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3 years ago

15

An 19-cm-long bicycle crank arm, with a pedal at one end, is attached to a 23-cm-diameter sprocket, the toothed disk around whic

h the chain moves. A cyclist riding this bike increases her pedaling rate from 65 rpm to 90 rpm in 10 s .

1 answer:

Cloud [144]3 years ago

4 0

Answer:

The tangential acceleration of the pedal is 0.0301 m/s².

Explanation:

Given that,

Length = 19 cm

Diameter = 23 cm

Time = 10 sec

Initial angular velocity = 65 rpm

Final velocity = 90 rpm

Suppose we need to find the tangential acceleration of the pedal

We need to calculate the tangential acceleration of the pedal

Using formula of tangential acceleration

$a_{t}=r\alpha$

$a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{\omega_{2}-\omega_{1}}{t}$

$a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{90\times\dfrac{2]pi}{60}-65\times\dfrac{2\pi}{60}}{10}$

$a_{t}=0.0301\ m/s^2$

Hence, The tangential acceleration of the pedal is 0.0301 m/s².

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The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 5 m/s then ho

Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

$F = \frac{MV}{t}$

solving this two equations together;

$\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}$

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

$t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9} \ s \ = 1.639 \ ns$

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

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2 years ago

A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting

rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is $\sqrt{725}$ ft/ sec

Explanation:

Given:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

Lets find the derivative of distance with respect to time

$\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}$

Substituting the values of h(t) and x(t) and simplifying we get,

$\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}$

$\frac{dh}{dt} = 25ft/sec$

$\frac{dx}{dt} = 10 ft/sec$

$\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}$ = $\frac{725}{\sqrt{725}}$ = $\sqrt{725}$ ft / sec

5 0

2 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

$U=-\frac{GMm}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

$\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

and so, substituting:

$R=6370 km\\h_A = 5970 km\\h_B = 21200 km$

We find

$\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448$

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

$K=\frac{1}{2}\frac{GMm}{r}$

So, the ratio between the two kinetic energies is

$\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

$E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}$

For satellite A, we have

$E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J$

For satellite B, we have

$E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J$

So, satellite B has the greater total energy (since the energy is negative).

(d) $-2.57\cdot 10^8 J$

The difference between the energy of the two satellites is:

$E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J$

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3 years ago

A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di

dimulka [17.4K]

Answer:

d₂ = 1.466 m

Explanation:

In this case we must use the rotational equilibrium equations

Στ = 0

τ = F r

we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive

+ W d₁ - w_cat d₂ = 0

d₂ = W / w d₁

d₂ = M /m d₁

d₂ = 5.00 /2.9 0.850

d₂ = 1.466 m

6 0

3 years ago

vodomira [7]

Given parameters:

Mass on earth = 50kg

Unknown:

Mass on planet Xenon = ?

Weight on planet Xenon = ?

Mass is the amount of matter contained in a particular substance.

Weight is the force on a body and it is derived from the product of mass and acceleration due to gravity.

Weight = mass x acceleration due to gravity

Planet Xenon has half the gravitational force of Earth.

This translated gives $\frac{9.8}{2}$ = 4.9m/s²

Now, mass is always the same every where if the amount of matter in a substance does not change.

In this problem, mass = 50kg on planet xenon.

Weight = mass x acceleration due to gravity = 50 x 4.9 = 245N

The weight on Xenon is 245N and the mass is 50kg

4 0

3 years ago