident light, with h being the Planck constant, c being the speed of light, and \lambda being the wavelength

\phi is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

\lambda=190 nm=1.9\cdot 10^{-7}m, so the energy of the incident light is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J

Converting in electronvolts,

E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

\phi = E-K=6.5 eV-4.0 eV=2.5 eV

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

\lambda=510 nm=5.1\cdot 10^{-7} m

So its energy is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J

Converting in electronvolts,

E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: \phi = 4.5 eV

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

E=\phi+K=4.5 eV+2.7 eV=7.2 eV

Then the copper is replaced with sodium, which has work function of

\phi = 2.3 eV

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

K=E-\phi = 7.2 eV-2.3 eV=4.9 eV

7 0
3 years ago
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