Water is traveling through a horizontal pipe with a speed of 1.7 m/s and at a pressure of 205 kPa. This pipe is reduced to a new pipe which has a diameter half that of the first section of pipe. Determine the speed and pressure of the water in the new, reduced in size pipe.

Answer:

E = 3.45*10^-19 N/C

Explanation:

a) The electric field between two parallel plates id given by the following formula:

[tex]E=\frac{\sigma}{\epsilon_o}[/tex]           (1)

where:

σ: surface charge density of the plates = 39.0nC/m^2

εo: dielectric permittivity of vacuum = 8.85*10^-12 C/Nm^2

You replace these values in the equation (1):

[tex]E=\frac{39.0*10^{-9}C/m^2}{8.85*10^{-12}C^2/Nm^2}\\\\E=3.45*10^{-19}\frac{N}{C}[/tex]

The electric field in between the parallel plates is 3.45*10^-19 N/C

Answer:

Maximum height of jump on Rhea is 37.16 times of that on Earth, i.e 37.16h

Maximum range of jump on Rhea is 37.16 of times that on Earth, i.e 37.16R

Explanation:

The acceleration due to gravity on Rhea = 0.264 m/s^2

Acceleration due to gravity on earth here = 9.81 m/s^2

this means that the acceleration due to gravity g on earth is 9.81/0.264 = 37.16 times that on Rhea.

maximum height that can be achieved by the froghopper is given by the equation;

h = [tex]\frac{u^{2}sin^{2} \alpha}{2g}[/tex]

let us put all the numerator of the equation as k, since the velocity of take off is the same for Earth and Rhea. The equation is simplified to

h = [tex]\frac{k}{2g}[/tex]

for earth,

h =  [tex]\frac{k}{2*9.81}[/tex] =  [tex]\frac{k}{19.62}[/tex]

for Rhea,

h  =  [tex]\frac{k}{2*0.264}[/tex] =  [tex]\frac{k}{0.528}[/tex]

therefore,

h on Rhea is [tex]\frac{k}{0.528}[/tex] ÷ [tex]\frac{k}{19.62}[/tex] = 37.16 times of that on Earth, i.e 37.16h

Equation for range R is given as

R =  [tex]\frac{u^{2}sin 2\alpha}{g}[/tex]

following the same approach as before,

R on Rhea will be [tex]\frac{k}{0.264}[/tex] ÷ [tex]\frac{k}{9.81}[/tex] = 37.16 of times that on Earth, i.e 37.16R

Answer:

The time taken is  [tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]

Explanation:

From the question we are told that

     The speed of the current is  [tex]v__{R}}[/tex]

     The speed of the swimmer in direction of current is [tex]v__{S}}[/tex]

      The distance traveled by the swimmer is  [tex]D[/tex]

       The time taken to travel this distance is  [tex]t_{out}[/tex]

      The speed of the swimmer against  direction of current is  [tex]v__{s}}[/tex]

The resultant speed for downstream current is

       [tex]V_{r} = v__{S}} +v__{R}}[/tex]

The time taken can be mathematically represented as

      [tex]t_{out} = \frac{D}{V_{r}}[/tex]

      [tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

[tex]K=qV[/tex]          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

[tex]K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J[/tex]

The kinetic energy of the particle is also:

[tex]K=\frac{1}{2}mv^2[/tex]       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}[/tex]

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

[tex]|F_B|=qvBsin(\theta)[/tex]

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

[tex]|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N[/tex]

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

[tex]r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm[/tex]

the radius of the trajectory of the electron is 10.1 cm

The speed, magnetic force and radius are respectively; 10.75 * 10⁶ m/s; 7.57 * 10⁻¹² N; 0.101 m

A) We know that the formula for kinetic energy can be expressed as;

K = qV

where;

q is charge of the particle = 2e = 2(1.6 × 10⁻¹⁹ C) = 3.2 × 10⁻¹⁹ C

V is potential difference = 1.2 × 10⁶ V

K = 3.2 × 10⁻¹⁹ *  1.2 × 10⁶

K = 3.84 × 10⁻¹³ J

Also, formula for kinetic energy is;

K = ¹/₂mv²

where v is speed

Thus;

v = √(2K/m)

v = √(2 * 3.84 × 10⁻¹³)/(6.64 * 10⁻²⁷)

v = 10.75 * 10⁶ m/s

B) The magnetic force is given by the formula;

F_b = qvB

F_b = (3.2 × 10⁻¹⁹ * 10.75 * 10⁶ * 2.2)

F_b = 7.57 * 10⁻¹² N

C) The formula to find the radius is;

r = mv/qB

r = (6.64 * 10⁻²⁷ * 10.75 * 10⁶)/(1.6 × 10⁻¹⁹ * 2.2)

r = 0.101 m

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Answer:

Explanation:

The work required to push will be equal to work done by friction . Let  d be the displacement required .

force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction

work done by force of friction

mg x μ x d   = 660

80 x 9.8 x .272 x d = 660

d = 3 .1 m .

Answer:

Since you would have to do work on the charge to bring it back to its original position, the charge moves to a position of lower potential and lower potential energy.

The positive charge is released from a point such that it will move along an uniform electric field to the position of lower potential and lower potential energy. Therefore, option (A) is correct,

When a positive charge (say +Q) is released from a point (say A) and moves in an uniform electric field to reach the point (say B), then some work is done on the charge. This work done is given as,

[tex]W=+Q(V_{A}-V_{B})[/tex]

Here, [tex]V_{A}[/tex] and [tex]V_{B}[/tex] are the potential differences between the points A and B respectively..

This means the charge is moving from higher potential to lower potential. And since it is moving along the uniform electric field, therefore the electric potential energy of charged system is decreased.

Thus, we conclude that on releasing the positive charge from a point, it starts moving along the electric field towards the direction of lower electric potential and lower electric potential energy. Hence, option (A) is correct.

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Answer:

a) When moving towards a high pressure center the pressure values ​​increase in the equipment

b) This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

When moving towards a high pressure center the pressure values ​​increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

[tex]U=(M+m)gh[/tex]       (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

[tex]h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m[/tex]

The potential energy is:

[tex]U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J[/tex]

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

[tex]U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}[/tex]

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

[tex]Mv_1+mv_2=(M+m)v[/tex]    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

[tex]M(0)+mv_2=(M+m)v[/tex]    

[tex]v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}[/tex]

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

[tex]I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}[/tex]

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

[tex]T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N[/tex]    

The force on the cord after the impact is 2.59N

Answer:

Explanation:

Apply the law of conversion of energy

[tex]V_f = \sqrt{2gh}[/tex]

where,

h = height of which the bullet and block rise after impact

[tex]h = L - Lcos\theta\\\\h = 2.2 - (2.2*cos60)\\\\h = 1.1m[/tex]

Therefore,

[tex]V_f = \sqrt{2gh}\\\\V_f = \sqrt{2*9.8*1.1}\\\\V_f = 4.64m/s[/tex]

From conservation of momentum principle, [tex]m_Bv_B = 0[/tex]

[tex]m_ov_o + m_Bv_B = (m_b+m_B)V_f\\\\0.006V_o = (0.006+1)*4.64\\\\V_o = 777.97m/s[/tex]

C) The force in the cable is due to the centrfugal force of the system, which is due to the motion of the system is a curved path and weight of the system

[tex]F = \frac{m_b+m_B}{L}V_f^2 + (m_b+m_B)g\\\\F = \frac{0.006+1}{2.2}*4.64^2 + (0.006+1)9.81\\\\F = 19.71N[/tex]

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Answer: [tex]x=\frac{2}{3}t^3+3t+1[/tex]

Explanation:

Given

velocity of object is given by

[tex]v(t)=3+2t^2[/tex]

and we know change of position w.r.t time is velocity

[tex]\Rightarrow \dfrac{dx}{dt}=v[/tex]

[tex]\Rightarrow \dfrac{dx}{dt}=3+2t^2[/tex]

[tex]\Rightarrow dx=(3+2t^2)dt[/tex]

Integrating both sides we get

[tex]\Rightarrow \int_{1}^{x}dx=\int_{0}^{t}(3+2t^2)dt[/tex]

[tex]\Rightarrow x\mid _{1}^{x}=(3t+\frac{2}{3}t^3)\mid _{0}^{t}[/tex]

[tex]\Rightarrow x-1=3(t-0)+\frac{2}{3}(t^3-0)[/tex]

[tex]\Rightarrow x=\frac{2}{3}t^3+3t+1[/tex]

Answer

The rate at which the magnetic field is changing is  [tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]

Explanation

From the question we are told that

   The electric field strength is [tex]E = 3.5mV/m = 3.5 *10^{-3} \ V/m[/tex]

    The radius is  [tex]r = 1.5 \ m[/tex]

The rate of change of the  magnetic  field  is mathematically represented as

        [tex]\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}[/tex]

Where [tex]dl[/tex] is change of a unit length

     [tex]\frac{d \phi}{dt} = A * \frac{dB}{dt}[/tex]

Where A is the area which is mathematically represented as

     [tex]A = \pi r^2[/tex]

    So

    [tex]E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )[/tex]  

  [tex]E L = ( \pi r^2) (\frac{dB}{dt} )[/tex]  

where L is the circumference of the circle which is mathematically represented as

     [tex]L = 2 \pi r[/tex]

So

     [tex]E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ][/tex]

      [tex]E = \frac{r}{2} [\frac{dB}{dt} ][/tex]

       [tex][\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }[/tex]

substituting values

      [tex][\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }[/tex]

      [tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]    

Answer:

WEIGHT ON MOON IS 0.2004N

Explanation:

mass of the body=120g=[tex]\frac{120}{1000}[/tex]kg=0.12kg (we will convert g into kg)

gravity on moon=1.67m/s²( to find the mass of anybody on another we should know its gravity)

as we know that (from the formula of weight)

weight=mass×gravity

w=mg

w=0.12kg²×1.67m/s²

w=0.2004N

Answer:

[tex]\tau = 1\ ms[/tex]

Explanation:

First we need to find the capacitance of the capacitor.

The capacitance is given by:

[tex]C = \epsilon_0 * area / distance[/tex]

Where [tex]\epsilon_0[/tex] is the air permittivity, which is approximately 8.85 * 10^(-12)

The radius is 12/2 = 6 mm = 0.006 m, so the area of the capacitor is:

[tex]Area = \pi * radius^{2}\\Area = \pi * 0.006^2\\Area = 113.1 * 10^{-6}\ m^2[/tex]

So the capacitance is:

[tex]C = \frac{8.85 * 10^{-12} * 113.1 * 10^{-6}}{0.001}[/tex]

[tex]C = 10^{-12}\ F = 1\ pF[/tex]

The time constant of a rc-circuit is given by:

[tex]\tau = RC[/tex]

So we have that:

[tex]\tau = 10^{9} * 10^{-12} = 10^{-3}\ s = 1\ ms[/tex]

Answer:

i) angle of incidence;i = 29.43°

ii) δm = 38.92°

Explanation:

Prism is equilateral so angle of prism (A) = 60°

Refractive index of glass; n_glass = 1.52

A) Let's assume the incident angle = i and Critical angle = θc

We know that, sin θc = 1/n

Thus;

sin θc = 1/n_glass

θc = sin^(-1) (1/n_glass)

θc = sin^(-1) (1/1.52)

θc = 41.14°

Now, the angle of prism will be the sum of external angle that is critical angle and reflected angle.

Thus;

A = r + θc

r = A - θc

So;

r = 60° - 41. 14°

r = 18.86°

From, Snell's law. If we apply it to this question, we will have;

(sin i)/(sin r) = n_glass

Where;

i is angle of incidence and r is angle of reflection.

Let's make i the subject;

i = sin^(-1) (n_glass × sin r)

i = sin^(-1) (1.52 × sin 18.86)

i = sin^(-1) 0.4914

i = 29.43°

B) The formula to calculate minimum deviation would be from;

μ = [sin ((A + δm)/2)]/(sin A/2)

Where;

μ is Refractive index

δm is minimum angle of deviation

A is angle of prism

Now Refractive index is given by a formula; μ = (sin i)/(sin r)

So; μ = (sin 29.43)/(sin 18.86)

μ = 1.52

Thus;

1.52 = [sin ((60 + δm)/2)]/(sin 60/2)

1.52 * sin 30 = sin ((60 + δm)/2)

0.76 = sin ((60 + δm)/2)

sin^(-1) 0.76 = ((60 + δm)/2)

49.46 × 2 = (60 + δm)

98.92 - 60 = δm

δm = 38.92°

Answer:

A. Final Floor = 15.5 = 15 (Considering downward portion of elevator)

B. T = 14859.5 N = 14.89 KN

Explanation:

A.

First we calculate distance covered by the elevator during downward motion. Downward motion consists of two parts. First one is uniformly accelerated. For that part we use 2nd equation of motion:

s₁ = Vi t + (0.5)at²

where,

s₁ = distance covered during accelerated downward motion = ?

Vi = initial speed = 0 m/s (since elevator is initially at rest)

t = time taken = 6 s

a = acceleration = 1.5 m/s²

Therefore,

s₁ = (0 m/s)(6 s) + (0.5)(1.5 m/s²)(6 s)²

s₁ = 4.5 m

also we find the final velocity using 1st equation of motion:

Vf = Vi + at

Vf = 0 m/s + (1.5 m/s²)(6 s)

Vf = 9 m/s

Now, the second part of downward motion is with constant velocity. So:

s₂ = vt

where,

s₂ = distance covered during constant speed downward motion = ?

v = Vf = 9 m/s

t = 6 s

Therefore,

s₂ = (9 m/s)(6 s)

s₂ = 54 m

Now for distance covered during upward motion is given by the 2nd equation of motion. Since the values of acceleration and time are same. Therefore, it will be equal in magnitude to s₁:

s₃ = s₁ = 4.5 m

Therefore, the total distance covered by elevator is given by following equation:

s = s₁ + s₂ - s₃      (Downward motion taken positive)

s = 4.5 m + 54 m - 4.5 m

s = 54 m

Therefore, net motion of the elevator was 54 m downwards.

So the final floor will be:

Final Floor = Initial Floor - Distance Covered/Length of a floor

Final Floor = 29 - 54 m/4m

Final Floor = 15.5 = 15 (Considering the downward portion or floor of elevator)

B.

The maximum tension will occur during the upward accelerated motion. It is given by the formula:

T = m(g + a)

where,

T = Max. Tension in Cable = ?

m = total mass of person and elevator = 115 kg + 1200 kg = 1315 kg

g = 9.8 m/s²

a = acceleration = 1.5 m/s²

Therefore,

T = (1315 kg)(9.8 m/s² + 1.5 m/s²)

T = 14859.5 N = 14.89 KN

Answer:

E = -4556.18 N/m

Explanation:

Given data

u = 3.6×10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5×10^-2 m  (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u×cosA×t

⇒t = x/(u×cos A)

We also know that force in an electric field is given as

F = qE

q= charge , E= strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u×sinA×t - 0.5×a×t^2

⇒y = u×sinA×t - 0.5×(qE/m)×t^2

if y = 0 then

⇒t = 2mu×sinA/(qE) = x/(u×cosA)

Also, E = 2mu^2×sinA×cosA/(x×q)

Now plugging the values we get

E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})

E = -4556.18 N/m

The value of Ey such that the particle will cross the x axis at x=1.5 cm is -4556.18 N/m.

The field developed when a charge is moved. In this field, a charge experiences an electrostatic force of attraction or repulsion depending on the nature of charge.

Given is a particle leaves the origin with a speed of 3.6 x 10⁶ m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis.

The distance x = 1.5 cm = 1.5×10⁻² m (assumed, not given in question)

The horizontal distance traveled by particle is

x = ucosθt

t = x/ucosθ

The force in an electric field is F = qE...................(1)

where, q is charge , E is the strength of electric field

From, newton 2nd law of motion, Force F = ma.................(2)

Equating both the equations, we get

ma = qE

a = qE/m..................(3)

The vertical distance, y =usinθt - 1/2at²

From equation 3, we have

y = usinθt  -  1/2 (qE/m) t²

if y = 0, t = 2musinθ/(qE) = x / (ucosθ)

The electric field is represented as

Also, E = 2mu²×sinθ×cosθ/(xq)

Plug the values, we get

E = 2×(9.1×10⁻³¹)×(3.6 x 10⁶)²×sin34°×cos34°/( 1.5×10⁻² ×(-1.6)×10⁻¹⁹)

E = -4556.18 N/m

Thus, the electric field of the particle is  -4556.18 N/m.

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Answer:

The average  emf induce is   [tex]V = 2.625 * 10^{-5} \ V[/tex]

Explanation:

From the question we are told that

  The radius of the coil is  [tex]r = 0.30 \ m[/tex]

   The number of turns is  [tex]N = 420 \ turns[/tex]

    The frequency of the transition radio wave is  [tex]f = 1.3\ MHz = 1.3 *10^{6} Hz[/tex]

     The magnetic field is  [tex]B_,{max} = 1.7 * 10^{-13} \ T[/tex]

     The time taken for the magnetic field to go from zero to maximum is [tex]\Delta T = \frac{T}{4}[/tex]

The period of the transmitted radio wave is  [tex]T = \frac{1}{f}[/tex]

    So  

              [tex]\Delta T = \frac{T}{4} = \frac{1}{4 f}[/tex]

 The potential difference can be mathematically represented as

               [tex]V = NA (\frac{\Delta B}{\Delta T} )[/tex]

           [tex]V = NA ([B_{max} - B_{min} ] * 4f)[/tex]

Where  [tex]B_{min} = 0T[/tex]

substituting values

                   [tex]V = 420 * (\pi *(0.30)^2) * (1.7 *10^{-13} * 4 * 1.3 *10^{6})[/tex]

                  [tex]V = 2.625 * 10^{-5} \ V[/tex]

Answer:

a. FALSE

b. FALSE

c. TRUTH

d. FALSE

e. FALSE

Explanation:

To determine which statements are truth or false you focus in the following formula, for the electric potential generated by a conducting sphere:

[tex]V=\frac{Q}{4\pi \epsilon_o R}[/tex]      inside the sphere

[tex]V'=\frac{Q}{4\pi \epsilon_o r}[/tex]      for r > R (outside the sphere)

R: radius of the sphere

ε0: dielectric permittivity of vacuum

Q: charge of the sphere

As you can notice, inside the sphere the potential is constant. Inside the sphere, the potential is the same. Outside the surface the potential decreases as 1/r, being r the distance to the center of the sphere.

Hence, you can conclude:

a. The potential at the center of the sphere is zero. FALSE

b.The potential is lowest, but not zero, at the center of the sphere. FALSE

c. The potential at the center of the sphere is the same as the potential at the surface. TRUTH

d. The potential at the center is the same as the potential at infinity. FALSE

e. The potential at the surface is higher than the potential at the center. FALSE

Answer:

ΔL = 1.011 mm

Explanation:

Let's begin by listing out the given information:

Length (L) = 600 mm = 0.6 m,

Diameter (D) = 40 mm = 0.04 m ⇒ Radius (r) = 20 mm = 0.2 m,

Area (cross sectional) = πr² = 3.14 x .02² = 0.001256 m²,

Modulus of Elasticity (E) = 85 GN/m²,

Compressive load (F) = 180 KN

Using the formula, Stress = Load ÷ Area

Mathematically,

σ = F ÷ A = 180 x 10³ ÷ 0.001256

σ = 143312.1 KN/m²

Modulus of elasticity = stress ÷ strain

E = σ ÷ ε

ε = ΔL/L

85 x 10⁹ = 143312.1 x 10³ ÷ (ΔL/L)

ΔL = 143312.1 x 10³ ÷ 85 X 10⁹ = 1686.02 * 10⁻⁶

ΔL = L x 1686.02 * 10⁻⁶

ΔL = 0.6 * 1686.02 * 10⁻⁶ = 1011.61 x 10⁻⁶

ΔL = 1.011 x 10⁻³ m

ΔL = 1.011 mm

∴The bar contracts by 1.011 mm

Complete Question

The complete question is shown on the first uploaded image

Answer:

The experimental value of density is   [tex]\rho = 1.05*10^{3} \ kg/m^3 \pm 101 \ kg/m^3[/tex]

Comparing it with the value of density of water ([tex]1.0*10^{3} \ kg/m^3[/tex]) we can see that the density of ice is greater

Explanation:

From the question we are told

    The mass is [tex]M = (425 \pm 25) \ g =(0.425 \pm 0.025) \ kg[/tex]

   The volume is [tex]V = (405 \pm 15 ) \ mL = (0.000405 \pm 1.5*10^{-5}) \ m^3[/tex]

The experimental value of density is mathematically evaluated as

        [tex]\rho = \frac{M}{V}[/tex]

       [tex]\rho = \frac{0.425}{0.000405}[/tex]

       [tex]\rho = 1.05 *10^{3} \ kg/m^3[/tex]

The possible error in this experimental value of density is mathematically evaluated as

        [tex]\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} +\frac{\Delta V}{V}[/tex]

substituting value

         [tex]\frac{\Delta \rho}{1.05*10^{3}} = \frac{0.025}{0.425} +\frac{1.5*10^{-5}}{0.000405}[/tex]

        [tex]\Delta \rho = 101 \ kgm^{-3}[/tex]

Thus the experimental value of density is

             [tex]\rho = 1.05*10^{3} \ kg/m^3 \pm 101 \ kg/m^3[/tex]

Answer with Explanation:

Concepts and reason

The concept to solve this problem is that if a capacitor is connected in a RC circuit then it allows the flow of charge through circuit only till it gets fully charged. Once the capacitor is charged it will not allow any charge or current to flow.

Opposite is the case with inductor in the RL circuit. According to Faraday's law an inductor develops an emf to oppose the voltage applied but once the flux change stops then the inductor behaves just like a normal wire as if no inductor is there.

In attached figure, resistor is connected in series to the capacitor.

As we considered [tex]V_{C}[/tex] the voltage across the capacitor and [tex]V_{s}[/tex] the voltage across the source.

Voltage across a resistor In RC circuit.

[tex]V_{R}=V_S\left ( e^{-\frac{t}{RC}} \right )[/tex]

Voltage across a resistor In RL circuit.

[tex]V_{R}=V_S\left (1- e^{-\frac{Rt}{L}} \right )[/tex]

The sketch of [tex]\mathbf{v_R(t)}[/tex] as a function of t for each of the circuits can be seen in the diagram attached below.

For the Pre-Laboratory exercise, based on the assumption that the RC circuit has a capacitor and a sensing resistor while the RL circuit has a sensing resistor and an inductor.

The input voltage for both circuits is regarded as the square wave and if the square wave is much larger than the time constant for each.

Therefore, we can conclude that the below diagram shows an appropriate sketch of  [tex]\mathbf{v_R(t)}[/tex] as a function of t for each of the circuits.

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Answer:

91.5 miles

Explanation:

61 miles per hour so 61(x amount of hours)

so 61 x 1.5 hours is 91.5 miles

Answer:

a

The total initial momentum of the two-block system is  [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]

b

The magnitude of the final velocity of the two-block system [tex]v_f = 1.9014 \ m/s[/tex]

c

 the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision is  

    [tex]\Delta KE =- 847.08 \ J[/tex]

Explanation:

From the question we are told that

    The mass of first  block  is [tex]m_1 = 2.50 \ kg[/tex]

      The initial velocity of first   block is [tex]u_1 = 27.0 \ m/s[/tex]

          The mass of second block is  [tex]m_2 = 33.0\ kg[/tex]

          initial velocity of second block is  [tex]u_2 = 0 \ m/s[/tex]

The magnitude of the of the total initial momentum of the two-block system is mathematically repented as

        [tex]p_i = (m_1 * u_1 ) + (m_2 * u_2)[/tex]

substituting values

        [tex]p_i = (2.50* 27 ) + (33 * 0)[/tex]

        [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]

According to the law of linear momentum conservation

        [tex]p_i = p_f[/tex]

Where  [tex]p_f[/tex] is the total final momentum of the system which is mathematically represented as

       [tex]p_f = (m_+m_2) * v_f[/tex]

Where [tex]v_f[/tex] is the final velocity of the system

      [tex]p_i = (m_1 +m_2 ) v_f[/tex]

substituting values

       [tex]67.5 = (2.50+33 ) v_f[/tex]

        [tex]v_f = 1.9014 \ m/s[/tex]

The change in kinetic energy is mathematically represented as

     [tex]\Delta KE = KE_f -KE_i[/tex]

Where [tex]KE_f[/tex] is the final kinetic energy of the two-body system  which is mathematically represented as

        [tex]KE_f = \frac{1}{2} (m_1 +m_2) * v_f^2[/tex]

substituting values

        [tex]KE_f = \frac{1}{2} (2.50 +33) * (1.9014)^2[/tex]

        [tex]KE_f =64.17 J[/tex]

While [tex]KE_i[/tex] is the initial kinetic energy of the two-body system

     [tex]KE_i = \frac{1}{2} * m_1 * u_1^2[/tex]

substituting values

       [tex]KE_i = \frac{1}{2} * 2.5 * 27^2[/tex]

        [tex]KE_i = 911.25 \ J[/tex]

So

    [tex]\Delta KE = 64.17 -911.25[/tex]

  [tex]\Delta KE =- 847.08 \ J[/tex]

Answer:

EACH SEX CELL WILL HAVE 10 CHROMOSOMES BECAUSE n+n=2n

means haploid parent cells join or fuse to form diploid zygote

Answer:

10

Explanation:

Answer:

  k = 6,547 N / m

Explanation:

This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is

         w = √ (k / m)

angular velocity and rel period are  related

         w = 2π / T

substitution

         T = 2π √(m / K)

in Experimental measurements give us the following data

  m (g)     A (cm)    t (s)   T (s)

  100        6.5         7.8    0.78

  150        5.5          9.8   0.98

   200      6.0        10.9    1.09

   250       3.5        12.4    1.24

we look for the period that is the time it takes to give a series of oscillations, the results are in the last column

        T = t / 10

To find the spring constant we linearize the equation

        T² = (4π²/K)    m

therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is

         m ’= 4π² / k

where m’ is the slope

           k = 4π² / m'

the equation of the line of the attached graph is

       T² = 0.00603 m + 0.0183

therefore the slope

       m ’= 0.00603  s²/g

    we calculate

         k = 4 π² / 0.00603

          k = 6547 g / s²

we reduce the mass to the SI system

         k = 6547 g / s² (1kg / 1000 g)

         k = 6,547 kg / s² =

         k = 6,547 N / m

let's reduce the uniqueness

         [N / m] = [(kg m / s²) m] = [kg / s²]

The spring-mass system forms a linear graph between the time period and mass. And the value of spring-constant from the given data is 6.46 N/m.

Given data:

Mass suspended by spring is, [tex]m=100 \;\rm g =0.1 \;\rm kg[/tex].

Number of oscillations is, [tex]n =10\;\rm oscillations[/tex].

Time period of oscillation is, [tex]T=7.8 \;\rm s[/tex].

The expression for the angular frequency of spring-mass system is,

[tex]\omega =\drac \sqrt{\dfrac{k}{m} }[/tex]  ......................................................(1)

Here, k is the spring constant.

Angular frequency is also expressed as,

[tex]\omega = 2 \pi f[/tex] .........................................................(2)

here, f is the linear frequency of spring-mass system.

And linear frequency is,

[tex]f=\dfrac{n}{T}\\f=\dfrac{10}{7.81}\\f=1.28 \;\rm cycles/sec[/tex]

Then substitute equation (2) in equation (1) as,

[tex]2 \pi f=\drac \sqrt{\dfrac{k}{m} }\\2 \pi \times 1.28=\drac \sqrt{\dfrac{k}{0.1} }\\(2 \pi \times 1.28)^{2}= \dfrac{k}{0.1}\\k = 6.46 \;\rm N/m[/tex]

Thus, the value of spring constant is 6.46 N/m. And the suitable graph for the spring-mass system is given below.

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Answer:

30 minutes

Explanation:

Energy per time is constant, so:

E₁ / t₁ = E₂ / t₂

m₁C₁ΔT₁ / t₁ = m₂C₂ΔT₂ / t₂

(1 kg) C (70°C − 25°C) / 15 min = (1.5 kg) C (80°C − 20°C) / t

(1 kg) (45°C) / 15 min = (1.5 kg) (60°C) / t

3/min = 90 / t

t = 30 min

Answer:

a population increase

Explanation:

During the 20th century, the world population increased from 1.65 billion to 6 billion. In 1970, the world's population was half that of today. In less than 15 years, 47% of the population will live in areas already under heavy water stress. In Africa, between 75 and 250 million people will face growing shortages in 2020 due to climate change. The scarcity of some arid and semi-arid regions will have a decisive impact on migration.

Answer:

Explanation:

During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy

mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .

mgr  = 1/2 x ( 1/2 m r²+ mr²) x ω²

gr  = 1/2 x 1/2  r² x ω² + 1/2r² x ω²

g = 1 / 4 x ω² r + 1 / 2 x ω² r

g = 3  x ω² r/ 4

ω² = 4g /3 r

= 4 x 9.8 /  3 x  .25

= 52.26

ω = 7.23  rad / s .

Answer:

1.95 x 10^8 m/s.

Explanation:

Answer:

the answer is 1.95 x 10^8 m/s

Explanation:

Answer:

The width of the river is  [tex]z = 60.62 \ m[/tex]

Explanation:

From the question we are told that

     The distance of the base line is D = 130 m

       The angle is  [tex]\theta = 25^o[/tex]

A diagram illustration the question is shown on the first uploaded image

    Applying Trigonometric Rules for Right-angled Triangle,

            [tex]tan 25 = \frac{z}{130}[/tex]

Now making  z the subject

           [tex]z = 130 * tan (25)[/tex]

          [tex]z = 60.62 \ m[/tex]