In pushing a heavy box across the floor, is the force you need to apply to start the box moving greater than, less than, or the same as the force needed to keep the box moving? On what are you basing your choice?

How do you think the force of friction is related to the weight of the box? Explain.

Answer: x, y (0, 1)

Explanation:

the X coordinate of the center mass is

X(c) = Σm(i)*x(i) / Σx(i)

X(c) = (0 + 0 + 0 + 0) / (1.93 + 3.06 + 2.41 + 3.96)

X(c) = 0 / 11.36

X(c) = 0

The y coordinate of the center mass is

Y(c) = Σm(i)y(i) / Σm(i)

Y(c) = [(1.93)(2.93) + (3.06)(2.58) + (2.41)(0) + (3.96)(-0.498)] / (1.93 + 3.06 + 2.41 + 3.96)

Y(c) = (5.6549 + 7.8948 + 0 - 1.97208) / 11.36

Y(c) = 11.57762 / 11.36

Y(c) = 1.02

Therefore, the center of masses is at x, y (0, 1)

Explanation:

It is given that mass at the right side of the pulley weighs 9 kg, mass at the center weighs 8 kg and mass at the left side weighs 5 kg.

Hence, starting from the right side

              [tex]T_{1}[/tex] = 9 kg

Now,  [tex]T_{2} = 2 \times 9 kg[/tex]

                      = 18 g

Hence, the force balance on 8 kg mass is as follows.

            [tex]T_{2} + mg = T_{3}[/tex]

     [tex]T_{3}[/tex] = [tex](18 + 8) \times g[/tex]

                 = 26 g

Now, force balance on 5 kg mass is as follows.

              5 g + T = [tex]2 \times 26 g[/tex]

                 T = 47 g

                   = [tex]47 \times 9.8[/tex]

                   = 460.6 N

Therefore, we can conclude that value of force T is 460.6 N.

The value of force T is 460.6N

Tension is a force developed by a rope, string, or cable when stretched under an applied force.

Given:

Mass (m₁) = 9 kg, m₂ = 8 kg, m₃ = 5kg, g = acceleration due to gravity = 9.8 m/s².

T₁ = 9 kg * 9.8 m/s² = 88.2N

T₂ = 2 * m₁ * g = 2 * 9 kg * 9.8 = 176.4N

T₂ + m₂g = T₃

176.4 + 8*9.8 = T₃

T₃ = 254.8N

Force balance on 5 kg mass:

T + 5(9.8) = 2 * T₃

T + 5(9.8) = 2 * 254.8

T = 460.6N

The value of force T is 460.6N

Find out more on Tension at: https://brainly.com/question/24994188

Answer:

The force developed in the link is 6.36 N.

Explanation:

Given that,

Mass of block A = 10 kg

Mass of block B = 6 kg

Coefficients of kinetic friction [tex]\mu_{A}= 0.1[/tex]

Coefficients of kinetic friction [tex]\mu_{B}= 0.3[/tex]

Suppose the angle is 30°

We need to calculate the acceleration

Using formula of acceleration

[tex]a=\dfrac{m_{A}g\sin\theta+m_{B}g\sin\theta-\mu_{A}m_{A}g\cos\theta-\mu_{A}m_{A}g\cos\theta}{m_{A}+m_{B}}[/tex]

Put the value into the formula

[tex]a=\dfrac{10\times9.8\sin30+6\times9.8\sin30-0.1\times10\times9.8\times\cos30-0.3\times6\times9.8\times\cos30}{16}[/tex]

[tex]a=3.415\ m/s^2[/tex]

We need to calculate the force developed in the link

For block A,

Using balance equation

[tex]ma=m_{A}g\sin\theta-\mu m_{A}g\cos\theta-T[/tex]

[tex]T=ma+\mu m_{A}g\cos\theta-m_{A}g\sin\theta[/tex]

Put the value into the formula

[tex]T=10\times3.415+0.1\times10\times9.8\times\cos30-10\times9.8\times\sin30[/tex]

[tex]T=-6.36\ N[/tex]

Negative sign shows the opposite direction of the force.

Hence, The force developed in the link is 6.36 N.

Answer:

a) Magnitude of maximum emf induced = 0.0714 V = 71.4 mv

b) Maximum current through the bulb = 0.00793 A = 7.93 mA

Explanation:

a) The induced emf from Faraday's law of electromagnetic induction is related to angular velocity through

E = NABw sin wt

The maximum emf occurs when (sin wt) = 1

Maximum Emf = NABw

N = 1

A = 4 cm² = 0.0004 m²

B = 6 T

w = (284/60) × 2π = 29.75 rad/s

E(max) = 1×0.0004×6×29.75 = 0.0714 V = 71.4 mV

Note that: since we're after only the magnitude of the induced emf, the minus sign that indicates that the induced emf is 8n the direction opposite to the change in magnetic flux, is ignored for this question.

b) Maximum current through the bulb

E(max) = I(max) × R

R = 9 ohms

E(max) = 0.0714 V

I(max) = ?

0.0714 = I(max) × 9

I(max) = (0.0714/9) = 0.00793 A = 7.93 mA

Hope this Helps!!

Answer:

1.2 A of current will send (3.744 × 10¹⁸) ions to the cathode per second.

Explanation:

According to Faraday's second law of electrolysis, the amount of ions/mass of substance deposited at an electrode depends on its equivalent weight.

For a divalent ion, it will require 2F of electricity per mole.

1 F = 96500 C

Amount of electricity that passes through the electrolyte per second = (magnitude of current) × (time) = It = (1.2 × 1) = 1.2 C

2F (2×96500C) of electricity will deposit 1 mole of Copper ions

That is,

193000 C of electricity will deposit 1 mole of Copper.

1.2 C will deposit (1.2×1/193000); 0.0000062176 mole of Copper.

1 mole of Copper contains (6.022 × 10²³) ions according to the Avogadro's constant.

0.0000062176 mole of Copper will contain (0.0000062176 × 6.022 × 10²³) ions = (3.744 × 10¹⁸) ions.

Therefore, 1.2 A of current will send (3.744 × 10¹⁸) ions to the cathode per second.

Hope this Helps!!!

Answer: No

Explanation:

The force F required is equal to the Force exerted in stretching the first material since conditions are the same

Answer:

No.

Explanation:

Both wires are made up of the same material which means they will have the same young modulus.

Y = F/A × L/ΔL

From the equation above, the force applied per cross sectional (same diameter) are is the same for both materials and is constant.

So

L1/ΔL1 = L2/ΔL2

Or L2/L1 = ΔL2/ΔL1

So the force needed to bring about a stretching the the breaking point is the same. The only difference is in by how much the longer wire would have to stretch before reaching its breaking point.

Answer:

Option A

498 Hz

Explanation:

Beat frequency is given by F1-F2 where F is the frequency of source while F2 is frequency that disappear as one moves towards the source. when a source moves towards observer the frequency increases

Substituting 500 for F1 and 2 for F2 then

Beat frequency is 500-2=498 Hz

Option A

Explanation:

The give data is as follows.

             C = 6 [tex]\mu F[/tex] = [tex]6 \times 10^{-6} F[/tex]

             V = 6 V

Now, we know that the relation between charge, voltage and capacitor for series combination is as follows.

              Q = CV

                  = [tex]6 \times 10^{-6} F \times 6 V[/tex]

                  = [tex]36 \times 10^{-6} C[/tex]

or,               = 36 [tex]\mu C[/tex]

Thus, we can conclude that maximum charge of the given capacitor is 36 [tex]\mu C[/tex].

Answer:

Explanation:

capacitance, C = 6 μF

Voltage, V =  6 V

Let the maximum charge is Q.

Q = C x V

Q = 6 x 6 = 36 μC

Answer:

[tex]3.2\times 10^{-7}\ m[/tex] or 0.32 μm.

Explanation:

Given:

The radiations are UV radiation.

The frequency of the radiations absorbed (f) = [tex]9.38\times 10^{14}\ Hz[/tex]

The wavelength of the radiations absorbed (λ) = ?

We know that, the speed of ultraviolet radiations is same as speed of light.

So, speed of UV radiation (v) = [tex]3\times 10^8\ m/s[/tex]

Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

[tex]v=f\lambda[/tex]

Now, expressing the above equation in terms of wavelength 'λ', we have:

[tex]\lambda=\frac{v}{f}[/tex]

Now, plug in the given values and solve for 'λ'. This gives,

[tex]\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m[/tex]

Therefore, the wavelength of the radiations absorbed by the ozone is nearly [tex]3.2\times 10^{-7}\ m[/tex] or 0.32 μm.

Answer:

2.51 m/s

Explanation:

Parameters given:

Angle, A = 33°

Mass, m = 90kg

Inclined distance, D = 2m

Force, F = 600N

Initial speed, u = 2.3m/s

From the relationship between work and kinetic energy, we know that:

Work done = change in kinetic energy

W = 0.5m(v² - u²)

We also know that work done is tẹ product of force and distance, hence, net work done will be the sum of the total work done by the force from the students and gravity.

Hence,

W = F*D*cosA - w*D*sinA

w = m*9.8 = weight

=> W = 600*2*cos33 - 90*9.8*2*sin33

W = 45.7J

=> 45.7 = 0.5*m*(v² - u²)

45.7 = 0.5*90*(v² - 2.3²)

45.7 = 45(v² - 5.29)

=> v² - 5.29 = 1.016

v² = 6.306

v = 2.51 m/s

The final velocity is 2.51 m/s

Answer: v = 3.26 m/s

Explanation:

Force applied (F) = 900 N,

mass of professor and chair (m) = 90kg,

g = acceleration due to gravity = 9.8 m/s²,

Fr = frictional force = 0 ( since from the question, the rollers the chair and professor are moving is frictionless),

θ = angle of inclination = 33°

a = acceleration of object =?

From newton's second law of motion, we have that

F - (mg sinθ +Fr) = ma

Where mg sinθ is the horizontal component of the weight of the mass of professor and chair due to the inclination of the ramp.

But Fr = 0

Hence, we have that

F - mg sinθ = ma

600 - (90×9.8×sin30) = 90 (a)

600 - 480.371= 90a

119.629 = 90a

a = 119.629/ 90

a = 1.33 m/s².

But the body started the morning (at the bottom of the ramp) with a velocity of 2.30 m/s²

Hence u = initial velocity = 2.30 m/s², a = acceleration = 1.33m/s², v = final velocity =?, s = distance covered = 2m

By using equation of motion for a constant acceleration, we have that

v² = u² + 2as

v ² = (2.3)² +2(1.33)×(2)

v² = 5.29 + 5.32

v² = 10.61

v = √10.61

v = 3.26 m/s

Answer:

h = 4.281 m

Explanation

Given Data,

Weight of disk = 2kg

Speed of water jet (V) = 10 m/s

diameter of water jet = 25 mm

Cal. the velocity of the water jet as function height by applying Bernoulli's  eqtn of water surface to the jet

[tex]\frac{P}{\rho } +\frac{V^{2}}{2} + gz[/tex] = Constant

[tex]\frac{V_{0}^{2}}{2} + g(0) = \frac{V^{2}}{2} + g[/tex]

[tex]V=\sqrt{V_{0}^{2}-2gh}[/tex]

Relation between  [tex]V_{0}[/tex] & V

[tex]m=\rho V_{0}A_{0}[/tex]

[tex]\rho VA=\rho V_{0}A_{0}[/tex]

[tex]VA= V_{0}A_{0}[/tex]

Momentum

[tex]F_{w}+F_{d}= \frac{\partial }{\partial t}\int_{cv} w\rho dA + \int_{cs} w\rho V dA[/tex]

[tex]-mg = w_{1}[-\rho VA] +w_{2}[\rho VA][/tex]

[tex]w_{1}[/tex] = V

[tex]w_{2}[/tex] = 0

[tex]mg = \rho V^{2}A[/tex]

[tex]mg = \rho VV_{0} A[/tex]

[tex]mg = \rho VV_{0} A_{0}[/tex]

[tex]mg = \rho V_{0} A_{0} \sqrt{V_{0} ^{2}-2gh }[/tex]

Solving for h

[tex]h = \frac{1}{2g}[V_{0}^{2}-\frac{mg}{\rho V_{0}A_{0}}][/tex]

g is gravitational acc.

[tex]= \frac{1}{2\times 9.81}[10^{2}-(\frac{2\times 9.81}{999\times10\times\frac{\Pi }{4}\times(0.025)^{2}})^{2} ][/tex]

[tex]= \frac{1}{19.62}[100-(\frac{19.68}{4.9038})^{2}][/tex]

[tex]= \frac{100-16.0078}{19.62}[/tex]

h = 4.281 m

h of disk on which it remains stationary.

Answer:

1. Charge on plates (plates remain connected to battery) increases.

2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases.

3.Capacitance (plates isolated from battery before inserting dielectric) increases.

4. Voltage between plates (plates remain connected to battery) remains the same.

5. Charge on plates (plates isolated from battery before inserting dielectric) remains the same.

6. Capacitance (plates remain connected to battery) increases.

7. Electric potential energy (plates remain connected to battery) increases.

8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases.

Explanation:

When a dielectric material is inserted between the plates of a capacitor, the capacitance is increase by a factor K, the dielectric constant.

[tex]C = KC_0[/tex]

By the capacitance formula, the other factors change accordingly.

1. Charge on plates (plates remain connected to battery) increases, because charge and capacitance are directly proportional.

2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases, because potential is inversely proportional to capacitance, and potential energy is given by the following formula

[tex]U = \frac{1}{2}CV^2[/tex]

3.Capacitance (plates isolated from battery before inserting dielectric) increases.

4. Voltage between plates (plates remain connected to battery) stays the same, because the voltage is applied by the battery.

5. Charge on plates (plates isolated from battery before inserting dielectric) remains constant. If the plates isolated from the battery, then the total charge is conserved.

6. Capacitance (plates remain connected to battery)  increases when a dielectric is inserted.

7. Electric potential energy (plates remain connected to battery)  increases, because when plates remain connected to battery the voltage remains the same. But the capacitance increases. Therefore, electric potential energy increases.

8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases, because voltage is inversely proportional to capacitance.

Final answer:

When a dielectric is inserted into a capacitor, the capacitance increases regardless of whether the plates are connected to a battery or isolated. If connected to a battery, charge on the plates increases while voltage remains the same; if isolated, charge remains the same while voltage decreases.

Explanation:

When inserting a dielectric material into a parallel-plate capacitor, the effect on different parameters depends on whether the capacitor is connected to a battery or isolated:

Charge on plates (plates remain connected to battery): The charge on the plates will increase.

Electric potential energy (plates isolated from  the battery before inserting the dielectric): The electric potential energy will decrease due to the polarization of the dielectric.

Capacitance (plates isolated from the battery before inserting the dielectric): The capacitance will increase.

Voltage between plates (plates remain connected to battery): The voltage will remain the same, as it is maintained by the battery.

Charge on plates (plates isolated from battery before inserting dielectric): The charge will remain the same as isolation prevents any charge exchange.

Capacitance (plates remain connected to battery): The capacitance will increase.

Electric potential energy (plates remain connected to the battery): The electric potential energy will increase, as more charge is accumulated due to the dielectric.

Voltage between plates (plates isolated from battery before inserting dielectric): The voltage will decrease because the same charge is now across a larger capacitance.

It is important to note that the introduction of a dielectric material affects the capacitance of a parallel-plate capacitor by reducing the effective electric field and permitting greater charge storage capability for a given voltage.

Answer:

(c) 10.29 J

(d) 113.19 J

(e) 113.19 J

(f) 10061 N/m

Explanation:

15 cm = 0.15 m

Let g = 9.8 m/s2

(c) The work done by gravitational force is the product of gravity force and the distance compressed

[tex]E_p = mgx = 7*9.8*0.15 = 10.29 J[/tex]

(d) By using law of energy conservation with potential energy reference being 0 at the maximum compression point. As the ball falls and come to a stop at the compression point, its potential energy is transferred to elastic energy, which is the work that the mattress does on the ball:

[tex]E_p = E_e[/tex]

[tex]E_e = mgh[/tex]

where h = 1.5 + 0.15 = 1.65 m is the vertical distance that it falls.

[tex]E_e = 7*9.8*1.65 = 113.19 J[/tex]

(e) Before the compression, the potential energy of the mattress is 0. After the compression, the potential energy is 113.19J. So it has increased by 113.19J due to the potential energy transferred from the falling ball.

(f) [tex]E_e = 113.19 = kx^2/2[/tex]

[tex]k0.15^2/2 = 113.19[/tex]

[tex]k = 10061 N/m[/tex]

Answer:

(C) Wg = 113.2J

(D) Wm = 10.3J

(E) E = 1/2kx² + mgh where h is the height above the mattress and x is the compressed distance in the mattress.

(F) k = 457N/m.

Explanation:

See attachment below.

Answer:

21.5 °C.

Explanation:

Given:

α = −5.00 × 10−4 °C−1

To = 4°C

Ro = 217 Ω

Rt = 215.1 Ω

Rt/Ro = 1 + α(T - To)

215.1/217 = 1 + (-5 × 10^−4) × (T - 4)

-0.00876 = -5 × 10^−4 × (T - 4)

17.5 = (T - 4)

T = 21.5 °C.

The temperature on a spring day can be calculated by using the relationship between temperature and resistance. Given the initial resistance, the change in resistance and the temperature coefficient of resistivity for carbon, the spring day temperature can be deduced to be 21.51 °C.

The change in temperature can be determined by using the equation ΔR = R0αΔT, where ΔR is the change in resistance, R0 is the initial resistance, α is the temperature coefficient of resistivity and ΔT is the change in temperature. The initial resistance of the carbon resistor is given as 217.0 Ω, the resistance on a spring day was 215.1 Ω, so the change in resistance ΔR is 215.1 - 217.0 = -1.9 Ω. The temperature coefficient for carbon is given as -5.00×10-4 C-1, and α is -5.00×10-4 C-1. We can rearrange the above equation to solve for ΔT: ΔT = ΔR / (R0 * α) = -1.9 Ω / (217.0 Ω * -5.00×10-4 C-1) = 17.51 C.

Thus, the temperature on the spring day would be the winter day temperature plus the calculated change in temperature: T_spring = T_winter + ΔT = 4.00 °C + 17.51 °C = 21.51 °C.

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Answer:

[tex]q = 2.067 \times 10^{-5}\ C[/tex]

Explanation:

Given,

mass = 1.41 g = 0.00141 Kg

Electric field,E = 670 N/C.

We know,

Force in charge due to Electric field.

F = E q

And also we know

F = m g

Equating both the equation of motion

m g = E q

[tex]q =\dfrac{mg}{E}[/tex]

[tex]q =\dfrac{0.00141 \times 9.81}{670}[/tex]

[tex]q = 2.067 \times 10^{-5}\ C[/tex]

Charge of the particle is equal to [tex]q = 2.067 \times 10^{-5}\ C[/tex]

Answer: [tex]V_{\epsilon}\propto \frac{d\phi_{B}}{dt}[/tex]

Explanation:

A direct proportionality means a linear relationship between two variables and rate of change means an application of derivatives. Hence, the mathematical model is:

[tex]V_{\epsilon}\propto \frac{d\phi_{B}}{dt}[/tex]

Faraday's Law of Induction, primarily focused on Physics, reveals that the induced electromotive force in a coil is proportional to the rate of change of magnetic flux through the coil and is also dependent on the number of coil turns.

Understanding Faraday's Law of Induction

Faraday's Law of Induction is a critical concept in electromagnetism, underlying the working principles of various electrical devices. The induced electromotive force (emf) is directly proportional to the time rate of change of magnetic flux. In simpler terms, when the magnetic environment of a circuit changes, an emf is induced. Faraday's Law is given by the equation EMF = -N(ΔΦ/Δt), where 'N' represents the number of turns in a coil, and (ΔΦ/Δt) is the rate of change of magnetic flux over time. This equation highlights three main factors influencing the induced EMF:

The negative sign in Faraday's Law is a reflection of Lenz's Law, which indicates that the induced emf always works to oppose the change in magnetic flux that produces it.

Answer: 23.92 ns

Explanation:

The capacitance, C = (e(0) * A) / d

C = [8.85*10^-12 * 2*10^-2 * 10*10^-2] / 1*10^-3

C = (1.77*10^-14) / 1*10^-3

C = 1.77*10^-11

C = 17.7 pF

Vc = V * [1 - e^-(t/CR)]

75 = 100 * [1 - e^-(t/CR)]

75/100 = 1 - e^-(t/CR)

e^-(t/CR) = 1 - 0.75

e^-(t/CR) = 0.25

If we take the log of both sides, we then have

Log e^-(t/CR) = Log 0.25

-t/CR = In 0.25

t = -CR In 0.25

t = - 17.7*10^-12 * 975 * -(1.386)

t = 2.392*10^-8 s

t = 23.92 ns

Answer:

[tex]F=800N[/tex]

Explanation:

The average force is defined as the mass of the body multiplied by its average velocity over the contact time. According to the third Newton's law,  the magnitude of the average force exerted on the wall by the ball is equal to  the magnitude of the average force exerted on the ball by the wall. Thus:

[tex]F=m\frac{v_f-v_i}{t}\\F=0.80kg\frac{25\frac{m}{s}-(-25\frac{m}{s})}{0.05s}\\F=800N[/tex]

Answer:

800 N

Explanation:

Force: This can be defined as the product of mass and acceleration. The S.I unit of force is Newton(N).

From the question,

F = m(v-u)/t ....................... Equation 1

Where F = force exerted on the wall by the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time.

Note: Assuming the direction of the initial motion to be negative

Given: m = 0.80 kg, v = 25 m/s ( bounce back), u = -25 m/s, t = 0.05 s

Substitute into equation 1

F = 0.8[25-(25)]/0.05

F = 0.8(50)/0.05

F = 0.8(1000)

F = 800 N

Answer:

Explanation:

Energy stored in a capacitor is given by the expression

E = Q² / 2C , where Q is charge on it and C  is its capacitance . As we put dielectric slab between the plates , capacitance increases , Due to it energy decreases as capacitance form the denominator in the formula above and Q is constant .

Hence energy decreases.

Answer:

[tex]a. m_i_c_e=54.6g\\b. m_i_c_e=48.7g\\m_c_o_l_d_w_a_t_e_r=900g[/tex]

Explanation:

First we need to state our assumptions:

Thermal properties of ice and water are constant, heat transfer to the glass is negligible, Heat of ice [tex]h_i_f=333.7KJkg[/tex]

Mass of water,[tex]m_w=\rho V =1\times0.3=0.3Kg[/tex].

Energy balance for the ice-water system is defined as

[tex]E_i_n-E_o_u_t=\bigtriangleup E_s_y_s\\0=\bigtriangleup U=\bigtriangleup U_i_c_e+\bigtriangleup U_w[/tex]

a.The mass of ice at [tex]0\textdegree C[/tex] is defined as:

[tex][mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[0+333.7+418\times(5-0)]+0.3\times4.18\times(5-20)=0\\m_i_c_e=0.0546Kg=54.6g[/tex]

b.Mass of ice at [tex]20\textdegree C[/tex] is defined as:

[tex][mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[2.11\times(0-(20))+333.7+4.18\times(5-0)]+0.3\times4.18\times(5-20)=0\\\\m_i_c_e=0.0487Kg=48.7g[/tex]

c.Mass of cooled water at [tex]T_c_w=0\textdegree C[/tex]

[tex]\bigtriangleup U_c_w+\bigtriangleup U_w=0[/tex]

[tex][mc(T_2-T_1)]_c_w+[mc(T_2-T_1)]_w=0\\m_c_w\times4.18\times(5-0)+0.3\times4.18\times(5-20)\\m_c_w=0.9kg=900g[/tex]

(a) The mass of ice needed if the ice is at 0°C is [tex]\( 56.4 \, \text{g} \).[/tex]

(b) The mass of ice needed if the ice is at -20°C is [tex]\( 50.1 \, \text{g} \).[/tex]

(c) The mass of cold water needed if the cooling is done with water at 0°C is [tex]\( 900 \, \text{g} \).[/tex]

To determine how much ice needs to be added to the water to cool it from 20°C to 5°C, we need to consider the heat transfer involved. We'll use the principles of heat balance, assuming no heat is lost to the surroundings.

Given Data:

Calculations:

(a) Ice at 0°C:

1. Heat required to cool the water from 20°C to 5°C:

[tex]\[ Q_{\text{water}} = m_w \cdot c_w \cdot (T_i - T_f) \][/tex]

  where:

  So,

[tex]\[ m_w = 1 \, \text{kg/L} \times 0.3 \, \text{L} = 0.3 \, \text{kg} \][/tex]

[tex]\[ Q_{\text{water}} = 0.3 \, \text{kg} \times 4.18 \, \text{kJ/kg}^\circ \text{C} \times (20^\circ \text{C} - 5^\circ \text{C}) \][/tex]

[tex]\[ Q_{\text{water}} = 0.3 \times 4.18 \times 15 \][/tex]

[tex]\[ Q_{\text{ice}} = m_{\text{ice}} \cdot L_f \][/tex]

2. Heat absorbed by ice to melt at 0°C:

[tex]\[ Q_{\text{ice}} = m_{\text{ice}} \cdot L_f \][/tex]

  To find the mass of ice needed:

[tex]\[ m_{\text{ice}} = \frac{Q_{\text{water}}}{L_f} \][/tex]

[tex]\[ m_{\text{ice}} = \frac{18.81 \, \text{kJ}}{333.7 \, \text{kJ/kg}} \][/tex]

[tex]\[ m_{\text{ice}} = 0.0564 \, \text{kg} = 56.4 \, \text{g} \][/tex]

(b) Ice at -20°C:

1. Heat required to cool ice from -20°C to 0°C:

[tex]\[ Q_{\text{cool}} = m_{\text{ice}} \cdot c_i \cdot (0^\circ \text{C} - (-20^\circ \text{C})) \][/tex]

[tex]\[ Q_{\text{cool}} = m_{\text{ice}} \cdot 2.1 \, \text{kJ/kg}^\circ \text{C} \cdot 20 \][/tex]

[tex]\[ Q_{\text{cool}} = m_{\text{ice}} \cdot 42 \, \text{kJ/kg} \][/tex]

2. Heat absorbed by ice to melt at 0°C:

[tex]\[ Q_{\text{melt}} = m_{\text{ice}} \cdot L_f = m_{\text{ice}} \cdot 333.7 \, \text{kJ/kg} \][/tex]

3. Total heat absorbed by the ice:

[tex]\[ Q_{\text{total}} = Q_{\text{cool}} + Q_{\text{melt}} \][/tex]

[tex]\[ Q_{\text{total}} = m_{\text{ice}} \cdot 42 \, \text{kJ/kg} + m_{\text{ice}} \cdot 333.7 \, \text{kJ/kg} \][/tex]

[tex]\[ Q_{\text{total}} = m_{\text{ice}} \cdot (42 + 333.7) \, \text{kJ/kg} \][/tex]

[tex]\[ Q_{\text{total}} = m_{\text{ice}} \cdot 375.7 \, \text{kJ/kg} \][/tex]

Using the heat required to cool the water:

[tex]\[ m_{\text{ice}} = \frac{Q_{\text{water}}}{Q_{\text{total}}} \][/tex]

[tex]\[ m_{\text{ice}} = \frac{18.81 \, \text{kJ}}{375.7 \, \text{kJ/kg}} \][/tex]

[tex]\[ m_{\text{ice}} = 0.0501 \, \text{kg} = 50.1 \, \text{g} \][/tex]

(c) Cooling with Water at 0°C:

1. Heat required to cool the water from 20°C to 5°C:

[tex]\[ Q_{\text{water}} = 18.81 \, \text{kJ} \][/tex]

2. Heat absorbed by the cold water to warm from 0°C to 5°C:

[tex]\[ Q_{\text{cold water}} = m_{\text{cold water}} \cdot c_w \cdot (5^\circ \text{C} - 0^\circ \text{C}) \][/tex]

[tex]\[ 18.81 \, \text{kJ} = m_{\text{cold water}} \cdot 4.18 \, \text{kJ/kg}^\circ \text{C} \cdot 5 \][/tex]

[tex]\[ m_{\text{cold water}} = \frac{18.81 \, \text{kJ}}{4.18 \, \text{kJ/kg}^\circ \text{C} \times 5} \][/tex]

[tex]\[ m_{\text{cold water}} = \frac{18.81}{20.9} \][/tex]

[tex]\[ m_{\text{cold water}} \approx 0.900 \, \text{kg} = 900 \, \text{g} \][/tex]

The solution is in the attachment

Answer:

P =1156mmHg

Explanation:

Given

Ptriple = 846mmHg

Ttriple = 273.16K

T = temperature of boiling water = 100°C = (273.16 +100)K = 373.16K

P/Ptriple = T/Ttriple

P = T/Ttriple × Ptriple

P = 373.16/×273.6 × 846

P = 1156mmHg.

The temperature used in this solution is the absolute temperature because the gas thermometer use the absolute temperature scale. On this scale (Kelvin temperature scale) the zero point, Absolute zero is 273.16K.

Answer:

Explanation:

Ideal gas obeys the law , [tex]PV = RT[/tex]

When volume remains constant

[tex]\frac{P2}{P1} = \frac{T2}{T1}[/tex]

Temperature of triple point of water is [tex]T1 = 273.16K[/tex]

Temperature of normal boiling point of water is [tex]T2 = 373.15 K[/tex]

therefore,

[tex]P2 = 846 * \frac{373.15}{273.16}\\\\P2 = 1155.68 mm[/tex]

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Answer:

Squids = 450 - 490 nm (Moderate Frequency) (Blue)

Bees = 300 - 650 nm (Lower Frequency Bands)

Frogs = 280 - 580 nm (Very Low Frequency)

Explanation:

All of the above mentioned ranges are compared to that of humans.

I'm just surprised a little bit in the imagination that how these organisms see the world through their unique eyes. On the other hands, they are evolved like this just like we do so that may not be surprising enough. SIKE

Answer:

Explanation:

mass of elephant, m1 = 5240 kg

mass of ball, m2 = 0.150 kg

initial velocity of elephant, u1 = - 4.55 m/s

initial velocity of ball, u2 = 7.81 m/s

Let the final velocity of ball is v2.

Use the formula of collision

[tex]v_{2}=\left ( \frac{2m_{1}}{m_{1}+m_{2}} \right )u_{1}+\left ( \frac{m_{2}-m_{1}}{m_{1}+m_{2}} \right )u_{2}[/tex]

[tex]v_{2}=\left ( \frac{2\times 5240}{5240+0.150} \right )(-4.55)+\left ( \frac{0.15-5240}{5240+0.150} \right )(7.81)[/tex]

v2 = - 16.9 m/s

The negative sign shows that the ball bounces back towards you.

(b) It is clear that the velocity of ball increases and hence the kinetic energy of the ball increases. This gain in energy is due to the energy from elephant.

Answer:

n = 1.7*10²² electrons.

Explanation:

       [tex]I =\frac{\Delta q}{\Delta t} \\ \\ \Delta q = I* \Delta t \\ \\ \Delta t = 5hs*\frac{3600s}{1h} = 18000 s[/tex]

       ⇒ Δq = I * Δt = 0.151 A * 18000 s = 2718 C

         Δq = n*e

        [tex]n = \frac{\Delta q}{e} =\frac{2718C}{1.6-19C} = 1.7e22 electrons.[/tex]

Explanation:

Below is an attachment containing the solution.

Answer:

0.28 ft

Explanation:

We are given that

Strength=m=[tex]36ft^2/s[/tex]

Distance between source and sink=15 ft

Distance between the sink of the source and origin=[tex]a=\frac{15}{2}[/tex] ft

Uniform velocity, U=12 ft/s

We have to find the length of the oval.

Formula to find the half length of the body

[tex]\frac{l}{a}=(\frac{m}{\pi Ua}+1)^{\frac{1}{2}}[/tex]

Where a=Distance between sink of source and origin

U=Uniform velocity

m=Strength

l=Half length

Using the formula

[tex]\frac{l}{\frac{15}{2}}=(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}[/tex]

[tex]l=\frac{2}{15}(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}[/tex]

[tex]l=0.14[/tex]

Length of oval=[tex]2l=2(0.14)=0.28 ft[/tex]

Answer:

80 - 32t

Explanation:

The height, h, in terms of time, t, is given as:

h(t) = 650 + 80t − 16t²

Velocity is the derivative of distance with respect to time:

v(t) = dh(t)/dt = 80 - 32t

The velocity of the object as a function of time is given by the derivative of the height function, which is v(t) = 80 - 32t.

The height h(t) of an object is given by the equation h(t) = 650 + 80t − 16t2. To find the velocity v(t), we need to take the derivative of h(t) with respect to time t. Using the power rule, we get:

v(t) = dh/dt = 0 + 80 - 32t.

So, the velocity of the object as a function of time t is v(t) = 80 - 32t.

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Answer:

Voltage= 9.4 V

Current= 1.3 A

Explanation:

Equivalent resistance will be the sum of two resistors hence

[tex]R_{equ}[/tex]=2+7.2=9.2Ω

From Ohm's law, V=IR where I is current, R is equivalent resistance and V is voltage

Terminal voltage will be given by V=e-Ir where r is internal resistance, I is current and e is electromotive force.

From Ohm's law, current is given by dividing emf of the battery by the equivalent resistance hence

[tex]I=\frac {e}{R_{equ}}\\I=\frac {12}{9.2}=1.304347826A\approx 1.3A[/tex]

This is the current through external resistor

Terminal voltage will be given by V=e-Ir hence V=12-(1.3*2)=9.4 V

Final answer:

The terminal voltage across the 7.2
Ω resistor when connected to a 12.0 V battery with 2.0
Ω internal resistance is approximately 9.392 V, and the current flowing through the resistor is approximately 1.304 A.

Explanation:

The student is asking about the calculation of terminal voltage and current through a resistor when connected to a battery that has internal resistance. This is a problem related to the topic of electrical circuits in Physics.

To find the current through the 7.2
Ω resistor, use Ohm's law which states that V = IR, where V is voltage, I is current, and R is resistance. Since the battery has an internal resistance, we need to consider the total resistance in the circuit which is the sum of the internal resistance and the resistance of the 7.2
Ω resistor.

Total resistance (Rtotal) = Internal resistance (r) + External resistance (R) = 2.0
Ω + 7.2
Ω = 9.2
Ω

Using Ohm's Law, the current (I) through the circuit can be calculated as follows:
I = V / Rtotal = 12.0 V / 9.2
Ω = 1.304
A (approximately).

To find the terminal voltage (Vterminal) across the 7.2
Ω resistor when the current is flowing, we account for the voltage drop across the internal resistance:
Vterminal = emf - I × r = 12.0 V - (1.304
A × 2.0
Ω) = 12.0 V - 2.608 V = 9.392 V (approximately).

The terminal voltage across the 7.2
Ω resistor when it is connected to the battery is 9.392 V, and the current flowing through the resistor is 1.304
A.

Explanation:

Below is an attachment containing the solution.

(a) In a parallel circuit, each bulb receives the full voltage of the household circuit. Therefore, the current running through each bulb is 1.2A.

(a) In a parallel circuit, the voltage across each component remains the same. Therefore, each bulb in the circuit will receive the full 120V supplied by the household circuit.

Using Ohm's Law (V = IR), we can calculate the current running through each bulb. The resistance of each bulb is given as 100 Ω. Therefore, the current through each bulb is:

I = V/R = 120V / 100 Ω = 1.2A

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I think your question should be:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is

[tex] S = 1.23*10^9 W/m^2 [/tex]

What is the rms value of (a) the electric field and

(b) the magnetic field in the electromagnetic wave emitted by the laser

Answer:

a) [tex] 6.81*10^5 N/c [/tex]

b) [tex] 2.27*10^3 T [/tex]

Explanation:

To find the RMS value of the electric field, let's use the formula:

[tex] E_r_m_s = sqrt*(S / CE_o)[/tex]

Where

[tex] C = 3.00 * 10^-^8 m/s [/tex];

[tex] E_o = 8.85*10^-^1^2 C^2/N.m^2 [/tex];

[tex] S = 1.23*10^9 W/m^2 [/tex]

Therefore

[tex] E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]} [/tex]

[tex] E_r_m_s= 6.81 *10^5N/c [/tex]

b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:

[tex] B_r_m_s = E_r_m_s / C [/tex];

[tex] = 6.81*10^5 N/c / 3*10^8m/s [/tex];

[tex] B_r_m_s = 2.27*10^3 T [/tex]

Complete Question:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is 1.38 * 10⁹ W/m². What is the rms value of the electric field in the electromagnetic wave emitted by the laser?

Answer:

E = 7.21 * 10⁶ N/C

Explanation:

S = 1.38 * 10⁹ W/m²..............(1)

The formula for the average intensity of light can be given by:

S = c∈₀E²

The speed of light, c = 3 * 10⁸ m/s²

Permittivity of air, ∈₀ = 8.85 * 10⁻¹²m-3 kg⁻¹s⁴ A²

Substituting these parameters into equation (1)

1.38 * 10⁹ = 3 * 10⁸ * 8.85 * 10⁻¹² * E²

E² = (1.38 * 10⁹)/(3 * 10⁸ * 8.85 * 10⁻¹²)

E² = 0.052 * 10¹⁷

E = 7.21 * 10⁶ N/C