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3 years ago

5

A body of mass 10 kg is pulled by a force 8 N. calculate the acceleration and the the final velocity of the body after 5 second.

1 answer:

Shtirlitz [24]3 years ago

3 0

Acceleration = force / mass = 8/10 m/s^2
After 5 s (from the start) velocity = 5 x 8/10= 4 m/s

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Who reported four “element” classifications, but included some substances that were combinations of elements rather than true el

marishachu [46]

Explanation :

Antoine-Laurent Lavoisier reported four "element" classifications but included some substances that were combinations of elements rather than true elements in his listing.

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3 years ago

A truck with 0.420-m-radius tires travels at 32.0 m/s. what is the angular velocity of the rotating tires in radians per second?

andrey2020 [161]

Angular velocity of the rotating tires can be calculated using the formula,

v=ω×r

Here, v is the velocity of the tires = 32 m/s

r is the radius of the tires= 0.42 m

ω is the angular velocity

Substituting the values we get,

32=ω×0.42

ω= 32/0.42 = 76.19 rad/s

= 76.19× $\frac{1}{2\pi} *60$ revolution per min

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Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.

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3 years ago

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3 years ago

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3 years ago

Read 2 more answers

The vapor pressure of ethanol at 293 K is 5.95 kPa and at 336.5 K it is 53.3 kPa. Calculate the enthalpy of vaporization of etha

denis-greek [22]

Answer:

$H=41.3kJmol^{-1}$

Explanation:

The equation relating the the enthalphy, pressure and temperature is expressed as

$ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\$

Where P is the pressure, H is the enthalphy, and T is the temperature.

since the given values are

$T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}$

if we insert values, we arrive at

$ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}$

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3 years ago