A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 21.6 m/s at an angle of 50.0° to the horizontal.

Required:
By how much does the ball clear or fall short (vertically) of clearing the crossbar?

Answer:

0.9432 m/s

Explanation:

We are given;

Mass of swimmer;m_s = 64.38 kg

Mass of log; m_l = 237 kg

Velocity of swimmer; v_s = 3.472 m/s

Now, if we consider the first log and the swimmer as our system, then the force between the swimmer and the log and the log and the swimmer are internal forces. Thus, there are no external forces and therefore momentum must be conserved.

So;

Initial momentum = final momentum

m_l × v_l = m_s × v_s

Where v_l is speed of the log relative to water

Making v_l the subject, we have;

v_l = (m_s × v_s)/m_l

Plugging in the relevant values, we have;

v_l = (64.38 × 3.472)/237

v_l = 0.9432 m/s

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 is a. Doubles

when the loops are increased in the coil then the magnetic field created doubles

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 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

Answer:

Vi = 4.8 m/s

Explanation:

First we need to find the magnitude of constant tangential acceleration. For that purpose we use the following formula between points A and C:

a = (Vf - Vi)/t

where,

a = constant tangential acceleration from A to C = ?

Vf = Final Velocity at C = 5 m/s

Vi = Initial Velocity at A = 4.5 m/s

t = time taken to move from A to C = 4 s

Therefore,

a = (5 m/s - 4.5 m/s)/4 s

a = 0.125 m/s²

Now, applying the same equation between points B and C:

a = (Vf - Vi)/t

where,

a = constant tangential acceleration from A to B = 0.125 m/s²

Vf = Final Velocity at C = 5 m/s

Vi = Initial Velocity at B = ?

t = time taken to move from B to C = 1.6 s

Therefore,

0.125 m/s² = (5 m/s - Vi)/1.6 s

Vi = 5 m/s - (0.125 m/s²)(1.6 s)

Vi = 4.8 m/s

Answer:

a. ac = 1844.66 m/s²

b. Fc = 265.63 N

Explanation:

a.

The centripetal acceleration of the ball is given as follows:

ac = v²/r

where,

ac = centripetal acceleration = ?

v = speed of ball = (87 mph)(1 h/ 3600 s)(1609.34 m / 1 mile) = 38.9 m/s

r = radius of path = 82 cm = 0.82 m

Therefore,

ac = (38.9 m/s)²/0.82 m  

ac = 1844.66 m/s²

b.

The centripetal force is given as:

Fc = (m)(ac)

Fc = (0.144 kg)(1844.66 m/s²)

Fc = 265.63 N

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.

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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:

Container A will weigh more

Explanation:

Both containers are identical, so we assume that they weigh the same.

They both have the same volume, and will contain an equal volume of a material.

Since they both contain water to the top, this means that their volume is fully occupied. But container B contain a block of wood floating in it.

The fact that the block of wood floats in the water in container B shows that it is less dense than the water around it, and in the container A, this same space is completely filled with water.

What we derive from this is that the portion of space contained by the block of wood in container B is occupied by water in container A, but, in container B, the density of this space is lesser now, since the wood block floats.

Since density is mass per unit volume, and weight is proportional to mass, then we can see that the weight of this volume portion in container B is lesser than that of container A. The consequence is that container A will weigh more than container B because of this extra weight.

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:

Explanation:

From the given information:

(a)

the average magnetic flux through each turn of the inner solenoid can be calculated by the formula:

[tex]\phi _ 1 = B_1 A[/tex]

[tex]\phi _ 1 = ( \mu_o \dfrac{N_i}{l} i_1)(\pi ( \dfrac{d}{2})^2)[/tex]

[tex]\phi _ 1 = ( 4 \pi *10^{-7} \ T. m/A ) ( \dfrac{310}{20*10^{-2} \ m }) (0.130 \ A) ( \pi ( \dfrac{2.20*10^{-2} \ m }{2})^ 2[/tex]

[tex]\phi_1 = 9.625 * 10^{-8} \ Wb[/tex]

(b)

The mutual inductance of the two solenoids is calculated by the formula:

[tex]M = 23 *\dfrac{9.625*10^{-8} \ Wb}{0.130 \ A}[/tex]

M = [tex]1.703 *10^{-5}[/tex] H

(c)

the emf induced in the outer solenoid by the changing current in the inner solenoid can be calculate by using the formula:

[tex]\varepsilon = -N_o \dfrac{d \phi_1}{dt}[/tex]

[tex]\varepsilon = -M \dfrac{d i_1}{dt}[/tex]

[tex]\varepsilon = -(1.703*10^{-5} \ H) * (1800 \ A/s)[/tex]

[tex]\varepsilon = -0.030654 \ V[/tex]

[tex]\varepsilon = -30.65 \ V[/tex]

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:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

Given that

D(i) = 200 mm

D(o) = 400 mm

q' = 24000 W/m³

k(r) = 0.5 W/m.K

k(s) = 4 W/m.K

k(h) = 25 W/m².K

The expression for heat generation is given by

q = πr²Lq'

q = π . 0.1² . L . 24000

q = 754L W/m

Thermal conduction resistance, R(cond) = 0.0276/L

Thermal conduction resistance, R(conv) = 0.0318/L

Using energy balance equation,

Energy going in = Energy coming out

Which is = q, which is 754L

From the attachment, we deduce that the temperature between the rod and the sleeve is 71.8° C

At the same time, we find out that the temperature on the outer surface is 51° C

Also, from the second attachment, the temperature at the center of the rod was calculated to be, 191.8° C

Explanation:

In the given question, the two metal spheres were hanged with the nylon thread.

When these two spheres were brought close together, they attracted each other. The attraction between these spheres is the result of the opposite charges between them.

The possible ways by which these two metal spheres can be charged are by induction that is touching the metal or by rubbing them.

During induction, the same charges are transferred to each sphere. In this case, either both the spheres will be negatively charged or positively charged.

It is not possible that after the sphere touch each other they will cling together because the same charge repels each other and during touching, if one sphere is neutral, then the charged one will transfer the same charge. And as we know that same charge repel each other therefore they will repel each other.

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:

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:

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 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:

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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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:

[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:

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:

Umax = 105.8nJ

Umin =-105.8nJ

Umax-Umin = 211.6nJ

Explanation:

Complete Question

The complete question is shown on the first uploaded image

Answer:

the change in energy of the block due to friction is  [tex]E = -126 \ J[/tex]

Explanation:

From the question we are told that

    The  frictional force is  [tex]F_f = 21 \ N[/tex]

    The mass of the block is  [tex]m_b = 8 \ kg[/tex]

    The length of the ramp is  [tex]l = 6 \ m[/tex]

    The height of the block is  [tex]h = 2 \ m[/tex]

The change in energy of the block due to friction is mathematically represented as

     [tex]\Delta E = - F_s * l[/tex]

The negative sign is to show that the frictional force is acting against the direction of the block movement

  Now substituting values

      [tex]\Delta E = -(21)* 6[/tex]

      [tex]\Delta E = -126 \ J[/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:

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.

To know more about friction, follow

https://brainly.com/question/24386803