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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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A projectile of mass 1.800 kg approaches a stationary target body at 4.800 m/s. The projectile is deflected through an angle of

Artyom0805 [142]

Answer:

P = 5.22 Kg.m/s

Explanation:

given,

mass of the projectile = 1.8 Kg

speed of the target = 4.8 m/s

angle of deflection = 60°

Speed after collision = 2.9 m/s

magnitude of momentum after collision = ?

initial momentum of the body = m x v

= 1.8 x 4.8 = 8.64 kg.m/s

final momentum after collision

momentum along x-direction

P_x = m v cos θ

P_x = 1.8 x 2.9 x cos 60°

P_x = 2.61 kg.m/s

momentum along y-direction

P_y = m v sin θ

P_y = 1.8 x 2.9 x sin 60°

P_y = 4.52 kg.m/s

net momentum of the body

$P = \sqrt{P_x^2 + P_y^2}$

$P = \sqrt{4.52^2 + 2.61^2}$

P = 5.22 Kg.m/s

momentum magnitude after collision is equal to P = 5.22 Kg.m/s

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3 years ago

How many minutes in 1 hour

fgiga [73]

Answer:

60

Explanation:

Every 60 minutes is an hour

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2 years ago

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Two forces are applied to a 17 kg box, as shown. The box is on a smooth surface. Which statement best describes the acceleration

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To the picture the answer is A. I can’t answer the typed question because I need the picture for the box

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3 years ago

Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob

pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

$T^2 = \frac{4\pi^2}{GM}a^3$

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

$T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s$

PART A) Replacing the values to find a, we have

$a^3= \frac{T^2 GM}{4\pi^2}$

$a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}$

$a^3 = 6.46632*10^{39}$

$a = 1.86303*10^{13}m$

Therefore the semimajor axis is $1.86303*10^{13}m$

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

$R = a(1-e)$

$R = 1.86303*10^{13}(1-0.997)$

$R= 5.58*10^{10}m$

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3 years ago

A knight moves on a chessboard two squares up, down, left, or right followedby one square in one of the two directions perpendic

Contact [7]

Answer:

Explanation:

Check attachment for solution.

Generally the movement of the knight is L, i.e (2,1), (1,2),(-1,2),(1,-2) etc.

So using Pythagoras theorem

x^2+y^2=1^2+2^2

x^2+y^2=5

Then, the knight will make one movement when the displacement is √5.

So let take a look at other positions

(1,0),(0,1),(-1,0), (0,-1).

Then, for us to have this kind of movement, the knight has to make 3 movements.

When the displacement is 1, then it will make 3 movement.

Let examine other positions

(2,2) or(-2,-2) or (2,-2) or (-2,2)

When the displacement is √8

Then, the movement of the knight is 4.

Let examine other points

(2,0) or (0,2) or (-2,0) or (0,-2)

When the displacement is 2.

The knight make 2 movement

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3 years ago