1. Consider the ball example in the introduction when a ball is dropped from 3 meters. After the ball bounces, it raises to a height of 2 meters. The mass of the ball is 0.5 kg a. Calculate the speed of the ball right before the bounce. b. How much energy was converted into heat after the ball bounced off the ground

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 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:

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:

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

Read more at; https://brainly.com/question/22150651

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

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:

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.

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

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]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:

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:

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.

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:

The 40g mass will be attached at 69 cm

Explanation:

First, make a sketch of the meterstick with the masses placed on it;

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

               ↓                    Δ                      ↓

             20 g.................50 cm.................40g

                         38 cm                  y cm  

Apply principle of moment;

sum of clockwise moment = sum of anticlockwise moment

40y = 20 (38)

40y = 760

y = 760 / 40

y = 19 cm

Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm

              12cm             50 cm              69cm

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

               ↓                    Δ                      ↓

             20 g.................50 cm.................40g

                         38 cm                 19 cm                                              

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:

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:

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:

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:

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:

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

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 is a. Doubles

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

Answer:

When the spring in the ballistic pendulum is compressed, energy is stored up in the spring as potential energy. When the steel ball is launched by the spring, the stored up potential energy of the compressed spring is transformed and transferred into the kinetic energy of the steel ball as it flies off to hit its target. On hitting the soft target, some of the kinetic energy of the steel ball is transferred to the soft target (since they stick together), and they both start to swing together. During the process of swinging, the system's energy is transformed between kinetic and potential energy. At the maximum  displacement of the ball from its point of rest, all the energy is converted to potential energy of the system. At the lowest point of travel (at the rest point), all the energy of the system is transformed into kinetic energy. In between these two points, energy the energy of the system is a combination of both kinetic and potential energy.

In the end, all the energy will be transformed and lost as heat to the surrounding; due to the air resistance around; bringing the system to a halt.

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:

Explanation:

In this interesting exercise we have that spring A is 3 cm longer, due to previous experiments if these experiments did not reach the non-linear elongation point, the cosecant Km of the spring must remain the same, therefore when we lengthen the two springs these the longitudinal are lengthened.

As a consequence of the above according to Hockey law, the prediction of lengthening is the same, therefore the outside is the same in two two systems

            F = K Δx

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

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]