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3 years ago

How fast must a 70 kg student be running to have a kinetic energy of 568 J?!

Physics

1 answer:

Dmitry [639]3 years ago

4 0

Answer: 4.03 m/s

Explanation: Kinetic energy is

KE = 1/2 m v²

So we derive the equation to have v.

v = √ 2KE / m

= 2 x 568 J / 70 kg

= 16.23

= √ 16.23

= 4.03 m/s

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An ideal gas occupies a volume of 0.60 m3 at 5.0 atm and 400. K. What volume does it occupy at 4.0 atm and a temperature of 200.

kotegsom [21]

V1 = 0.60 m3, P1 = 5.00atm, T1=400K
V2 = ??m3, P2 = 4.00atm, T2=200K
P1V1/T1 = P2V2/T2
V2 = T2P1V1/T1P2 = 200×5×0.60/400×4.00 = 600/1600
V2 = 0.38 m3

7 0

3 years ago

A mass of 0.1 kg of helium fills a 0.2 m3 rigid tank at 350 kPa. The vessel is heated until the pressure is 700 kPa. Calculate t

QveST [7]

Explanation:

(a) First, we will calculate the number of moles as follows.

No. of moles = $\frac{\text{mass}}{\text{molar mass}}$

Molar mass of helium is 4 g/mol and mass is given as 0.1 kg or 100 g (as 1 kg = 1000 g).

Putting the given values into the above formula as follows.

No. of moles = $\frac{\text{mass}}{\text{molar mass}}$

= $\frac{\text{100 g}}{4 g/mol}$

= 25 mol

According to the ideal gas equation,

PV = nRT

or, $(P_{2} - P_{1})V = nR (T_{2} - T_{1})$

$(6.90 atm - 3.45 atm) \times 200 L = 25 \times 0.0821 L atm/mol K \Delta T$

$\Delta T$ = 336.17 K

Hence, temperature change will be 336.17 K.

(b) The total amount of heat required for this process will be calculated as follows.

q = $mC \Delta T$

= $100 g \times 5.193 J/g K \times 336.17 K$

= 174573.081 J/K

or, = 174.57 kJ/K (as 1 kJ = 1000 J)

Therefore, the amount of total heat required is 174.57 kJ/K.

3 0

3 years ago

Astone is thrown directly upward with an initial speed of 9.6 m/s from a height of 12.8 m. After what time interval (in s) does

lara31 [8.8K]

Answer:

1.89 seconds

Explanation:

t = Time taken

u = Initial velocity = 9.6 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=9.6-9.8\times t\\\Rightarrow \frac{-9.6}{-9.8}=t\\\Rightarrow t=0.97 \s$

Time taken to reach maximum height is 0.97 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=9.6\times 0.97+\frac{1}{2}\times -9.8\times 0.97^2\\\Rightarrow s=4.7\ m$

So, the stone would travel 4.7 m up

So, total height ball would fall is 4.7+12.8 = 17.5 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 17.5=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{17.5\times 2}{9.8}}\\\Rightarrow t=1.89\ s$

Time taken by the stone to travel 17.5 m is 1.89 seconds

3 0

3 years ago

A 9.0-kg bowling ball on a horizontal, frictionless surface experiences a net force of 6.0 n. what will be its acceleration?

Vladimir [108]

This question involves the concepts of Newton's Second Law of Motion.

The acceleration of the bowling ball will be "0.67 m/s²".

<h3>Newton's Second Law of Motion</h3>

According to Newton's Second Law of Motion, when an unbalanced force is applied on an object, it produces an acceleration in it, in the direction of the applied force. This acceleration is directly proportional to the force applied and inversely proportional to the mass of the object. Mathematically,

$F=ma\\\\a=\frac{F}{m}$