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3 years ago

10

. Calculate the efficiency of a bicycle if the input work to turn the pedal is 45J and the output work is 20J. * 1 point 2.25 2.

25% 44 44.4%

1 answer:

cestrela7 [59]3 years ago

8 0

20/45=0.4*100= 44.4 so the answer is..................................................

Answer: 44.4%

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The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of

Oksi-84 [34.3K]

Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Explanation: To find the answer, we need to know about the different equations of planetary motion.

<h3>How to find the initial speed of the rock as it left the astronaut's hand?</h3>

We have the expression for the initial velocity as,

$v=\sqrt{2gh}$

Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,

$g_g=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2}= 0.132$

Now, the velocity will become,

$v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s$

<h3>How to find the speed of the satellite?</h3>

As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,

$v=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10^{-11}*7.88*10^{18}}{1.45*10^5} } =3.624km/s$

Thus, we can conclude that,

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Learn more about the equations of planetary motion here:

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2 years ago

Read 2 more answers

A sphere moves in simple harmonic motion with a frequency of 4.80 Hz and an amplitude of 3.40 cm. (a) Through what total distanc

geniusboy [140]

Answer:

a) the total distance traveled by the sphere during one cycle of its motion = 13.60 cm

b) The maximum speed is = 102.54 cm/s

The maximum speed occurs at maximum excursion from equilibrium.

c) The maximum magnitude of the acceleration of the sphere is = 30.93 $m/s^2$

The maximum acceleration occurs at maximum excursion from equilibrium.

Explanation:

Given that :

Frequency (f) = 4.80 Hz

Amplitude (A) = 3.40 cm

a)

The total distance traveled by the sphere during one cycle of simple harmonic motion is:

d = 4A (where A is the Amplitude)

d = 4(3.40 cm)

d = 13.60 cm

Hence, the total distance traveled by the sphere during one cycle of its motion = 13.60 cm

b)

As we all know that:

$x = Asin \omega t$

Differentiating the above expression with respect to x ; we have :

$\frac{d}{dt}(x) = \frac{d}{dt}(Asin \omega t)$

$v = A \omega cos \omega t$

Assuming the maximum value of the speed(v) takes place when cosine function is maximum and the maximum value for cosine function is 1 ;

Then:

$v_{max} = A \omega$

We can then say that the maximum speed therefore occurs at the mean (excursion) position where ; x = 0 i.e at maximum excursion from equilibrium

substituting $2 \pi f$ for $\omega$ in the above expression;

$v_{max} = A(2 \pi f)$

$v_{max} = 3.40 cm (2 \pi *4.80)$

$v_{max} = 102.54 \ cm/s$

Therefore, the maximum speed is = 102.54 cm/s

The maximum speed occurs at maximum excursion from equilibrium.

c) Again;

$v = A \omega cos \omega t$

By differentiation with respect to t;

$\frac{d}{dt}(v) = \frac{d}{dt}(A \omega cos \omega t)$

$a =- A \omega^2 sin \omega t$

The maximum acceleration of the sphere is;

$a_{max} =A \omega^2$

where;

$w = 2 \pi f$

$a_{max} = A(2 \pi f)^2$

where A= 3.40 cm = 0.034 m

$a_{max} = 0.034*(2 \pi *4.80)^2$

$a_{max} = 30.93 \ m/s^2$

The maximum magnitude of the acceleration of the sphere is = 30.93 $m/s^2$

The maximum acceleration occurs at maximum excursion from equilibrium where the oscillating sphere will have maximum acceleration at the turning points when the sphere has maximum displacement of $x = \pm A$

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3 years ago

How can an object become charged

morpeh [17]

Answer:

An object gets charged when it's atoms lose or gain an electron to become an ion. For example: ... This means that the fur loses it's electrons to the plastic rod and both objects are now charged. The fur is positively charged because it lost electrons and the rod is negatively charged because it gained electrons

Explanation:

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3 years ago

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Draw velocity-ime graph for uniform moion of an object , when initially body is at rest.

Klio2033 [76]

When the body is at rest, its speed is zero, and the graph lies on the x-axis.

When the body is in uniform motion, the speed is constant, and the graph is a horizontal line, parallel to the x-axis and some distance above it.

It's impossible to tell, based on the given information, how these two parts of the
graph are connected. There must be some sloping (accelerated) portion of the graph
that joins the two sections, but it cannot be accounted for in either the statement
that the body is at rest or that it is in uniform motion, since acceleration ... that is,
any change of speed or direction ... is not 'uniform' motion'.

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3 years ago

A child pulls a wagon at a constant velocity along a level sidewalk. The child does this by applying a 22 newton force to the wa

Jlenok [28]

Answer:

The answer is 18 N.

Explanation:

A force can be divided into components x and y components. The component along the x-axis is called the horizontal component and along the y-axis is called the vertical component. In this case, as the force is in a horizontal direction and is also known as x-component of force. The x- component of force is

Fx = Fcosθ

Fx = 22(cos 35°)

Fx = 22 x 0.819

Fx = 18 N

Child's horizontal pull forces are equal to that of frictional resistance force on the wagon.

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3 years ago