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jok3333 [9.3K]
3 years ago
14

you push a sled of mass 15 kg across the snow with a force of 180 N for a distance of 2.5 m. There is no friction. if the sled s

tarted at rest. what is the velocity of the sled after you push it?
Physics
1 answer:
Oxana [17]3 years ago
5 0
<span>7.7 m/s First, determine the acceleration you subject the sled to. You have a mass of 15 kg being subjected to a force of 180 N, so 180 N / 15 kg = 180 (kg m)/s^2 / 15 kg = 12 m/s^2 Now determine how long you pushed it. For constant acceleration the equation is d = 0.5 A T^2 Substitute the known values getting, 2.5 m = 0.5 12 m/s^2 T^2 2.5 m = 6 m/s^2 T^2 Solve for T 2.5 m = 6 m/s^2 T^2 0.41667 s^2 = T^2 0.645497224 s = T Now to get the velocity, multiply the time by the acceleration, giving 0.645497224 s * 12 m/s^2 = 7.745966692 m/s After rounding to 2 significant figures, you get 7.7 m/s</span>
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A satellite originally moves in a circular orbit of radius R around the Earth. Suppose it is moved into a circular orbit of radi
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3 0
3 years ago
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Aleks [24]

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Explanation:

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3 0
3 years ago
A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou
Setler [38]

Answer:

a. wavelength of the sound, \vartheta = 1.315\vartheta_{o}

b. observed frequecy, \lambda = 0.7604\lambda_{o}

Given:

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speed of sound in air or vacuum, v_{a} = 343 m/s

speed of sound observed, v_{o} = 0 m/s

Solution:

From the relation:

v = \vartheta \lambda        (1)

where

v = velocity of sound

\vartheta = observed frequency of sound

\lambda = wavelength

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\lambda = \frac{v_{a}}{\vartheta }         (2)

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\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}

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Using eqn (2) and (3):

\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}

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4 0
3 years ago
What is the accletation of a 1500kg with a net force of 7500 N
matrenka [14]
Acceleration formulae is:
a=Fnet/mass
According to the question
a=7500N/1500kg
a=5m/s sq.
3 0
3 years ago
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