If f(x) = 27+1, what is f\x) when x = 3?
1

Answer:

v = Square Root of StartFraction 2 times K E over m EndFraction.

Explanation:

Kinetic Energy ,K.E = 1/2 × mass × velocity squared

K.E = 1/2 × mv2

Hence v2 = (2× K.E)/m

v = √ 2K.E/m

Expressing it in computer language we have :

v = Square Root of StartFraction 2 times K E over m EndFraction.

Answer:

v = Square Root of StartFraction 2 times K E over m EndFraction

Explanation:

also d on edgenuity:)

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:

0.645 N/M

Explanation:

Given

Mass=50.00g

We have to convert into the kg

So Mass =0.050 Kg

[tex]Time\ = \frac{14}{8}\ = 1.75\ sec[/tex]

We know that

[tex]T\ =2\ PI\sqrt{\frac{M}{K} }[/tex]........................Eq(1)

Where T= time

and M= Mass

K= Stiffness constant

On squaring both side we get

[tex]K=\frac{4\pi^{2} M}{T^{2} }[/tex]....Eq(2)

Putting the value of M ,T and π in Eq(2) we get

K=0.645 N/M

Answer:

A) v = 148,242.72 m/s

B) v = 6,328,025.58 m/s

Explanation:

To solve this, we will equate electric potential to kinetic energy.

Formula for Electric potential is qV where q is charge and V is potential difference.

While formula for kinetic energy is ½mv² where m is mass and v is velocity

Thus;

qV = ½mv²

Let us make the velocity the formula;

v = √(2qV/m)

A) PROTON

Charge of proton has a constant value of 1.6 × 10^(-19) C

Mass of proton has a constant value of 1.66 × 10^(-27) kg

We are given that potential difference = 114 V.

So, v = √(2qV/m)

Thus; v = √(2*1.6 × 10^(-19)*114/(1.66 × 10^(-27)))

v = 148,242.72 m/s

B) ELECTRON

Charge of electron has a constant value of 1.6 × 10^(-19) C

Mass of electron has a constant value of 9.11 × 10^(-31) kg

v = √(2qV/m)

Thus;

v = √(2*1.6*10^(-19)*114)/(9.11 × 10^(-31)))

v = 6,328,025.58 m/s

Answer:

The mass of the lead will be "1.127 kg".

Explanation:

The given values are:

(Ice) m₁ = 50 g i.e.,

0.050 kg

(Water) m₂ = 195 g i.e.,

0.190 kg

(Copper cup) m₃ = 100 g i.e.,

0.100 kg

m₁, m₂ and m₃ at temperature,

t₁ = 0°C

Temperature of lead,

t₂ = 96°C

Temperature of Final equilibrium,

t₃ = 12°C

Let m₄ be the mass of the lead.

On applying formula, we get

⇒  [tex]m_{1}L+m_{1}s_{1} \Delta t+m_{2}s_{2} \Delta t+m_{2}s_{2} \Delta t=m_{4}s_{4} \Delta t[/tex]

On putting the estimated values, we get

⇒  [tex](0.050)(334)+(0.050)(4186)(12-0)+(0.190)(4186)(12-0)+(0.100)(387)(12-0)=m_{4} (128)(96-12)[/tex]

⇒  [tex]16.7+2511.6+9544.08+50.7=10752\times m_{4}[/tex]

⇒  [tex]12,123.08=10752\times m_{4}[/tex]

⇒  [tex]m_{4}=\frac{12,123.08}{10752}[/tex]

⇒  [tex]m_{4}=1.127 \ kg[/tex]

Answer: The speed of the ball is 7.64 m/s.

Explanation:

The distance between the player and the pins is 16.5m

if the velocity of the ball is V, then the time in which the ball reaches the pins is:

T = 16.5/V

Now, after this point, the sound needs

T' = 16,5/340 = 0.049 seconds to reach the player, this means that the time in that the ball needs to reach te pins is:

2.65 s - 0.49s = 2.16s

Then we have:

T = 2.16s = 16.5/V

V = 16.5/2.16 m/s = 7.64 m/s

Water requires more energy to raise its temperature than sand does.  In fact, of all the common substances that we see around us every day, water is one of the BEST at storing heat energy.

This is a big part of the reason why we use frozen water to cool our soda, instead of cold wood or cold steel balls.  

It's also a big part of the reason why we warm up the bed in the Winter with a hot water bag, instead of a bag of hot rocks or hot BBs.

On a hot summer day, the temperature of the sand is much higher than the temperature of the water. The same amount of energy was supplied by the sun to both the sand and the water, but the water required more energy to raise its temperature than the sand.

The specific heat of any substance is explained by the amount of heat required to increase the temperature by 1 degree; here, the specific heat of water is much higher than that of sand. The sand needs 670 joules of energy to raise the temperature, while the water needs nearly 3800 joules of energy to raise one degree of temperature.

Despite the fact that the sun cast the same amount of light on both water and sand, sand heated up faster than water. The water has a high latent heat of vaporization, which means it needs more energy to vaporize. The animal body maintains homeostasis as a result of this water.

Hence, water requires more energy to raise the temperature due to its high specific heat.

Learn more about the specific heat, here

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#SPJ2

Answer: -254.51°O

Explanation:

Ok, in our scale, we have:

-182.9°C corresponds to 100° O

-218.3°C corresponds to 0°

Then we can find the slope of this relation as:

S = (100° - 0°)/(-182.9°C - (-218.3°C)) = 2.82°O/°C

So we can have the linear relationship between the scales is:

Y = (2.82°O/°C)*X + B

in this relation, X is the temperature in Celcius and Y is the temperature in the new scale.

And we know that when X = -182.9°C, we must have Y = 0°O

then:

0 = (2.82°O/°C)*(-182.9°C) + B

B = ( 2.82°O/°C*189.9°C) = 515.778°O.

now, we want to find the 0 K in this scale, and we know that:

0 K = -273.15°C

So we can use X =  -273.15°C in our previous equation and get:

Y = (2.82°O/°C)*(-273.15°C) + 515.778°O = -254.51°O

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:

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:

[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:

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:

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:

B) Make sure spheres are not in contact with each other. Rub the glass rod on the silk to charge the rod positively (hence to charge piece of silk negatively). Touch one of the spheres with the glass rod then touch another one with the piece of silk.

Explanation:

In static electricity, frictional rubbing can excite electrons, and make them jump from one material to another. Rubbing the glass rod and the silk together makes electron flow from the glass rod to the silk piece. On separation, the rod and silk will be seen to have equal amount of charge but, the glass rod will be positively charged from losing electron, and the silk piece will be negatively charged from gaining extra electrons. When they are each made to touch the metal spheres that are not in contact, the spheres will acquire equal amount of charges opposite the charge on the material that was used to charge them.

Answer:

Friction

Explanation:

Answer:

gravitational force = m*g

= 9.1*10^-31 * 9.8

= 8.92*10^-30 N

acts downward

---------------------

electric force = q*E

= (1.6*10^-19)*(90)

= 1.44*10^-17 N

DIRECTION : upward

----------------------

magnetic force = q*V*B

= (1.6*10^-19)*(8*10^6)*(50*10^-6)

= 6.4*10^-17 N

direction = downward

Answer:

This is an inelastic collision. This means, unfortunately, that KE cannot save you, at least in the problem's current form.  

Let's see what conservation of momentum in both directions does ya:

Conservation in the x direction:

Only 1 object here has a momentum in the x direction initally.  

m1v1i + 0 = (m1 + m2)(vx)

3.09(5.10) = (3.09 + 2.52)Vx

Vx = 2.81 m/s

Explanation:

Conservation in the y direction:

Again, only 1 object here has initial velocity in the y:

0 + m2v2i = (m1 +m2)Vy

(2.52)(-3.36) = (2.52 + 3.09)Vy

Vy = -1.51 m/s

++++++++++++++++++++

Now that you have Vx and Vy of the composite object, you can find the final velocity by doing Vf = √Vx^2 + Vy^2)

Vf = √(2.81)^2 + (-1.51)^2

Vf = 3.19 m/s

Answer:

Umax = 105.8nJ

Umin =-105.8nJ

Umax-Umin = 211.6nJ

Explanation:

Answer:

The values of the forces are

      [tex]F_1 = 10.6 \ N[/tex] ,  [tex]F_2 = 21.33 \ N[/tex]

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

      The magnitude of  F is  [tex]F = 32 \ N[/tex]

Generally at equilibrium the torque is mathematically evaluated as

          [tex]\sum \tau = 0[/tex]

From the diagram we have

         [tex]r * F_1 - [\frac{r}{2} ] F_2 + 0 F = 0[/tex]

=>       [tex]F_1 = 0.5 F_2[/tex]

Generally at equilibrium the Force is  mathematically evaluated  as

         [tex]\sum F = 0[/tex]

 From the diagram

        [tex]F - F_ 1 - F_2 = 0[/tex]

substituting values

     [tex]32 - (0.5F_2 ) - F_2 = 0[/tex]

      [tex]F_2 = 21.33 \ N[/tex]

So  

      [tex]F_1 = 0.5 * 21.33[/tex]

       [tex]F_1 = 10.6 \ N[/tex]

Answer:

The correct answer is option E

Explanation:

Relative density of the different phases of the same compound like water are basically determined by their number of molecules per volume when each of the molecules have the same mass in each of their phases.

Now, for the water vapor phase, it's molecules have very little interaction with themselves and so they are at large distance apart, whereas in ice(solid), molecules are in continuous contact with each other because they are at close distance between each other. Therefore, it's obvious that there are less molecules per liter in water vapour than in ice, and thus the density is smaller.

The correct answer is option E

Answer:

Radius at liftoff 8.98 m

Explanation:

At the working altitude;

maximum radius = 24 m

air pressure = 0.030 atm

air temperature = 200 K

At liftoff;

temperature = 349 K

pressure = 1 atm

radius = ?

First, we assume balloon is spherical in nature,

and that the working gas obeys the gas laws.

from the radius, we can find the volume of the balloon at working atmosphere.

Volume of a sphere = [tex]\frac{4}{3} \pi r^{3}[/tex]

volume of balloon = [tex]\frac{4}{3}[/tex] x 3.142 x [tex]24^{3}[/tex] = 57913.34 m^3

using the gas equation,

[tex]\frac{P1V1}{T1}[/tex] = [tex]\frac{P2V2}{T2}[/tex]

The subscript 1 indicates the properties of the gas at working altitude, and the subscript 2 indicates properties of the gas at liftoff.

imputing values, we have

[tex]\frac{0.03*57913.34}{200}[/tex] = [tex]\frac{1*V2}{349}[/tex]

0.03 x 57913.34 x 349 = 200V2

V2 = 606352.67/200 = 3031.76 m^3  this is the volume occupied by the gas in the balloon at liftoff.

from the formula volume of a sphere,

V =  [tex]\frac{4}{3} \pi r^{3}[/tex] =  [tex]\frac{4}{3}[/tex] x 3.142 x [tex]r^{3}[/tex] = 3031.76

4.19[tex]r^{3}[/tex]  = 3031.76

[tex]r^{3}[/tex] = 3031.76/4.19  

radius r of the balloon on liftoff = [tex]\sqrt[3]{723.57}[/tex] = 8.98 m

Answer:

0.9 cm

Explanation:

The computation in the increase in the length of the joined rod is shown below:

As we know that

Increase in length = increase in the length of aluminum rod + increase in The length of steel rod

[tex]= 10cm \times 2.4e - 5\times (90-15) + 80cm\times 1.2e - 5\times (90-15)[/tex]

= 0.9 cm

We simply added the length of aluminium rod and length of steel rod so that the length of the joined rod could come and the same is to be considered  

We have that for the Question, it can be said that

the increase in the length of the joined rod

[tex]\triangle L= 0.9cm[/tex]

From the question we are told

An aluminum rod is 10.0 cm long and a steel rod is 80.0 cm long when both rods are at a temperature of 15°C. Both rods have the same diameter.

The rods are now joined end-to-end to form a rod 90.0 cm long.

If the temperature is now raised from 15°C to 90°C,

what is the increase in the length of the joined rod?

The coefficient of linear expansion of aluminum is 2.4 × 10-5 K-1 and that of steel is 1.2 × 10-5 K-1.

Generally the equation for the change length  is mathematically given as

[tex]\triangle L=change in L of aluminium rod +change in L of steel rod\\\\Therefore\\\\\triangle L= 10*2.4*10^{-5}*(90-15) + 80cm*1.2*10^{-5}*(90-15)\\\\\triangle L= 0.9cm\\\\[/tex]

Hence

the increase in the length of the joined rod

[tex]\triangle L= 0.9cm[/tex]

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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:

So work done is 1.2J

Explanation:

Applied force=F=4N

Displacement covered=S=0.3m

Work done=?

As we know that

Work done = Force applied * Displacement covered

Work done=4N*0.3m

Work done =1.2 joules

I hope it will help you

Answer:

See the answer below.

Explanation:

"Latent Heat", also called the "Heat of Vaporization", is the amount of energy necessary to change a liquid to a vapour at constant temperature and pressure.

Best Regards!

Answer:

The capillary rise of the glycerin is most nearly  [tex]y = 0.0204 \ m[/tex]

Explanation:

From the question we are told that

  The diameter of the glass tube is  [tex]d = 1 \ mm = 0.001 \ m[/tex]

   The density of glycerin is  [tex]\rho = 1260 \ kg /m^3[/tex]

   The surface tension of the glycerin is [tex]\sigma = 6.3 *10^{-2} \ N /m[/tex]

The capillary rise of the glycerin is mathematically represented as

       [tex]y = \frac{4 * \sigma * cos (\theta )}{ \rho * g * d}[/tex]

substituting value  

       [tex]y = \frac{4 * 6.3 *10^{-2} * cos (0 )}{ 1260 * 9.8 * 0.001}[/tex]

      [tex]y = 0.0204 \ m[/tex]

Therefore the height  of the glass tube  the glycerin was able to cover is

[tex]y = 0.0204 \ m[/tex]  

Question: The coefficient of linear expansion of steel is 11 x 10⁻⁶ per °c . A steel ball has a volume of

exactly 100 cm³ at 0 C. When heated to 100 C, its volume becomes:

Answer:

100.11 cm³

Explanation:

From the question,

γ = (v₂-v₁)/(v₁Δt)...................... Equation 1

Where γ = coefficient of volume expansion, v₂ = final volume, v₁ = initial volume, Δt = change in temperature.

make v₂ the subject of the equation

v₂ = v₁+γv₁Δt..................... Equation 2

Given: v₁ = 100 cm³, γ = 11×10⁻⁶/°C, Δt = 100 °C.

Substitute into equation 2

v₂ = 100+100(11×10⁻⁶)(100)

v₂ = 100+0.11

v₂ = 100.11 cm³

Answer:

[tex]I_e = \frac{1}{3}*m*L^2[/tex]

Explanation:

Solution:-

- Here we are given the moment of inertia of a uniform slender rod with mass ( m ) and length ( L ). The thickness / radius / diameter of the rod is considered to be insignificant.

- The moment of inertia ( Ir ) of a rod with an axis perpendicular to it at its center is given as:

                                   [tex]I_r = \frac{1}{12}*m*L^2[/tex]

- We are to determine the moment of inertia of the rod at any one of its ends using the parallel axis theorem.

- The theorem is mostly used to translate the pivotal axis to any point on the mass or in space. With respect to that point the moment of inertia is determined using the parallel axis theorem. The moment of inertia of the object at its center of mass must be known to apply the theorem.

- The theorem is expressed as:

                                  [tex]I_e = I_r + m*d^2[/tex]

Here,

                       d: Is the distance between the center of mass and the arbitrary point.

- Since we are asked to determine the moment of inertial at one of the rod's ends. We can evaluate the distance " d " from its center of mass to its end. The center lies at " L / 2 " distance from either of its ends. Hence, d = L / 2.

- We will plug in the parameters in the theorem and evaluate:

                               [tex]I_e = \frac{1}{12}*m*L^2 + m*[\frac{L}{2} ]^2 \\\\I_e = \frac{1}{12}*m*L^2 + m*\frac{L^2}{4} \\\\I_e = m*L^2 * [ \frac{1}{12}+ \frac{3}{12} ] = m*L^2 *\frac{4}{12} \\\\I_e = \frac{1}{3}*m*L^2[/tex]

Answer:

A) See Annex

B) Fq₁₂ = K *  Q₁*Q₂ /16 [N] (repulsion force)

C)  Fq₃₂  = K * Q₃*Q₂ /16 [N] (repulsion force)

D) Fq₄₂ = K * Q₄*Q₂ /32 [N] (attraction force)

E) Net force (its components)

Fnx = (2,59/64 )* K*Q²  [N] in direction of original Fq₃₂

Fny =(2,59/64 )* K*Q² [N] in direction of original Fq₁₂

Explanation:

For calculation of d (diagonal of the square, we apply Pythagoras Theorem)

d² = L² + L²    ⇒  d² = 2*L²     ⇒ d = √2*L²   ⇒ d= (√2 )*L

d = 4√2 units of length   (we will assume meters, to work with MKS system of units)

B) Force of Q₁ exerts on charge Q₂

Fq₁₂  = K * Q₁*Q₂ /(L)²     Fq₁₂ = K *  Q₁*Q₂ /16 (repulsion force in the direction indicated in annex)

C) Force of Q₃ exerts on charge Q₂

Fq₃₂  = K * Q₃*Q₂ /(L)²     Fq₃₂  = K * Q₃*Q₂ /16  (repulsion force in the direction indicated in annex)

D) Force of -Q₄ exerts on charge Q₂

Fq₄₂ = K * Q₄*Q₂ / (d)²      Fq₄₂ = K * Q₄*Q₂ /32 (Attraction force in the direction indicated in annex)

E) Net force in the case all charges have the same magnitude Q (keeping the negative sign in Q₄)

Let´s take the force that  Q₄ exerts on Q₂  and Q₂ = Q  ( magnitude) and

Q₄ = -Q

Then the force is:

F₄₂ = K * Q*Q / 32       F₄₂  = K* Q²/32  [N]

We should get its components

F₄₂(x) = [K*Q²/32 ]* √2/2   and so is F₄₂(y)  =  [K*Q²/32 ]* √2/2

Note that this components have opposite direction than forces  Fq₁₂  and

Fq₃₂  respectively, and that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and  F₄₂(y) respectively

In new conditions

Fq₁₂ = K *  Q₁*Q₂ /16    becomes  Fq₁₂ = K * Q²/ 16 [N]   and

Fq₃₂ = K* Q₃*Q₂ /16      becomes   Fq₃₂ = K* Q² /16  [N]

Note that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and  F₄₂(y) respectively

Then over x-axis we subtract Fq₃₂ - F₄₂(x)  = Fnx

and over y-axis, we subtract   Fq₁₂ - F₄₂(y) = Fny

And we get:

Fnx = K* Q² /16 - [K*Q²/32 ]* √2/2  ⇒  Fnx =  K*Q² [1/16 - √2/64]

Fnx = (2,59/64 )* K*Q²

Fny has the same magnitude  then

Fny =(2,59/64 )* K*Q²

The fact that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and  F₄₂(y) respectively, means that Fnx and Fny remains as repulsion forces

Answer:

The energy dissipated is  [tex]E = 9.8 J[/tex]

Explanation:

From the question we are told that

    The distance covered at constant velocity d =  2 m  

      The velocity is  [tex]v = 1.5 \ m/s[/tex]

Generally at constant velocity the magnetic  force on the bar is mathematically represented as

         [tex]F = m * g[/tex]

substituting values

        [tex]F = 0.5 * 9.8[/tex]

        [tex]F = 4.9 \ N[/tex]

The energy dissipated is mathematically evaluate as

         [tex]E = F * d[/tex]

 substituting the value

         [tex]E = 4.9 * 2[/tex]

        [tex]E = 9.8 J[/tex]