At a pressure of one atmosphere oxygen boils at −182.9°C and freezes at −218.3°C. Consider a temperature scale where the boiling point of oxygen is 100.0°O and the freezing point is 0°O. Determine the temperature on the Oxygen scale that corresponds to the absolute zero point on the Kelvin scale.

Answer:

d) Both blocks have had equal losses of energy to friction

Explanation:

As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces

Moreover, the block A is twice as fast than block B at the time of crossing the finish line

So based on the above information,  it contains the losses of identical friction

And we also know that

Friction energy loss is

[tex]= \mu \times m \times g \times D[/tex]

It would be the same for both the blocks

hence, the option d is correct

The correct answer will be both blocks have had equal losses of energy to friction.

Friction is defined as when any object is slides on a surface by means of any external force then the force in the opposite direction generated between the surface and the body restrict the motion of the body this force is called as the friction.

As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces.

Moreover, the block A is twice as fast as block B at the time of crossing the finish line.

So based on the above information,  it contains the losses of identical friction.

And we also know that

Friction energy loss is

[tex]E_f=\mu m g D[/tex]

It would be the same for both the blocks

Hence both blocks have had equal losses of energy to friction.

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Answer:

1.1ohms

Explanation:

According to ohms law E = IR

If potential difference of a battery is 2.2 V when it is connected across a resistance of 5 ohm and if suddenly the voltage Falls to 1.8V then the current in the 5ohms resistor I = V/R = 1.8/5

I = 0.36A (This will be the load current).

Before we can calculate the value of the internal resistance, we need to know the voltage drop across the internal resistance.

Voltage drop = 2.2V - 1.8V = 0.4V

Then we calculate the internal resistance using ohms law.

According to the law, V = Ir

V= voltage drop

I is the load current

r = internal resistance

0.4 = 0.36r

r = 0.4/0.36

r = 1.1 ohms

Answer:

The deceleration of the car is [tex]\approx -0.065m/s^{2}[/tex]

Explanation:

to solve this, we will have to apply the knowledge that will be got from the equations of motion.

There are several equations of motion, and depending on the parameters given in the problem, we can choose the perfect equation that can best be used to solve the problem.

In this case, since we are given the velocity and time, and we are solving for the acceleration, we will use this formula

[tex]v = u +at[/tex]

where v= final velocity = 0

u = initial velocity = 135Km/h [tex]\approx 0.278 m/s[/tex]

t= time = 4.29 seconds.

[tex]a = \frac{v - u}{t}[/tex]

[tex]a =\frac{0-0.278}{4.29} \approx 0.065m/s^{2}[/tex]

Hence, the deceleration of the car is [tex]\approx -0.065m/s^{2}[/tex]

Answer:

Explanation:

Given that:

The mass of the cell is 9.0 x 10^-14 kg

The charges of the cell is -2.5pC and the other -3.30 pC

[tex]q_1=-2.5\times10^{-12}C \ \ and \ \ q_2=-3.75\times10^{-12}C[/tex]

Radius is  3.75 × 10-6 m

The final distance is twice the radius

i.e [tex]2*(3.75 \times 10^{-6}) = 7.5*10^{-6}m[/tex]

The formula for the velocity of the cell is

[tex]mv^2=\frac{q_1q_2}{4\pi \epsilon 2 r} \\[/tex]

[tex]v=\sqrt{\frac{q_1q_2}{4\pi \epsilon 2 r} }[/tex]

[tex]=\sqrt{\frac{(-2.5\times10^{-12})(-3.3\times10^{-12}}{4(3.14)(8.85\times10^{-112}(2\times3.75\times10^{-6})(9\times10^{-14})} } \\\\=\sqrt{\frac{(-8.25\times10^{-24})}{(7503.03\times10^{-32})} } \\\\=\sqrt{109955.5779} \\\\=331.60m/s[/tex]

The maximum acceleration of the cells as they move toward each other and just barely touch is

[tex]ma= \frac{q_1q_2}{4\pi \epsilon (2r)^2} \\\\a= \frac{q_1q_2}{4\pi \epsilon (2r)^2(m)}[/tex]

[tex]=\frac{(-2.5\times10^{-12})(-3.3\times10^{-12})}{4(3.14)(8.85\times10^{-12})(2\times3.75\times10^{-6})^2(9\times10^{-14})}[/tex]

[tex]=\frac{(-8.25\times10^{-24})}{(56272.725\times10^{-38})} \\\\=1.47\times10^{10}m/s^2[/tex]

The answers obtained are;

1. The initial speed of each of the red blood cells is [tex]v= 331.66\,m/s[/tex].

2. The maximum acceleration of the cells is [tex]a=1.47\times 10^{10}\,m/s^2[/tex].

The answer is explained as shown below.

We have, the mass of the red blood cell;

Also, the charges of the cells are;

The distance between the charges when they barely touch will be two times the radius of each charge.

1. As both the cells are negatively charged they will repel each other.

Now, substituting all the known values in the equation, we get;

[tex]v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m\times(9\times 10^{-14}\,kg)} =110000\,m^2/s^2[/tex]

2. Also as the force between them is repulsive, there must be an acceleration to make them barely touch each other.

Substituting the known values, we get;

[tex](9\times 10^{-14}\,kg)\times a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2}[/tex]

[tex]\implies a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2\times(9\times 10^{-14}\,kg) }[/tex]

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Answer:

[tex]v_B=3.78\times 10^5\ m/s[/tex]

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is [tex]6.63\times 10^{-27}\ kg[/tex]

Speed of nucleus at A is [tex]v_A=6.2\times 10^5\ m/s[/tex]

Potential at point A, [tex]V_A=1.5\times 10^3\ V[/tex]

Potential at point B, [tex]V_B=4\times 10^3\ V[/tex]

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

[tex]\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s[/tex]

So, the speed at point B is [tex]3.78\times 10^5\ m/s[/tex].

Answer:

The ration of the two wavelength is  [tex]\frac{\lambda_1}{\lambda_2} = \frac{8}{3}[/tex]

Explanation:

Generally two slit constructive interference can be mathematically represented as

      [tex]\frac{y}{L} = \frac{m * \lambda}{d}[/tex]

Where  y is the distance between fringe

           d  is the distance between the two slit

           L is the distance between the slit and the wall

           m is the order of the fringe

given that  y , L  , d  are constant  we have that

     [tex]\frac{m }{\lambda } = constant[/tex]

So  

    [tex]\frac{m_1 }{\lambda_1 } = \frac{m_2 }{\lambda_2 }[/tex]

So     [tex]m_1 = 8[/tex]

  and  [tex]m_2 = 3[/tex]

=>     [tex]\frac{m_2}{m_1} = \frac{\lambda_1}{\lambda_2}[/tex]

=>     [tex]\frac{8}{3} = \frac{\lambda_1}{\lambda_2}[/tex]

So

     [tex]\frac{\lambda_1}{\lambda_2} = \frac{8}{3}[/tex]

Answer:

v = 349.23 m/s

Explanation:

It is required to find the orbital speed for a satellite [tex]3.5\times 10^8\ m[/tex] from the center of mass.

Mass of Mars, [tex]M=6.4\times 10^{23}\ kg[/tex]

The orbital speed for a satellite is given by the formula as follows :

[tex]v=\sqrt{\dfrac{GM}{r}} \\\\v=\sqrt{\dfrac{6.67\times 10^{-11}\times 6.4\times 10^{23}}{3.5\times 10^8}} \\\\v=349.23\ m/s[/tex]

So, the orbital speed for a satellite is 349.23 m/s.

Answer:

v = 4.08m/s₂

Explanation:

D. (1m, 0.5m)

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

Answer:

[tex]d_2=3.16cm[/tex]

Explanation:

So, in order to solve this problem, we must start by building a diagram of the problem itself. (See attached picture) And together with the diagram, we must build a free body diagram, which will include the forces that are being applied on the given charged particle together with their directions.

In this case we only care about the x-direction of the force, since the y-forces cancel each other. So if we do a sum of forces on the x-direction, we get the following:

[tex]\sum{F_{x}}=0[/tex]

so:

[Tex]-F_{12}+F_{13x}+F_{14x}=0[/tex]

Since [tex]F_{13x}=F_{14x}[/tex] we can simplify the equation as:

[tex]-F_{12}+2F_{13x}=0[/tex]

we can now solve this for [tex]F_{12}[/tex] so we get:

[tex]F_{12}=2F_{13x}[/tex]

Now we can substitute with the electrostatic force formula, so we get:

[tex]k_{e}\frac{q_{1}q_{2}}{r_{12}^{2}}=2k_{e}\frac{q_{1}q_{3}}{r_{13}^{2}}cos \theta[/tex]

We can cancel [tex]k_{e}[/tex] and [tex]q_{1}[/tex]

so the simplified equation is:

[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{r_{13}^{2}}cos \theta[/tex]

From the given diagram we know that:

[tex]cos \theta = \frac{d_{1}}{r_{13}}[/tex]

so when solving for [tex]r_{13}[/tex] we get:

[tex]r_{13}=\frac{d_{1}}{cos\theta}[/tex]

and if we square both sides of the equation, we get:

[tex]r_{13}^{2}=\frac{d_{1}^{2}}{cos^{2}\theta}[/tex]

and we can substitute this into our equation:

[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{d_{1}^{2}}cos^{3} \theta[/tex]

so we can now solve this for [tex]r_{12}[/tex] so we get:

[tex]r_{12}=\sqrt{\frac{d_{1}^{2}q_{2}}{2q_{3}cos^{3}\theta}}[/tex]

which can be rewritten as:

[tex]r_{12}=d_{1}\sqrt{\frac{q_{2}}{2q_{3}cos^{3}\theta}}[/tex]

and now we can substitute values.

[tex]r_{12}=(3cm)\sqrt{\frac{7x10^{-19}C}{2(2x10^{-19}C)cos^{3}(20^{o})}}[/tex]

which solves to:

[tex]r_{12}=6.16cm[/tex]

now, we must find [tex]d_{2}[/tex] by using the following equation:

[tex]r_{12}=d_{1}+d_{2}[/tex]

when solving for [tex]d_{2}[/tex] we get:

[tex]d_{2}=r_{12}-d_{1}[/tex]

when substituting we get:

[tex]d_{2}=6.16cm-3cm[/tex]

so:

[tex]d_{2}=3.16cm[/tex]

Answer:

   h = 2 R (1 +μ)

Explanation:

This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the

let's use the mechanical energy conservation agreement

starting point. Lower, just at the curl

       Em₀ = K = ½ m v₁²

final point. Highest point of the curl

        [tex]Em_{f}[/tex] = U = m g y

Find the height y = 2R

      Em₀ = Em_{f}

      ½ m v₁² = m g 2R

       v₁ = √ 4 gR

Any speed greater than this the body remains in the loop.

In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law

X axis

    -fr = m a                      (1)

Y Axis  

      N - W = 0

      N = mg

the friction force has the formula

     fr = μ  N

     fr = μ m g

    we substitute 1

    - μ mg = m a

     a = - μ g

having the acceleration, we can use the kinematic relations

    v² = v₀² - 2 a x

    v₀² = v² + 2 a x

the length of this zone is x = 2R

    let's calculate

     v₀ = √ (4 gR + 2 μ g 2R)

     v₀ = √4gR( 1 + μ)

this is the speed so you must reach the area with fricticon

finally have the third part we use energy conservation

starting point. Highest on the ramp without rubbing

     Em₀ = U = m g h

final point. Just before reaching the area with rubbing

     [tex]Em_{f}[/tex] = K = ½ m v₀²

      Em₀ = Em_{f}

     mgh = ½ m 4gR(1 + μ)

       h = ½ 4R (1+ μ)

       h = 2 R (1 +μ)

Answer:

The temperature is  [tex]T = 168.44 \ K[/tex]

Explanation:

From the question ewe are told that

   The rate of heat transferred is    [tex]P = 13.1 \ W[/tex]

     The surface area is  [tex]A = 1.55 \ m^2[/tex]

      The emissivity of its surface is  [tex]e = 0.287[/tex]

Generally, the rate of heat transfer is mathematically represented as

           [tex]H = A e \sigma T^{4}[/tex]

=>         [tex]T = \sqrt[4]{\frac{P}{e* \sigma } }[/tex]

where  [tex]\sigma[/tex] is the Boltzmann constant with value  [tex]\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.[/tex]

substituting value  

             [tex]T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }[/tex]

            [tex]T = 168.44 \ K[/tex]

Answer:

According to Newtons 2nd law of motion ;

  The acceleration an object experiences is as a result of the net force which is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

Explanation:

This law is simply saying ;

Force = Mass ×Acceleration

I Hope It Helps  :)

Answer:

46648 J

Explanation:

mass m= 85 Kg

velocity v = 56 m/s

distance covered s =40 m

According to Question,

frictional drag force to him that matched his weight

[tex]\Rightarrow F_d =mg\\=85\times9.81=833 N[/tex]

Therefore, work done by practometer against the drag force = heat was generated by this frictional drag force in J

W=Q= F_d×s

=833×56 = 46648 J

Answer:

Umax = 105.8nJ

Umin =-105.8nJ

Umax-Umin = 211.6nJ

Explanation:

Answer:

[tex]\theta=65.18rad[/tex]

Explanation:

The angle in rotational motion is given by:

[tex]\theta=\frac{w_o+w_f}{2}t[/tex]

Recall that the angular speed is larger than regular frequency (in rpm) by a factor of [tex]2\pi[/tex], so:

[tex]\omega_f=2\pi f\\\omega_f=2\pi*250rpm\\\omega_f=1570.80 \frac{rad}{min}[/tex]

The wheel spins from rest, that means that its initial angular speed is zero([tex]\omega_o[/tex]). Finally, we have to convert the given time to minutes and replace in the first equation:

[tex]t=5s*\frac{1min}{60s}=0.083min\\\theta=\frac{\omega_f}{2}t\\\theta=\frac{1570.800\frac{rad}{min}}{2}(0.083min)\\\theta=65.18rad[/tex]

Answer:

The magnitude of the electric force on the particle in terms of the variables given is, F = qE

The magnitude of the magnetic force on the particle in terms of the variables given is, F = q (v x B)

Explanation:

Given;

a charged particle, q

magnitude of electric field, E

magnitude of magnetic field, B

The magnitude of the electric force on the particle in terms of the variables given;

F = qE

The magnitude of the magnetic force on the particle in terms of the variables given;

F = q (v x B)

where;

v is the constant velocity of the charged particle

Answer:

The magnitude of the electric force acting on a charged particle moving through an electric field = |qE|

The magnitude of the magnetic force of a charged particle moving at a particular velocity through a magnetic field = |qv × B|

Explanation:

The electric force acting on a charged particle, q, moving through an electric field, E, is given as a product of the charge on the particle (a scalar quantity) and the electric field (a vector quantity).

Electric force = qE

The magnitude of the electric force = |qE|

That is, magnitude of the product of the charge and the electric field vector.

The magnetic force acting on a charged particle, q, moving with a velocity, v, through a magnetic field, B is a vector product of qv [a product of the charge of the particle (a scalar quantity) and the velocity of the particle (a vector quantity)] and B (a vector quantity).

It is given mathematically as (qv × B)

The magnitude of the magnetic force is the magnitude of the vector product obtained.

Magnitude of the magnetic force = |qv × B|

Hope this Helps!!!

Answer:

The astronauts are separated by 28 m.

Explanation:

The separation of the astronauts can be found by conservation of linear momentum:

[tex] p_{i} = p_{f} [/tex]

[tex] m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f} [/tex]

[tex] m_{1}*0 + m_{2}*0 = m_{1}v_{1f} + m_{2}v_{2f} [/tex]

[tex] m_{1}v_{1f} = -m_{2}v_{2f} [/tex]

[tex] v_{1f} = -\frac{m_{2}v_{2f}}{m_{1}} = -\frac{80v_{2f}}{60} [/tex]

Now, the distance (x) is:      

[tex] x = \frac{v}{t} [/tex]  

The distance traveled by the astronaut 1 is:

[tex] x_{1} = v_{1f}*t = -\frac{80v_{2f}}{60}*t [/tex]    (1)

And, the distance traveled by the astronaut 2 is:

[tex] x_{2} = v_{2f}*t [/tex]  (2)

From the above equation we have:  

[tex] t = \frac{x_{2}}{v_{2f}} [/tex]    (3)                                    

By entering equation (3) into (1) we have:    

[tex] x_{1} = -\frac{80v_{2f}}{60}*(\frac{x_{2}}{v_{2f}}) [/tex]

[tex] x_{1} = -\frac{4*12}{3} = -16 m [/tex]    

The minus sign is because astronaut 1 is moving in the opposite direction of the astronaut 2.      

Finally, the separation of the astronauts is:

[tex] x_{T} = |x_{1}| + x_{2} = (16 + 12)m = 28 m [/tex]

Therefore, the astronauts are separated by 28 m.

I hope it helps you!

The total separation between the two astronauts is 28m.

The given parameters:

Apply the principle of conservation of momentum to determine the final velocity of each astronauts as follows;

[tex]m_1u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\\\\60(0) + 80(0) = 60(v_1) + 80(v_2)\\\\0 = 60v_1 + 80v_2\\\\-60v_1 = 80v_2\\\\v_1 = \frac{-80v_2}{60} \\\\v_1 = -1.333v_2[/tex]

Let the time when astronaut 2 moved 12 m = t

The distance traveled by astronaut 1 is calculated as;

[tex]x_1 = v_1 t\\\\x_1 = -1.333v_2t[/tex]

The  distance traveled by astronaut 2 is calculated as;

[tex]x_2 = v_2 t\\\\12 = v_2t\\\\t = \frac{12}{v_2}[/tex]

Now solve for the distance of astronaut 1

[tex]x_1 = - 1.333v_2 \times t\\\\x_1 = -1.333 v_2 \times \frac{12}{v_2} \\\\x_1 = -16 \ m[/tex]

The total separation between the two astronauts is calculated as follows;

[tex]d = |x_1| + x_2\\\\d = 16 + 12\\\\d = 28 \ m[/tex]

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Answer:

about 1 meter

Explanation:

The distance past the edge that the man will get before the plank starts to tip is; 0.4285 m

We are given;

Mass of plank; m = 30 kg

Length of plank; L = 6m

Mass of man; M = 70 kg

Since the plank has 2 supports which are the deck of the ship, then it means that, we can take moments about the right support before the 2m stick out of the plank.

Thus;

Moment of weight of plank about the right support;

τ_p = mg((L/2) - 2)

τ_p = 30 × 9.8((6/2) - 2)

τ_p = 294 N.m

Moment of weight of man about the right support;

τ_m = Mgx

where x is the distance past the edge the man will get before the plank starts to tip.

τ_m = 70 × 9.8x

τ_m = 686x

Now, moment of the board is counterclockwise while that of the man is clockwise. Thus;

τ_m = τ_p

686x = 294

x = 294/686

x = 0.4285 m

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Complete question:

Two conductors made of the same material are connected across the same potential difference. Conductor A has seven times the diameter and seven times the length of conductor B. What is the ratio of the power delivered to A to power delivered to B.

Answer:

The ratio of the power delivered to A to power delivered to B is 7 : 1

Explanation:

Cross sectional area of a wire is calculated as;

[tex]A = \frac{\pi d^2}{4}[/tex]

Resistance of a wire is calculated as;

[tex]R = \frac{\rho L}{A} \\\\R = \frac{4\rho L}{\pi d^2} \\\\[/tex]

Resistance in wire A;

[tex]R = \frac{4\rho _AL_A}{\pi d_A^2}[/tex]

Resistance in wire B;

[tex]R = \frac{4\rho _BL_B}{\pi d_B^2}[/tex]

Power delivered in wire;

[tex]P = \frac{V^2}{R}[/tex]

Power delivered in wire A;

[tex]P = \frac{V^2_A}{R_A}[/tex]

Power delivered in wire B;

[tex]P = \frac{V^2_B}{R_B}[/tex]

Substitute in the value of R in Power delivered in wire A;

[tex]P_A = \frac{V^2_A}{R_A} = \frac{V^2_A \pi d^2_A}{4 \rho_A L_A}[/tex]

Substitute in the value of R in Power delivered in wire B;

[tex]P_B = \frac{V^2_B}{R_B} = \frac{V^2_B \pi d^2_B}{4 \rho_B L_B}[/tex]

Take the ratio of power delivered to A to power delivered to B;

[tex]\frac{P_A}{P_B} = (\frac{V^2_A \pi d^2_A}{4\rho_AL_A} ) *(\frac{4\rho_BL_B}{V^2_B \pi d^2_B})\\\\ \frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{\rho_AL_A} )*(\frac{\rho_BL_B}{V^2_B d^2_B})\\\\[/tex]

The wires are made of the same material, [tex]\rho _A = \rho_B[/tex]

[tex]\frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{L_A} )*(\frac{L_B}{V^2_B d^2_B})\\\\[/tex]

The wires are connected across the same potential; [tex]V_A = V_B[/tex]

[tex]\frac{P_A}{P_B} = (\frac{ d^2_A}{L_A} )* (\frac{L_B}{d^2_B} )[/tex]

wire A has seven times the diameter and seven times the length of wire B;

[tex]\frac{P_A}{P_B} = (\frac{ (7d_B)^2}{7L_B} )* (\frac{L_B}{d^2_B} )\\\\\frac{P_A}{P_B} = \frac{49d_B^2}{7L_B} *\frac{L_B}{d^2_B} \\\\\frac{P_A}{P_B} =\frac{49}{7} \\\\\frac{P_A}{P_B} = 7\\\\P_A : P_B = 7:1[/tex]

Therefore, the ratio of the power delivered to A to power delivered to B is

7 : 1

Answer:

resulting angular speed = 3.6 rev/s

Explanation:

We are given;

Initial angular speed; ω_i = 1.2 rev/s

Initial moment of inertia;I_i = 6 kg/m²

Final moment of inertia;I_f = 2 kg/m²

From conservation of angular momentum;

Initial angular momentum = Final angular momentum

Thus;

I_i × ω_i = I_f × ω_f

Making ω_f the subject, we have;

ω_f = (I_i × ω_i)/I_f

Plugging in the relevant values;

ω_f = (6 × 1.2)/2

ω_f = 3.6 rev/s

Answer:

F = 103.54N

Explanation:

In order to calculate the magnitude of the applied force, you take into account that the forces on the box are the applied force F and the weight of the box W.

The box moves with a constant velocity. By the Newton second law you have that the sum of forces must be equal to zero.

Furthermore, you have that the sum of forces are given by:

[tex]F-Wsin\theta=0[/tex]                (1)

F: applied force = ?

W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N

θ: degree of the incline = 25°

You solve the equation (1) for F:

[tex]F=Wsin\theta=(245N)sin(25\°)=103.54N[/tex]          (2)

The applied force on the box is 103.54N

Answer:

Q = 5.9 nC (Approx)

Explanation:

Given:

Further distance = 10 cm

Electric potential(V) = 190 v

Potential difference(V1) = 140 v

Find:

Magnitude of the charge Q

Computation:

V = KQ / r

190 = KQ / r.............Eq1

V1  = KQ / (r+10)

140 = KQ / (r+10) ............Eq2

From Eq2 and Eq1

r = 28 cm = 0.28 m

So,

190 = KQ / r

190 = (9×10⁹)(Q) / 0.28

53.2 = (9×10⁹)(Q)

5.9111 = (10⁹)(Q)

Q = 5.9 nC (Approx)

Answer:

We know that the density of the ice is smaller than the density of the water (and this is why the ice floats in water).

Dw > Di

Da is the density of the water and Di is the density of the ice

Since in Bowl A we have a volume V, only of water, then the mass of the bowl A is:

Dw*V.

Now, in the bowl B we have a combination of water and ice, suppose that Vw is the volume of water and Vi is the volume of ice, and we know that:

Vw + Vi = V.

Then the mass in this second bowl is:

Dw*Vw + Di*Vi = Dw*(V - Vi) + Di*Vi = Dw*V + (Di - Dw)*Vi

and we know that Dw > Di, then the left term is a negative term, then the mass of bowl B is smaller than the mass of bowl A.

Answer:

a

  [tex]KE = 7.17 *10^{7} \ J[/tex]

b

 [tex]t = 6411.09 \ s[/tex]

Explanation:

From the question we are told that

    The radius of the flywheel is  [tex]r = 1.50 \ m[/tex]

      The mass of the flywheel is [tex]m = 430 \ kg[/tex]

          The rotational speed of the flywheel is [tex]w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec[/tex]

      The power supplied by the motor is  [tex]P = 15.0 hp = 15 * 746 = 11190 \ W[/tex]

     Generally the moment of inertia of the flywheel is  mathematically represented as

       [tex]I = \frac{1}{2} mr^2[/tex]

substituting values

       [tex]I = \frac{1}{2} ( 430)(1.50)^2[/tex]

       [tex]I = 483.75 \ kgm^2[/tex]

The kinetic energy that is been stored is  

       [tex]KE = \frac{1}{2} * I * w^2[/tex]

substituting values

        [tex]KE = \frac{1}{2} * 483.75 * (544.61)^2[/tex]

        [tex]KE = 7.17 *10^{7} \ J[/tex]

Generally power is mathematically represented as

          [tex]P = \frac{KE}{t}[/tex]

=>      [tex]t = \frac{KE}{P}[/tex]

substituting the value

        [tex]t = \frac{7.17 *10^{7}}{11190}[/tex]

        [tex]t = 6411.09 \ s[/tex]

Answer:

Solution,

Mass=100 kg

Acceleration due to gravity(g)=24.79 m/s^2

Now,.

[tex]weight = m \times g \\ \: \: \: \: \: \: \: \: \: \: = 100 \times 24.79 \\ \: \: \: \: \: \: = 2479 \: newton[/tex]

hope this helps ..

Good luck on your assignment..

Answer:

88.29 N

Explanation:

mass of the ball = 4.5 kg

weight of the ball will be = mass x acceleration due to gravity(9.81 m/s^2)

weight W = 4.5 x 9.81 = 44.145 N

centrifugal forces Tc act on the ball as it swings.

At the top point of the vertical swing,

Tension on the rope = Tc - W.

At the bottom point of the vertical swing,

Tension on the rope = Tc + W

therefore,

difference in tension between these two points will be;

Net tension = tension at bottom minus tension at the top

= Tc + W - (Tc - W) = Tc + W -Tc + W

= 2W

imputing the value of the weight W, we have

2W = 2 x 44.145 = 88.29 N

Answer:

When the Net Forces are equal to 0

Explanation:

Momentum of a body can be defined as product of mass and velocity. It is in the same direction as in velocity. When the momentum of a body doesn't change, it is said to be conserved. If the momentum of a body is constant, the the net forces acting on a body becomes zero. When net forces acting on a body is zero, it means that no kinetic energy is being lost or gained, hence the kinetic energy is also conserved. If no energy is being gained or lost, it means that the body will experience no energy.