When the legal speed limit for the New York Thruway was increased from 55 mi/h to 65 mi/h,
how much time was saved by a motorist who drove the 700 km between the Buffalo entrance and
the New York City exit at the legal speed limit?

Answer:

47.17 degree C

Explanation:

mg = 3.5 kg, T1 = 94 degree C, sg = 129 J/kg C

mw = 0.2 kg, T2 = 22 degree C, sw = 4200 J/kg C

Let T be the temperature at equilibrium.

Heat given by the gold = Heat taken by water

mg x sg x (T1 - T) = mw x sw x (T - T2)

3.5 x 129 x (94 - T) = 0.2 x 4200 x (T - 22)

42441 - 451.5 T = 840 T - 18480

60921 = 1291.5 T

T = 47.17 degree C

Answer:

[tex]v_f = 1.6 m/s[/tex]

Explanation:

As per momentum conservation we know that initial momentum of one box car must be equal to the momentum of all box cars together

here we know that there is no external force on the system of all box cars so there momentum conservation is applicable

[tex]m_1v_{i} = (m_1 + m_2 + m_3 + m_4 + m_5)v_{f}[/tex]

so above equation is momentum of one box car is equal to the momentum of all box cars together

So here we will have

[tex]v_f = \frac{m_1 v_i}{m_1 + m_2 + m_3 + m_4 + m_5}[/tex]

[tex]v_f = \frac{m(8)}{5m}[/tex]

[tex]v_f = \frac{8}{5} = 1.6 m/s[/tex]

By applying the conservation of momentum, the final speed of a boxcar linking with four stationary boxcars would be 1.6 m/s. The mass of each boxcar is 18,537 kg and the initial speed of the moving boxcar is 8 m/s before the collision.

The student's question involves finding the final speed of a set of linked boxcars immediately after a collision. This problem deals with the conservation of momentum, a fundamental concept in physics. According to the law of conservation of momentum, the total momentum of a closed system of objects (which does not interact with external forces) remains constant. In this case, because the boxcars are linking together and moving as one unit post-collision, they must conserve momentum.

To compute the final speed after the collision, we will use the formula:

m1 * v1 + m2 * v2 = (m1 + m2 + m3 + m4 + m5) * vf

where

Given that the initial velocity of the stationary boxcars is 0 (since they are at rest) and knowing all the masses are equal:

m = 18,537 kg

v1 = 8 m/s

v2 = 0 m/s (stationary)

We can calculate the final velocity (vf) by the following:

(18,537 kg * 8 m/s) + (4 * 18,537 kg * 0 m/s) = (5 * 18,537 kg) * vf

Solving this for vf gives us:

vf = (18,537 kg * 8 m/s) / (5 * 18,537 kg)

vf = 1.6 m/s

Therefore, the final speed of the boxcars after the collision would be 1.6 m/s.

There is no illustration of the problem provided but I'll attempt to provide an answer.

The relationship between the electric potential difference between two points and the average strength of the electric field between those two points is given by:

║E║ = ΔV/d

║E║ is the magnitude of the average electric field, ΔV is the potential difference between A and B, and d is the distance between A and B.

We are given the following values:

║E║= 10N/C

d = 3m

Plug these values in and solve for ΔV

10 = ΔV/3

ΔV = 30V

Answer:

Work done, W = 401 kJ

Explanation:

It is given that,

Mass of the car, m = 1100 kg

Initial, velocity of the car, u = 27 m/s

Finally it stops, v = 0

According to work energy theorem, the change in kinetic energy is equal to the work done i.e.

[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]

[tex]W=\dfrac{1}{2}\times 1100\ kg\times (0-(27\ m/s)^2)[/tex]

W = -400950 J

or

[tex]W=-401\ kJ[/tex]

-ve sign shows the direction of force and displacement are in opposite direction.

So, 401 kJ work is required to stop a car. Hence, this is the required solution.

To stop an 1100 kg car traveling at 27 m/s, the work required is calculated using the kinetic energy formula KE = ½ mv² and is 401 kJ.

The question asks about the amount of work required to stop a car with a mass of 1100 kg that is traveling at a velocity of 27 m/s. To solve this, we can use the equation for kinetic energy (KE), which is the work needed to accelerate a body of a given mass from rest to its stated velocity. Using the equation KE = ½ mv², we find that KE = ½ (1100 kg) * (27 m/s)² = 399,150 J, which is approximately 401 kJ. Thus, the correct amount of work required to stop the car is 401 kJ.

Explanation:

The electric field is given by :

[tex]E=k\dfrac{Q}{r^2}[/tex]

Where

k is electrostatic constant, [tex]k=\dfrac{1}{4\pi \epsilon_o}=9\times 10^9\ Nm^2/C^2[/tex]

Q is the magnitude of charge

r is the distance from the charge

The gravitational field equation is given by :

[tex]g=\dfrac{GM}{r^2}[/tex]

G is universal gravitational constant

M is the mass

r is the distance

Final answer:

The equivalent gravitational field equation to E = Q/r² is g = G×M/r², where g is the gravitational field strength, G is the gravitational constant, M is the mass, and r is the distance from the mass.

Explanation:

If an electric field is defined by the equation E = Q/r², where E represents the electric field, Q is the charge, and r is the distance from the point charge, the analogous equation for the gravitational field would be g = G×M/r². In this equation, g is the gravitational field strength, G is the gravitational constant, M is the mass creating the gravitational field, and r is the distance from the center of mass.

The concept of the gravitational field is similar to that of the electric field. Both fields decrease with the square of the distance from the source (charge or mass). The field strength represents the force applied per unit charge for the electric field, measured in newtons per coulomb (N/C), and force applied per unit mass for the gravitational field, measured in newtons per kilogram (N/kg).

Answer:

135°.

Explanation:

R = 75 ohm, L = 0.01 H, C = 4 micro F = 4 x 10^-6 F

Frequency is equal to the half of resonant frequency.

Let f0 be the resonant frequency.

[tex]f_{0}=\frac{1}{2\pi \sqrt{LC}}[/tex]

[tex]f_{0}=\frac{1}{2\times 3.14 \sqrt{0.01\times 4\times 10^{-6}}}[/tex]

f0 = 796.2 Hz

f = f0 / 2 = 398.1 Hz

So, XL = 2 x 3.14 x f x L = 2 x 3.14 x 398.1 x 0.01 = 25 ohm

[tex]X_{c}=\frac{1}{2\pi fC}[/tex]

Xc = 100 ohm

[tex]tan\phi = \frac{X_{L}-_{X_{C}}}{R}[/tex]

tan Ф = (25 - 100) / 75 = - 1

Ф = 135°

Thus, the phase difference is 135°.

Answer:

option (c)

Explanation:

The frequency ranges for infrasonic is below 20 Hz.

The frequency ranges for audible is 20 Hz to 20000 Hz.

The frequency ranges for ultrasonic is above 20000 Hz.

So, infrasonic have frequency below the audible range. We cannot hear this frequency.

Answer:

Ratio of the centripetal accelerations = 16

Explanation:

We have ω₁ = 440 rev/min, ω₂ = 110 rev/min

We have centripetal acceleration

         [tex]a=\frac{v^2}{r}=\frac{(r\omega )^2}{r}=r\omega ^2[/tex]

Ratio of centripetal acceleration

         [tex]\frac{a_1}{a_2}=\frac{r\omega_1 ^2}{r\omega_2 ^2}=\left ( \frac{440}{110}\right )^2=4^2=16[/tex]

Ratio of the centripetal accelerations = 16

The centripetal acceleration ratio of a ceiling fan at different angular speeds can be found using the formula a = rω². The ratio a₁/a₂ for angular velocities of 440 rev/min and 110 rev/min is 16.

The ratio of centripetal accelerations a₁/a₂; can be found using the formula a = rω², where a is the centripetal acceleration, r is the distance from the axis of rotation, and ω is the angular velocity.

For ω₁ = 440 rev/min and ω₂ = 110 rev/min, the ratio a₁/a₂ = (rω₁²)/(rω₂²) = (ω₁/ω₂)² = (440/110)² = 16.

Therefore, the ratio of centripetal accelerations a₁/a₂ for the two angular speeds is 16.

Answer:

option (b) 4900 N

Explanation:

m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R

F = G Me x m / (R + h)^2

F = G Me x m / 2R^2

F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2

F = 4900 N

Answer:

624.5 N

Explanation:

m = mass of the truck = 2150 kg

v₀ = initial speed of the truck = 55 km/h = 15.3 m/s

v = final speed of the truck = 33 km/h = 9.2 m/s

t = time interval = 21 s

F = magnitude of the average net force

Using Impulse-Change in momentum equation

F t = m (v - v₀ )

F (21) = (2150) (9.2 - 15.3)

F = - 624.5 N

hence the magnitude of net force is 624.5 N

The magnitude of the average net force acting on the truck during the 21.0 s interval is about 626 N

[tex]\texttt{ }[/tex]

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

mass of truck = m = 2150 kg

initial speed of truck = u = 55.0 km/h = 15⁵/₁₈ m/s

final speed of truck = v = 33.0 km/h = 9¹/₆ m/s

time taken = t = 21.0 s

Asked:

average net force = ∑F = ?

Solution:

[tex]\Sigma F = m a[/tex]

[tex]\Sigma F = m ( v - u ) \div t[/tex]

[tex]\Sigma F = 2150 ( 9\frac{1}{6} - 15 \frac{5}{18} ) \div 21.0[/tex]

[tex]\Sigma F = 2150 ( -6\frac{1}{9} ) \div 21.0[/tex]

[tex]\Sigma F = -626 \texttt{ N}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Kinematics

Answer:

[tex]3.6\cdot 10^{-8} N[/tex]

Explanation:

The electrostatic force between the proton and the electron is given by:

[tex]F=k\frac{q_p q_e}{r^2}[/tex]

where

[tex]k=9.00\cdot 10^9 Nm^2 C^{-2}[/tex] is the Coulomb constant

[tex]q_p = 1.6\cdot 10^{-19} C[/tex] is the magnitude of the charge of the proton

[tex]q_e = 1.6\cdot 10^{-19}C[/tex] is the magnitude of the charge of the electron

[tex]r=8\cdot 10^{-11}m[/tex] is the distance between the proton and the electrons

Substituting the values into the formula, we find

[tex]F=(9\cdot 10^9 ) \frac{(1.6\cdot 10^{-19})^2}{(8\cdot 10^{-11})^2}=3.6\cdot 10^{-8} N[/tex]

Answer:

Density of unknown material = 3.86 g/cm³

Explanation:

Mass of unknown material, M = 54 gm

The object occupied 14 gm of water.

Volume of unknown material = Volume of 14 gm of water.

Density of water = 1 gm/cm³

Volume of 14 gm of water  

                 [tex]V=\frac{14}{1}=14cm^3[/tex]

Volume of unknown material, V = Volume of 14 gm of water = 14 cm³

Density of unknown material    

                              [tex]\rho =\frac{M}{V}=\frac{54}{14}=3.86g/cm^3[/tex]    

Answer:

18.4

Explanation:

Q1 = 4390 J

Q2 = 3582.2 J

The efficiency of heat engine is given by

n = 1 - Q2 / Q1

n = 1 - 3582.2 / 4390

n = 0.184

n = 18.4 %

Answer:

T would be two times the temperature That it was before the double so if the initial temperature was 20 it would now be 40

Explanation:

If the car initially had twice the speed ΔT would be 4 times the original change in temperature of the breaks.

If we assume no heat loss to the surrounding. Then the loss in kinetic energy (KE) of the car after applying the breaks is completely converted into heat energy (Q) of the breaks.

Loss in KE = Gain in Heat Energy

      ΔKE = ΔQ

Originally,    ΔKE = [tex]\frac{1}{2}mv^{2}[/tex] = ΔQ

Now,  ΔQ is directly proportional to change in temperature

           [tex]\frac{1}{2}mv^{2}[/tex] = ΔQ ∝ ΔT

If the speed is doubled, v' = 2v, then

  ΔKE = [tex]\frac{1}{2}mv^{'} ^{2}[/tex] = [tex]\frac{1}{2}m(4v^{2})[/tex] = 4 ×   [tex]\frac{1}{2}mv^{2}[/tex] = 4ΔQ ∝ 4ΔT

Hence, the temperature increases 4 times is speed is doubled.

Learn more about transformation of energy:

https://brainly.com/question/11505666

Answer:

b = 1084.5 kg/s

Explanation:

As we know that the amplitude of damped oscillation is given as

[tex]A = A_o e^{-\frac{bt}{2m}}[/tex]

here we know that

[tex]A = \frac{A_o}{2}[/tex]

after time t = 4.9 minutes

also we know that

[tex]m = 2.30 \times 10^5 kg[/tex]

now we will have

[tex]\frac{A_o}{2} = A_o e^{-\frac{bt}{2m}}[/tex]

[tex]\frac{bt}{2m} = ln2[/tex]

[tex]b = \frac{ln2 (2m)}{t}[/tex]

[tex]b = \frac{2(ln2)(2.30 \times 10^5)}{4.9 \times 60}[/tex]

[tex]b = 1084.5 kg/s[/tex]

Answer:

Given values of Planck Constant are equivalent in English system and metric system.

Explanation:

Value of Planck's constant is given in English system as 4.14 x 10⁻¹⁵eV s.

Converting this in to metric system .

We have 1 eV = 1.6 x 10⁻¹⁹ J

Converting

     4.14 x 10⁻¹⁵eV s = 4.14 x 10⁻¹⁵x 1.6 x 10⁻¹⁹ = 6.63 x 10⁻³⁴ Joule s

So Given values of Planck Constant are equivalent in English system and metric system.

The angular acceleration α of the CD is 1.590 rev/min².

The angular acceleration α can be calculated using the formula:

α = [tex]\frac{\Delta \omega}{\Delta t}[/tex]

Where [tex]\(\Delta \omega\)[/tex] is the change in angular velocity and [tex]\(\Delta t\)[/tex] is the change in time. Angular velocity [tex](\(\omega\))[/tex] is related to angular displacement [tex](\(\theta\))[/tex] and time (\(t\)) by the equation:

[tex]\[ \omega = \frac{\theta}{t} \][/tex]

Given that the CD accelerates from rest [tex](\(\omega_0 = 0\))[/tex] to a constant angular velocity of 477 rev/min, and rotates through an angular displacement of 0.250 rev}, we can use the following relationships:

[tex]\[ \omega = \frac{\theta}{t} \][/tex]

[tex]\[ \omega_f = \frac{477}{1} \, \text{rev/min} \][/tex]

[tex]\[ \omega_0 = 0 \, \text{rev/min} \][/tex]

Substituting these values into the first equation, we get:

[tex]\[ \frac{477}{1} = \frac{0.250}{t} \][/tex]

Solving for t, we find t = 0.525 min.

Now, substitute these values into the angular acceleration formula:

α = [tex]\frac{\frac{477}{1} - 0}{0.525}[/tex]

α = 1.590 rev/min²

Therefore, the angular acceleration of the CD is 1.590 rev/min².

Answer:

39.0 m/s

Explanation:

Linear velocity = angular velocity × radius

v = ωr

v = (30.0 rad/s) (1.30 m)

v = 39.0 m/s

The linear velocity of the ball is 39 m/s.

The given parameters;

The velocity of the ball is the tangential velocity of the ball along the circular path of the elbow joint.

v =  ωr

where;

Substitute the given parameters and solve for the value of v;

v = (30 rad/s) x (1.3 m)

v = 39 m/s

Thus, the linear velocity of the ball is 39 m/s.

Learn more here: https://brainly.com/question/15154527

Answer:

222.22 N

Explanation:

r = 5 cm = 0.05 m, R = 1.5 m, f = ? , F = 200,000 N

Use pascal's law

F / A = f / a

F / 3.14 x R^2 = f / 3.14 x r^2

F / R^2 = f / r^2

f = F x r^2 / R^2

f = 200000 x 0.05 x 0.05 / (1.5 x 1.5)

f = 222.22 N

The total energy is less than the potential energy because of energy conservation. Because you can solve with different linear solutions, I would say it depends on the nature of the system.

Answer:

magnetic field will allow the electron to go through 2 x [tex]10^{4}[/tex]  T k

Explanation:

Given data in question

velocity = 5.0 × [tex]10^{7}[/tex]

electric filed = [tex]10^{4}[/tex]

To find out

what magnetic field will allow the electron to go through, undeflected

solution

we know if electron move without deflection i.e. net force is zero on electron and we can say both electric and magnetic force equal in magnitude and opposite in directions

so we can also say

F(net) = Fe + Fb i.e. = 0

q V B +  q E = 0

q will be cancel out

[tex]10^{4}j[/tex] + 5e + 7i × B =  0

B = 2 x [tex]10^{4}[/tex]  T k

Answer:

m = 77.75 g

Explanation:

Here we know that at equilibrium the temperature of the system will be 10 degree C

so heat given by hot latte = heat absorbed by the ice

now we have

heat given by latte = [tex]m s\Delta T[/tex]

[tex]Q_1 = (200)(4.186)(45 - 10)[/tex]

[tex]Q_1 = 29302 J[/tex]

now heat absorbed by ice is given as

[tex]Q_2 = mL + ms\Delta T[/tex]

[tex]Q_2 = m(335 + 4.186(10 - 0))[/tex]

[tex]Q_2 = m(376.86)[/tex]

now by heat balance we have

[tex]Q_1 = Q_2[/tex]

[tex]29302 = m(376.86)[/tex]

[tex]m = 77.75 g[/tex]

To lower the temperature of the hot latte to 10°C, you need to add approximately 124.87 g of ice.

To find out how much ice you need to add to a hot latte to lower its temperature to 10°C, you can use the formula Q = m × Lf, where Q is the heat transferred, m is the mass of the ice, and Lf is the latent heat of fusion of ice. First, calculate the heat that needs to be transferred:

Q = (200 mL) × (0.997 g/mL) × (45°C - 10°C) × (4.184 J/g°C)

Q = 41,817 J

Next, divide the heat transferred by the latent heat of fusion of ice to find the mass of ice needed:

m = 41,817 J ÷ 335 J/g

m ≈ 124.87 g

Therefore, you need to add approximately 124.87 g of ice to the hot latte to lower its temperature to 10°C.

Answer:

[tex]F = 2.2 \times 10^{-3} N[/tex]

Explanation:

Force of gravitation between Astronaut and Moon is given by the Universal law of gravitation

[tex]F = \frac{Gm_1m_2}{r^2}[/tex]

now we will have

[tex]m_1 = 7.355 \times 10^{22} kg[/tex]

[tex]m_2 = 65 kg[/tex]

[tex]r = 3.81 \times 10^8 m[/tex]

now from above formula we have

[tex]F = \frac{(6.67 \times 10^{-11})(65)(7.355 \times 10^{22})}{(3.81 \times 10^8)^2}[/tex]

[tex]F = 2.2 \times 10^{-3} N[/tex]

Explanation:

Electric force is the force acting between two charged particles. Electric force between electron and proton is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Distance between them, [tex]r=5.3\times 10^{-11}\ m[/tex]

[tex]F=9\times 10^9\times \dfrac{-(1.6\times 10^{-19})^2}{(5.3\times 10^{-11}\ m)^2}[/tex]

[tex]F=-8.2\times 10^{-8}\ N[/tex]

Gravitational force is the force acting between two masses. The gravitational force between the electron and proton is given by :

[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]

Distance between them, [tex]r=5.3\times 10^{-11}\ m[/tex]

[tex]F=6.67\times 10^{-11}\times \dfrac{9.11\times 10^{-31}\ kg\times 1.67\times 10^{-27}\ kg}{(5.3\times 10^{-11}\ m)^2}[/tex]

[tex]F=3.61\times 10^{-47}\ N[/tex]

Hence, this is the required solution.

Answer:

1119.1 K

Explanation:

From Clausius-Clapeyron equation:

[tex]\frac{dP}{dT}=[/tex]Δ[tex]\frac{h_{v} }{R} (\frac{1}{T^{2} } )dT[/tex]

The equation may be integrated considering the enthalpy of vaporization constant, and its result is:

[tex]ln(\frac{P_{2} }{P_{1} } )=-[/tex]Δ[tex]\frac{h_{v} }{R}*(\frac{1}{T_{2} }-\frac{1}{T_{1} })[/tex]

Isolating the temperature [tex]T_{2}[/tex]

[tex]T_{2}=\frac{1}{\frac{-R}{dhv}*ln(\frac{P_{2} }{P_{1}}) +\frac{1}{T_{1}} }[/tex]

[tex]T_{2}=\frac{1}{\frac{-8.314}{1.00*10^5}*ln(\frac{500}{40}) +\frac{1}{906.15}}[/tex]

[tex]T_{2}=1119.1K[/tex]

Note: Remember to change the units of the enthalpy vaporization to J/mol; and the temperatures must be in Kelvin units.

There is a format mistake with the enthalpy of vaporization, each 'Δ' correspond to that.

Answer:

a) 10462341.6×10⁻¹³ A/m²

b) 133709238.907 A

Explanation:

n = Density of the protons in the solar wind = 11.6 cm⁻³ = 11.6×10⁶ m⁻³

v = Velocity of the protons = 563 km/s = 563000 m/s

e = Charge of a proton = 1.602×10⁻¹⁹ coulombs

R = Radius of Earth = 6.3781×10⁶ m

A = Area of Earth = πR² = π(6.3781×10⁶)²=127.8×10¹² m²

a) Current density

J = nev

⇒J = 11.6×10⁶×1.602×10⁻¹⁹×563000

⇒J = 10462341.6×10⁻¹³ A/m²

∴ Current density of these protons is 10462341.6×10⁻¹³ A/m²

b) Current

I = JA

⇒I = 10462341.6×10⁻¹³×127.8×10¹²

⇒I = 1337092389.07×10⁻¹

⇒I = 133709238.907 A

∴ Total current Earth receives is 133709238.907 A

Answer:

i = 0.5 A

Explanation:

As we know that magnetic flux is given as

[tex]\phi = NBA[/tex]

here we know that

N = number of turns

B = magnetic field

A = area of the loop

now we know that rate of change in magnetic flux will induce EMF in the coil

so we have

[tex]EMF = NA\frac{dB}{dt}[/tex]

now plug in all values to find induced EMF

[tex]EMF = (20)(50 \times 10^{-4})(\frac{6 - 2}{2})[/tex]

[tex]EMF = 0.2 volts[/tex]

now by ohm's law we have

[tex]current = \frac{EMF}{Resistance}[/tex]

[tex]i = \frac{0.2}{0.40} = 0.5 A[/tex]

Using Faraday's Law of electromagnetic induction and Ohm's Law, you can calculate that the magnitude of the induced current in the coil is 0.025 amperes.

The magnitude of the induced current can be calculated using Faraday's Law of electromagnetic induction, which states that the induced electromotive force (EMF) in a closed circuit is equal to the time rate of change of magnetic flux through the circuit. Mathematically, it is denoted as:

ε =-N (ΔΦ/Δt)

where

In this problem, N=20, the initial magnetic field (B1) is 2.0 T, the final magnetic field (B2) is 6.0 T, and the time interval (Δt) is 2.0 s.

Magnetic flux (Φ) is given by the product of the magnetic field and the area each loop encloses:

Φ = BA

The change in magnetic flux (ΔΦ) is then:

ΔΦ = N * A * ΔB

where A=50 cm²=5 x 10⁻⁴ m² (converted to m² from cm²), and ΔB = B2 - B1 = 4.0 T.

So,

ΔΦ = 20 * 5 x 10⁻⁴ * 4 = 0.02 Wb (webers)

Now we can calculate the induced EMF using Faraday's Law:

ε = -(0.02 Wb) / 2.0 s = -0.01 V (volts)

The negative sign merely indicates that the induced current flows in a direction to oppose the change in the magnetic field.

Finally, we calculate the magnitude of the induced current (I) by dividing the induced EMF by the total resistance (R) of the coil using Ohm's law:

I = ε / R = 0.01V/0.40 Ω = 0.025 A (amperes)

https://brainly.com/question/12993127

#SPJ11

Answer:

P = 34034.2 Watt

Explanation:

As we know that the slope angle is given as

[tex]\theta = 15.6^0[/tex]

now the weight of the rider along the slope is given as

[tex]W = mgsin\theta[/tex]

[tex]W = 69(9.81)sin15.6[/tex]

[tex]W = 182 N[/tex]

now total weight of all 53 riders along the slope is given as

[tex]F = 53 W[/tex]

[tex]F = 9647.5 N[/tex]

now the speed of the rope is given as

[tex]v = 12.7 km/h = 3.53 m/s[/tex]

now the power required is given as

[tex]P = F.v[/tex]

[tex]P = 9647.5(3.53) [/tex]

[tex]P = 34034.2 Watt[/tex]

The power required to operate the tow is 34020.45 W.

The weight of a single rider is calculated as follows;

[tex]W = mgsin(\theta)\\\\W = (69 \times 9.8) sin(15.6)\\\\W = 181.84 \ N[/tex]

W = 53 x 181.84

W = 9637.52 N

The power required to operate the tow is calculated as follows;

P = Fv

where;

P = 9637.52 x 3.53 = 34020.45 W

Learn more about power here: https://brainly.com/question/25263760

Answer:

0.1029 Kg

Explanation:

[tex]Given\ data[/tex]

[tex]mass\of \ice\ cube =50gm[/tex]

[tex]Latent\ heat \of\ vapourization\ of nitrogen(L) =200kJ/kg[/tex]

[tex]specific\ heat\ of\ ice=2100 J/(kgc)[/tex]

[tex]boiling\ point\ of\ nitrogen=77K\approx-169^{\circ}c[/tex]

[tex]now\ Equating\ heat \ absorb\ by \ ice\ from\ liquid\ nitrogen [/tex]

[tex]{m_{ice}}c[{\Delta T}]={m_{nitrogen}}L[/tex]

[tex]{m_{ice}}2100[0-(-196)}=m_{nitrogen}{200\times 1000}[/tex]

[tex]m_{nitrogen}=0.1029Kg[/tex]

Answer:

[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]

Explanation:

Let the linear charge density of the charged wire is given as

[tex]\frac{q}{L} = \lambda[/tex]

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

[tex]\int E. dA = \frac{q}{\epsilon_0}[/tex]

here we have

[tex]E \int dA = \frac{\lambda L}{\epsilon_0}[/tex]

[tex]E. 2\pi r L = \frac{\lambda L}{\epsilon_0}[/tex]

so we have

[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]