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3 years ago

A force acts on a 5kg object at rest. How fast will the object accelerate on a frictionless surface?

Physics

1 answer:

MakcuM [25]3 years ago

8 0

Answer: The answer is C.) 25 m/s^2.

Explanation: If you input 5 as s, you would have to use the exponent 2. This means that you have to multiply 5 by 5. 5 x 5= 25.

Edit: Also, because the surface is frictionless, it will make the object go faster too. Nothing can really slow it down unless something blocks it.

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A rigid tank initially contains 3kg of carbon dioxide (CO2) at a pressure of 3bar.The tank is connected by a valve to a friction

marissa [1.9K]

Answer:

Part a: The total amount of energy transfer by the work done is 54.81 kJ.

Part b: The total amount of energy transfer by the heat is 54.81 kJ

Explanation:

Mass of Carbon Dioxide is given as m1=3 kg

Pressure is given as P1=3 bar =300 kPA

Volume is given as V1=0.5 m^3

Pressure in tank 2 is given as P2=2 bar=200 kPa

T=290 K

Now the Molecular weight of $CO_2$ is given as

M=44 kg/kmol

the gas constant is given as

$R=\frac{\bar{R}}{M}\\R=\frac{8.314}{44}\\R=0.189 kJ/kg.K$

Volume of the tank is given as

$V=\frac{mRT}{P_1}\\V=\frac{3 \times 0.189 \times 290}{300 }\\V=0.5481 m^3$

Final mass is given as

$m_2=\frac{P_2V}{RT}\\m_2=\frac{200\times 0.5481}{0.189\times 290}\\m_2=2 kg$

Mass of the CO2 moved to the cylinder

$m=m_1-m_3\\m=3-2=1 kg$

The initial mass in the cylinder is given as

$m_{(cyl)_1}=\frac{P_{(cyl)_1}V_1}{RT}\\m_{(cyl)_1}=\frac{200\times 0.5}{0.189 \times 290}\\m_{(cyl)_1}=1.82 kg$

The mass after the process is

$m_{(cyl)_2}=m_{(cyl)_1}+m\\m_{(cyl)_2}=1.82+1\\m_{(cyl)_2}=2.82\\$

Now the volume 2 of the cylinder is given as

$V_{(cyl)_2}=\frac{m_{(cyl)_2}RT}{P_2}\\m_{(cyl)_2}=\frac{2.82\times 0.189\times 290}{200}\\m_{(cyl)_1}=0.774 m^3$

Part a:

So the Work done is given as

$W=P(V_2-V_1)\\W=200(0.774-0.5)\\W=54.81 kJ$

The total amount of energy transfer by the work done is 54.81 kJ.

Part b:

The total energy transfer by heat is given as

$Q=\Delta U+W\\Q=0+W\\Q=54.81 kJ$

As the temperature is constant thus change in internal energy is 0.

The total amount of energy transfer by the heat is 54.81 kJ

7 0

3 years ago

Three equal charges of magnitude 'e' are located at the vertices of an equilateral triangle of side 1m. Where should you place a

Naily [24]

Answer:

At centroid

Explanation:

In the given equilateral triangle ABC with side of 1 m. The three equal charges e,e,e are placed at the A,B and C.

And the fourth charge 2e is put at point O which is called centroid.

Now we can calculate the distance AD by applying pythagorean theorem as,

$AD^{2}=AB^{2}+BD^{2}$

Put the values and get.

$AD^{2}=1^{2}+(\frac{1}{2} )^{2}\\AD=\sqrt{\frac{3}{4} } \\AD=\frac{\sqrt{3}}{2}$

Now calculate AO as,

$AO=\frac{2}{3}\times \frac{\sqrt{3} }{2}\\AO=\frac{1}{\sqrt{3} }$

And the sides BO=CO=AO.

Now Force can be calculated as

$F_{1}=\frac{2ke^{2} }{\frac{1}{\sqrt{3} } ^{2} }\\F_{1}=6ke^{2}$

And similarly,

$F_{2}=F_{3}=6ke^{2}$

Now we can calculate resultant of $F_{2}andF_{3}$ in upward direction. as,

$F_{net}=\sqrt{F_{2}^{2}+F_{3}^{2}+2F_{2}F_{3}cos120 } \\F_{net}=\sqrt{F_{2}^{2}+F_{2}^{2}+2F_{2}F_{2}(-\frac{1}{2})}\\F_{net}=6ke^{2}$

Therefore the resultant force on centroid O.

$F=F_{1}-F_{net}\\F=6ke^{2}-6ke^{2}\\F=0$

Therefore the fourth charge 2e should be placed on centroid so that it experience zero force.

8 0

3 years ago

A plane has a takeoff speed of 89.3m/s and requires 1465m to reach that speed determine the acceleration of the plane and the ti

Tems11 [23]

Answer:

a = 2.72 ms⁻²

32.83 s

Explanation:

By using the kinematic equations you get,

v² = u² +2as and v = u + at where all terms in usual meaning

Using 1st equation,

89.3² = 0² + 2a×1465 ⇒ a = 2.72 ms⁻²

By 2nd equation,

89.3 = 0 + 2.72×t ⇒ t = 32.83 s

4 0

3 years ago

A kickoff sends a football with an initial velocity of 25 m/s at an angle of 50

eduard

Answer:

The x-component of the velocity is 16.07 m/s

The y-component of the velocity is 19.15 m/s

Explanation:

Given;

initial velocity of the football, u = 25 m/s

angle of projection, θ = 50⁰

The x-component of the velocity (horizontal component) is given as;

$V_x = Vcos(50^0)\\\\V_x = 25 \ m/s \ \ \times \ \ 0.6428\\\\V_x = 16.07 \ m/s$

The y-component of the velocity (vertical component) is given as;

$V_y = Vsin(50^0)\\\\V_y = 25 \ m/s \ \ \times \ \ 0.766\\\\V_y = 19.15 \ m/s$

7 0

3 years ago

A ball is thrown vertically upward with an initial velocity of 29.4 m/s². What is the maximum height reached by the ball? How lo

Elena L [17]

Answer:

Most likely it will reach 29.4 up in the air before coming back down.

Explanation:

The force of the ball will be the same on how high it will go.

4 0

3 years ago