An object of inertia 0.5kg is hung from a spring, and causes it to extend 5cm. In an elevator accelerating downward at 2 m/s^2 , how far will the spring extend if the same object is suspended from it? Draw the free body diagrams for both the accelerating and non-accelerating situations.

Explanation:

It is given that,

Mass of car A, [tex]m_A=1379\ kg[/tex]

Mass of car B, [tex]m_B=1902\ kg[/tex]

In a performance test, each of two cars takes 9.0 s to accelerate from rest to 28 m/s. Their acceleration is given by :

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{28\ m/s}{9}[/tex]

a = 3.12 m/s²

(a) The net force of car A, [tex]F_A=m_Aa[/tex]

[tex]F_A=1379\ kg\times 3.12\ m/s^2[/tex]

[tex]F_A=4302.48\ N[/tex]

(b) The net force of car B, [tex]F_B=m_Ba[/tex]

[tex]F_B=1902\ kg\times 3.12\ m/s^2[/tex]

[tex]F_A=5934.24\ N[/tex]

Hence, this is the required solution.

Final answer:

A gravitational lens occurs when a massive object like a galaxy causes the light from a more distant object to bend and magnify, much like how an optical lens works. Einstein predicted this based on general relativity, and its discovery has provided critical verification of this theory. The first gravitational lens was found in 1979, showing double images of a distant quasar.

Explanation:

A gravitational lens is a phenomenon where the light from a distant object, such as a quasar or galaxy, is bent and magnified by the gravity of a massive object, like a galaxy or galaxy cluster, that lies between the observer and the distant object. This effect is analogous to the way a conventional lens, like those found in glasses, bends light. Albert Einstein predicted this effect based on his theory of general relativity but thought it would be very unlikely for us to observe it. However, astronomers have since observed numerous instances of gravitational lensing, which provide significant evidence for general relativity. These observations include multiple images, arcs, or even rings of the same astronomical object, as gravity distorts the space around the massive intervening object.

The first gravitational lens was discovered in 1979 and presented dual images of the same distant object. This discovery and subsequent observations using powerful telescopes like the Hubble Space Telescope have allowed astronomers to study the universe's most distant objects and the distribution of dark matter. The bend of light due to massive objects can create spectacular effects such as Einstein rings, where a perfect alignment between the observer, massive object, and distant source results in a ring-like structure of light.

Answer:

C. 2.7 m/s

Explanation:

Consider the motion of the marble along the vertical direction

y = vertical displacement = height of tabletop = 1.0 m

a = acceleration = 9.8 m/s²

t = time taken to hit the floor = ?

v₀ = initial speed of the marble along vertical direction = 0 m/s

Using the equation

y = v₀ t + (0.5) a t²

1.0 = (0) t + (0.5) (9.8) t²

t = 0.45 s

Consider the motion of the marble along the horizontal direction :

v = speed with which the marble leaves the table

x = horizontal displacement = 1.2 m

t = 0.45 s

horizontal displacement  is given as

x = v t

1.2 = v (0.45)

v = 2.7 m/s

Answer:

The smallest microwave frequency is 1.5 GHz.

Explanation:

Given that,

Distance d= 20 cm

We need to calculate the wavelength

Using formula for maximum diffraction

[tex]d\sin\theta=\lambda[/tex]

For maximum, [tex]\sin\thea=1[/tex]

[tex]d=\lambda[/tex]

[tex]\lambda=20 cm=0.2 m[/tex]

We need to calculate the frequency

[tex]f =\dfrac{c}{\lambda}[/tex]

[tex]f=\dfrac{3\times10^{8}}{0.2}[/tex]

[tex]f=1.5\times10^{9}\ Hz[/tex]

[tex]f=1.5 GHz[/tex]

Hence, The smallest microwave frequency is 1.5 GHz.

Answer:

a) 2.5 m/s²

b) 6.12 m/s

Explanation:

Tension of rope = T = 356N

Weight of material = W = 478 N

Distance from the ground = s = 7.5 m

Acceleration due to gravity = g = 9.81 m/s²

Mass of material = m = 478/9.81 = 48.72

Final velocity before the bundle hits the ground = v

Initial velocity = u = 0

Acceleration experienced by the material when being lowered = a

a) W-T = ma

⇒478-356 = 48.72×a

[tex]\Rightarrow \frac{122}{48.72} = a[/tex]

⇒a = 2.5 m/s²

∴ Acceleration achieved by the material is 2.5 m/s²

b) v²-u² = 2as

⇒v²-0 = 2×2.5×7.5

⇒v² = 37.5

⇒v = 6.12 m/s

∴ Velocity of the material before hitting the ground is 6.12 m/s

Answer:

The wavelenght is λ= 6 * 10⁻⁴m.

Explanation:

v= 1500 m/s

f= 2.5 MHz = 2.5 *10⁶ Hz

λ=v/f

λ= 6 * 10⁻⁴ meters

Answer:

9

Explanation:

As the net force on Q is zero, so the force on Q due to q1 is balanced by the force on Q due to q2.

Let the force on Q due to q1 is F1 and force on Q due to q2 is F2.

F1 = F2

K Q q1 / (0.5 a)^2 = K Q q2 / (1.5 a)^2

q1 / 0.25 = q2 / 2.25

q2 / q1 = 2.25 / 0.25

q2 / q1 = 9

Answer:

[tex]F_g = 8.9 \times 10^{-30} N[/tex]

[tex]F_e = 1.6 \times 10^{-17} N[/tex]

[tex]F_m = 4.72 \times 10^{-17} N[/tex]

Explanation:

For gravitational force we know that

F = mg

now we have

[tex]m = 9.1 \times 10^{-31} kg[/tex]

[tex]F_g = (9.1 \times 10^{-31})(9.8)[/tex]

[tex]F_g = 8.9 \times 10^{-30} N[/tex]

Now electrostatic force

[tex]F = qE[/tex]

here we have

[tex]F = (1.6 \times 10^{-19})(100)[/tex]

[tex]F = 1.6 \times 10^{-17} N[/tex]

Now magnetic force on it is given by

[tex]F_{m} = qvB[/tex]

[tex]F_m = (1.6 \times 10^{-19})(5.90 \times 10^6)(50 \times 10^{-6})[/tex]

[tex]F_m = 4.72 \times 10^{-17} N[/tex]

The gravitational force on the electron is 8.90 x 10^-31 N downward, the electric force on the electron is 1.60 x 10^-17 N upward, and the magnetic force on the electron is 5.90 x 10^-25 N northward.

The gravitational force on an electron near the surface of the Earth can be calculated using the formula Fg = m * g, where m is the mass of the electron and g is the acceleration due to gravity. The electric force on the electron can be calculated using the formula Fe = q * E, where q is the charge of the electron and E is the magnitude of the electric field. The magnetic force on the electron can be calculated using the formula Fm = q * v * B, where v is the velocity of the electron and B is the magnitude of the magnetic field.

Given that the electron has a mass of 9.11 x 10^-31 kg, a charge of -1.60 x 10^-19 C, and a velocity of 5.90 x 10^6 m/s, the gravitational force on the electron is approximately 8.90 x 10^-31 N downward. The electric force on the electron is approximately 1.60 x 10^-17 N upward. The magnetic force on the electron is approximately 5.90 x 10^-25 N northward.

Answer:

Frequency

Explanation:

The number of cycles completed in one second by an oscillating object is called its frequency. The term frequency shows number of times an oscillation is occurring. It is denoted by [tex]\nu[/tex]. The reciprocal of frequency is called the time period of the wave i.e.

[tex]\nu=\dfrac{1}{T}[/tex]

T = time period

So, the correct option is (a) "frequency".

Final answer:

Frequency is the number of oscillations per second, measured in hertz (Hz), and is inversely related to the period (T), which is the time it takes to complete one oscillation.

Explanation:

Frequency is defined as the number of cycles completed in one second by an oscillating object. In physics, frequency (f) is the number of oscillations that take place per unit of time. The SI unit for frequency is the hertz (Hz), which is equivalent to one oscillation per second. The relationship between frequency and period (T) is given by f = 1/T, meaning that as the period increases, the frequency decreases, and vice versa.

Periodic motion, such as that of a mass suspended by a wire undergoing simple harmonic motion, has a constant time to complete one cycle, known as the period (T). Its units are usually seconds. Period refers to the time it takes to complete one oscillation, while frequency is the number of these oscillations within a given timeframe.

Without stepping up the voltage, 50% of the power is lost in the transmission due to the high I^2R losses. Stepping up the voltage using a transformer significantly reduces these losses and makes power transmission more effective.

The subject question pertains to the percentage power loss in a given transmission line when the voltage is not stepped up. Intuitively, power loss will be considerably greater when voltage is not stepped up due to a relatively high current flowing through the lines that enormously increases the I2R (current squared times the resistance) losses.

Firstly, calculate the initial power, PInitial using the formula P = IV, where I is current and V is voltage. In this case, PInitial = 60 A * 120 V = 7200 W. If the voltage is not stepped up, this power is transmitted at 120 V.

The power dissipated due to resistance (Ploss) can be calculated using the formula P = I2R, where I is current and R is resistance. So: Ploss = (60 A)2 * 1 Ω = 3600 W. The percentage of power loss is thus (Ploss / PInitial) * 100 = (3600 W / 7200 W) * 100 = 50%.

In contrast, when voltage is stepped up to 4500 V using a transformer, the current decreases to maintain the same power, greatly reducing power losses. This is eminently beneficial over long distances, making power transmission more efficient.

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Final answer:

the percentage power lost in the transmission line if the voltage is not stepped up is 50%

Explanation:

To calculate the percentage power lost in the transmission line if the voltage is not stepped up, we first calculate the power loss in the transmission line using the formula P = I²R, where I is the current and R is the resistance of the transmission line.

We are given that the current I is 60 A and the total resistance R is 1.0 Ω. The power loss in the transmission line would be:

P = (60 A)² × 1.0 Ω = 3600 W or 3.6 kW

The total power transmitted without stepping up the voltage is calculated as P = IV, where V is the voltage. Here, V is 120 V, so the total power transmitted is:

P = 60 A × 120 V = 7200 W or 7.2 kW

To find the percentage power lost, we divide the power lost by the total power transmitted and multiply by 100:

Percentage power lost = (3600 W / 7200 W) × 100% = 50%

Without the step-up transformer, 50% of the power would be lost in the transmission line, which is a significant loss compared to transmission at a higher voltage. This example illustrates the importance of high-voltage transmission in reducing power losses.

Answer:

Part a)

W = 5928 J

Part b)

P = 592.8 Watt

Explanation:

As we know that force of friction on the sled is given by

[tex]F_f = \mu F_n[/tex]

here we know that

[tex]\mu = 0.78[/tex]

also we know that normal force on the sled is counter balanced by the weight of the object

[tex]F_n = mg = 760 N[/tex]

now we have

[tex]F_f = (0.78)(760) = 592.8 N[/tex]

Now the applied force must be equal to this friction force so that it will start sliding on the grass

now if we push it by 10 m distance then work done to slide it given by

[tex]W = F.d[/tex]

[tex]W = (592.8)(10) = 5928 J[/tex]

Part B)

Power required to push the sled in 10 s

[tex]Power = \frac{work}{time}[/tex]

[tex]Power = \frac{5928}{10} = 592.8 Watt[/tex]

Answer:

The current pass through the coil is 6.25 A

Explanation:

Given that,

Diameter = 25 cm

Magnetic field = 1.0 mT

Number of turns = 100

We need to calculate the current

Using the formula of magnetic field

[tex]B =\dfrac{\mu_{0}NI}{2\pi r}[/tex]

[tex]I=\dfrac{B\times2\pi r}{\mu N}[/tex]

Where, N = number of turns

r = radius

I = current

Put the value into the formula

[tex]I=\dfrac{1.0\times10^{-3}\times2\pi\times12.5\times10^{-2}}{4\pi\times10^{-7}100}[/tex]

[tex]I=6.25\ A[/tex]

Hence, The current passes through the coil is 6.25 A

Answer:

[tex]f_{B}: f_{A} = \sqrt{\frac{2}{1}}[/tex]

Explanation:

For pendulum A: Length = L and gravity = g

The frequency of pendulum A is given by

[tex]f = \frac{1}{2\pi }\sqrt{\frac{g}{L}}[/tex]

Here, f is the frequency, L be the length

[tex]f_{A} = \frac{1}{2\pi }\sqrt{\frac{g}{L}}[/tex]     ... (1)

For pendulum B: Length = 2L, gravity = g

The frequency of pendulum B is given by

[tex]f_{B} = \frac{1}{2\pi }\sqrt{\frac{g}{2L}}[/tex]   .... (2)

Divide equation (1) by (2)

[tex]f_{B}: f_{A} = \sqrt{\frac{2}{1}}[/tex]

As per the details given, the capacitance of the pair of circular plates is approximately [tex]\(1.11 \times 10^{-11}\) Farads.[/tex]

A capacitor is a device made to store electric charge, and one of its basic electrical characteristics is capacitance.

When a voltage difference (potential difference) exists between a capacitor's two conducting plates, the capacity of the capacitor to store and hold electrical energy is measured.

Capacitance is a measurement of how much charge can be held on a capacitor's plates for a specific voltage.

The capacitance (C) of a parallel plate capacitor is given by the formula:

[tex]\[ C = \dfrac{\varepsilon_0 \cdot A}{d} \][/tex]

Where:

[tex]\( \varepsilon_0 \)[/tex] is the vacuum permittivity ([tex]\(8.85 \times 10^{-12} \, \text{F/m}\)[/tex])

A is the area of the plates (in square meters)

d is the distance between the plates (in meters)

Given:

Radius of the circular plates (r) = 5.0 cm

= 0.05 m

Distance between the plates (d) = 3.2 mm

= 0.0032 m

Calculate the area of one of the plates:

[tex]\[ A = \pi \cdot r^2 \\\\= \pi \cdot (0.05 \, \text{m})^2 \][/tex]

Calculate the capacitance (C):

[tex]\[ C = \dfrac{\varepsilon_0 \cdot A}{d} \\\\= \dfrac{8.85 \times 10^{-12} \, \text{F/m} \cdot \pi \cdot (0.05 \, \text{m})^2}{0.0032 \, \text{m}} \][/tex]

[tex]\[ C \approx 1.11 \times 10^{-11} \, \text{F} \][/tex]

Thus, the capacitance of the pair of circular plates is approximately [tex]\(1.11 \times 10^{-11}\)[/tex] Farads.

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The capacitance of a pair of circular plates with a radius of 5.0 cm separated by 3.2 mm of mica is 10.3 pF. This is calculated using the formula for the capacitance of a parallel plate capacitor and the area of a circle.

To determine the capacitance of a pair of circular plates of a parallel plate capacitor with a given radius, separated by a specific dielectric material, we use the formula:

C = ε0 εr (A/d)

Where:

C is the capacitance,

ε0 is the vacuum permittivity (8.85 × 10-12 F/m),

εr is the relative permittivity of the dielectric (for mica, it's approximately 6),

A is the area of one of the plates, and

d is the separation between the plates.

To find the area A of a circular plate with radius r, we use the formula A = πr2. For a radius of 5.0 cm (or 0.05 m), A = π(0.05 m)2 = 7.85 × 10-3 m2.

Substituting into the capacitance formula, with d = 3.2 mm (or 0.0032 m), we get:

C = (8.85 × 10-12 F/m) × 6 × (7.85 × 10-3 m2/0.0032 m) = 1.03 × 10-11 F

This result is usually expressed in picofarads (pF), so the capacitance is 10.3 pF.

Answer:

B) 2.18 m/s²

Explanation:

M = Original mass of earth

R = Original radius of earth

g = original acceleration due to gravity of earth = 9.8 m/s²

M' = New mass of earth = 2 M

R' = New radius of earth = 3 R

g = original acceleration due to gravity of earth = 9.8 m/s²

Original acceleration due to gravity of earth is given as

[tex]g=\frac{GM}{R^{2}}[/tex]

[tex]9.8 =\frac{GM}{R^{2}}[/tex]                               eq-1

g' = new acceleration due to gravity of earth

New acceleration due to gravity of earth is given as

[tex]g' =\frac{GM'}{R'^{2}}[/tex]

[tex]g' =\frac{G(2M)}{(3R)^{2}}[/tex]

[tex]g'=\left ( \frac{2}{9} \right )\frac{GM}{R^{2}}[/tex]

Using eq-1

[tex]g'=\left ( \frac{2}{9} \right )(9.8)[/tex]

g' = 2.18 m/s²

Final answer:

When the Earth's radius is tripled and its mass is doubled, the new surface gravitational acceleration would be calculated using the gravitational acceleration formula and would result in a new acceleration of B) 2.18 m/s².

Explanation:

To find the new surface gravitational acceleration (g') when the radius of the Earth is tripled and its mass is doubled, we use the formula for gravitational acceleration:

g' = (G × new mass) / (new radius)²

Where:

Substituting the new mass and radius into the formula, we get:

g' = (G × 2M) / (3R)²

g' = (2G × M) / (9R²)

g' = (2/9) × (G × M / R²)

Since G × M / R² is the original gravitational acceleration (g) of Earth, which is 9.8 m/s², we can now substitute:

g' = (2/9) × 9.8 m/s²

g' = 2.18 m/s²

So the new surface gravitational acceleration would be 2.18 m/s², which matches option B.

Answer:

8.64 m/s

Explanation:

v1 = 8 m/s, v2 = ?, P2 - P1 = 5338 Pa, density of water, d = 1000 kg/m^3

By the use of Bernoulli's theorem

P 1 + 1/2 x d x v1^2 = P2 + 1/2 x d x v2^2

P2 - P1 = 1/2 x d x (v2^2 - v1^2)

5338 = 0.5 x 1000 x (v2^2 - 64)

10.676 = v2^2 - 64

v2^2 = 74.676

v2 = 8.64 m/s  

Answer:

W = 36064 J

Explanation:

Work done by the force of gravity is given as

[tex]W_g = F_g \times d[/tex]

here we know that

[tex]F_g = mg[/tex]

also we have

[tex]d[/tex] = vertical displacement of the car in downward direction

now we have

[tex]d = 8 sin30[/tex]

[tex]d = 4 m[/tex]

now work done is given as

[tex]W = (mg) d[/tex]

[tex]W = (920 \times 9.81)(4)[/tex]

[tex]W = 36064 J[/tex]

Answer:

Speed of another player, v₂ = 1.47 m/s

Explanation:

It is given that,

Mass of football player, m₁ = 88 kg

Speed of player, v₁ = 2 m/s

Mass of player of opposing team, m₂ = 120 kg

The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :

[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]

V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.

[tex]v_2=-\dfrac{m_1v_1}{m_2}[/tex]

[tex]v_2=-\dfrac{88\ kg\times 2\ m/s}{120\ kg}[/tex]

[tex]v_2=-1.47\ m/s[/tex]

So, the speed of another player is 1.47 m/s. Hence, this is the required solution.

Answer:

The elastic deformation is 0.00131.

Explanation:

Given that,

Force F = 44500 N

Cross section [tex]A =15.2mm\times19.1\ mm[/tex]

We Calculate the stress

Using formula of stress

[tex]\sigma=\dfrac{F}{A}[/tex]

Where, F = force

A = area of cross section

Put the value into the formula

[tex]\sigma=\dfrac{44500}{15.2\times10^{-3}\times19.1\times10^{-3}}[/tex]

[tex]\sigma=153.27\times10^{6}\ N/m^2[/tex]

We need to calculate the strain

Using formula of strain

[tex]Y=\dfrac{\sigma}{\epsilon}[/tex]

[tex]epsilon=\dfrac{\sigma}{Y}[/tex]

Where,

[tex]\sigma[/tex]=stress

Y = young modulus of copper

Put the value into the formula

[tex]\epsilon=\dfrac{153.27\times10^{6}}{117\times10^{9}}[/tex]

[tex]\epsilon =0.00131[/tex]

Hence, The elastic deformation is 0.00131.

Bathroom scales measure weight by compressing springs in proportion to the applied force. Using Hooke's law, the spring constant can be determined to calculate the weight. If the scale is compressed by 0.21 inches, a 210-lb person is standing on it. When compressed by 0.06 inches, the weight is calculated to be 60 lb. The work done in compressing the scale by 0.06 inches is 3.6 lb-in or 0.3 ft-lb.

Bathroom scales measure weight. When you stand on a bathroom scale, it compresses slightly due to the weight placed on it. The scale contains springs that compress in proportion to the weight, similar to how rubber bands expand when pulled. Using Hooke's law, we can calculate the weight of someone based on how much the scale is compressed.

If a 210-lb person compresses the scale by 0.21 inches, and assuming the scale behaves like a spring, we can use Hooke's law (F = kx) to find the spring constant. The spring constant is the amount of force required to compress the spring by a certain distance. Rearranging the equation for Hooke's law, we get k = F/x. Plugging in the values, k = 210 lb / 0.21 in = 1000 lb/in.

To find the weight of someone who compresses the scale by 0.06 inches, we can again use Hooke's law. Rearranging the equation to solve for F, we have F = kx. Plugging in the values, F = 1000 lb/in * 0.06 in = 60 lb.

To calculate the work done in compressing the scale by 0.06 inches, we can use the formula for work: W = Fd, where W is work, F is force, and d is distance. Plugging in the values, W = 60 lb * 0.06 in = 3.6 lb-in or 0.3 ft-lb.

Answer:

67.5 cm

Explanation:

u = - 90 cm, v = 3 x u = 3 x 90 = 270 cm

let f be the focal length

Use lens equation

1 / f = 1 / v - 1 / u

1 / f = 1 / 270 + 1 / 90

1 / f = 4 / 270

f = 67.5 cm

To determine the focal length of the lens, we use the lens formula and set up an equation based on the given information. Solving for the image distance, we find that it is zero, indicating the image is formed at infinity. Therefore, the focal length of the lens is 90 cm.

To determine the focal length of the lens, we can use the lens formula:

1/f = 1/v - 1/u

Where f is the focal length, v is the image distance, and u is the object distance.

Given that the image distance triples when the object is moved from infinity to 90 cm in front of the lens, we can set up the following equation:

1/f = 1/(3v) - 1/(90)

Multiplying through by 90*3v, we get:

90*3v/f = 270v - 90*3v

90*3v/f = 270v - 270v

90*3v/f = 0

Simplifying further, we find that: v = 0

When the image distance is zero, it means the image is formed at infinity, so the lens is focused at the focal point. Therefore, the focal length of the lens is 90 cm.

The greatest slope the shoulder can have without causing the car to slip or tip over is 22.3⁰.

The given parameters;

The greatest slope of the shoulder is determined by taking net force on the car as shown below;

[tex]\Sigma F= ma \\\\Wsin\theta - \mu_s Wcos \theta = ma[/tex]

at constant velocity, acceleration, a, = 0

[tex]Wsin\theta - \mu_s W cos \theta = 0\\\\W sin\theta = \mu_s Wcos \theta \\\\\mu_s = \frac{Wsin\theta }{W cos \theta } \\\\\mu_s = tan \theta \\\\\theta = tan^{-1}(\mu_s)\\\\\theta = tan^{-1}(0.41)\\\\\theta =22.3\ ^0[/tex]

Thus, the greatest slope the shoulder can have without causing the car to slip or tip over is 22.3⁰.

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The greatest slope the car can be without slipping or tipping over is approximately 22.4 degrees where weight due to gravity is balanced by static friction.

This problem involves the concept of static friction, which prevents an object from sliding down a slope. Here, the car is held in place on a slope due to static friction between the tires and the road. First, we need to calculate the force of gravity acting on the car, which is the weight of the car mg. The mass of the car is given as 1.6 Mg = 1.6 x 10^6 grams. Therefore, the weight of the car is 1.6 x 10^6 g * 9.8 m/s² = 1.568 x 10^7 N.

Next, the force of static friction is given by the equation μsN, where N is the normal force and μs is the coefficient of static friction which is 0.41. Since the car is not moving, the net force on it must be zero. This implies that the force of static friction must balance out the force of gravity. So, we equate the forces and solve for the slope angle θ, getting a relationship: tan θ = μs.

So, θ = tan^-1 (0.41) ≈ 22.4°. Therefore, the greatest slope the shoulder can have without causing the car to slip or tip over is approximately 22.4 degrees.

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Answer:

4.999 × 10⁴ MeV

Explanation:

Kinetic energy is related to the rest energy by the Lorentz transformation equation as shown below.

Kinetic energy in the lab frame K= ( γ -1 ) Eo

The Lorentz constant γ = [tex]1/\sqrt{1-v^2/c^2}[/tex]

So γ =  [tex]1/\sqrt{1- (0.99217 c)^2/c^2}[/tex]= 8.0067

⇒ K = (8.0067-1)(7134) = 49986 MeV

                                       =49,999 MeV ( 4 significant figures)

Answer:

The magnetization of a small sample of aluminum is 40.6 A/m

Explanation:

Given that,

Magnetic susceptibility [tex]\chi_{1}=1.7\times10^{-5}[/tex] at 300 K

Magnetic field = 1.5 T

We need to calculate the magnetic susceptibility at 150 K

[tex]\dfrac{\chi_{2}}{\chi_{1}}=\dfrac{T_{1}}{T_{2}}[/tex]

Put the value into the formula

[tex]\dfrac{\chi_{2}}{1.7\times10^{-5}}=\dfrac{300}{150}[/tex]

[tex]\chi_{2}=\dfrac{300\times1.7\times10^{-5}}{150}[/tex]

[tex]\chi_{2}=3.4\times10^{-5}[/tex]

We need to calculate the magnetization

Using formula of magnetization

[tex]I=\chi_{2}\times H[/tex]

Where, H = magnetic intensity

Formula of magnetic intensity

[tex]H=\dfrac{B}{\mu_{0}}[/tex]

Where, B = magnetic field

Put the value of H into the formula of magnetization

[tex]I=\chi_{2}\times\dfrac{B}{\mu_{0}}[/tex]

[tex]I=3.4\times10^{-5}\times\dfrac{1.5}{4\pi\times10^{-7}}[/tex]

[tex]I=40.6\ A/m[/tex]

Hence, The magnetization of a small sample of aluminum is 40.6 A/m

Answer:

The pupillary light reflex (PLR) or photopupillary reflex is a reflex that controls the diameter of the pupil, in response to the intensity (luminance) of light that falls on the retinal ganglion cells of the retina in the back of the eye, thereby assisting in adaptation of vision to various levels of lightness/darkness.

A greater intensity of light causes the pupil to constrict (miosis/myosis; thereby allowing less light in), whereas a lower intensity of light causes the pupil to dilate (mydriasis, expansion; thereby allowing more light in)

Explanation:

Answer:

23.63 °C

Explanation:

[tex]m_{w}[/tex] = mass of water = 0.250 kg

[tex]T_{wi}[/tex] = initial temperature of water = 20.0 °C

[tex]c_{w}[/tex] = Specific heat of water = 4186 J/(kg °C)

[tex]m_{a}[/tex] = mass of aluminum = 0.400 kg

[tex]T_{ai}[/tex] = initial temperature of aluminum = 26.0 °C

[tex]c_{a}[/tex] = Specific heat of aluminum = 900 J/(kg °C)

[tex]m_{c}[/tex] = mass of copper = 0.100 kg

[tex]T_{ci}[/tex] = initial temperature of copper = 100 °C

[tex]c_{c}[/tex] = Specific heat of copper = 386 J/(kg °C)

[tex]T_{f}[/tex] = Final temperature of mixture

Using conservation of heat

[tex]m_{w}[/tex] [tex]c_{w}[/tex] ([tex]T_{f}[/tex] - [tex]c_{w}[/tex]) + [tex]m_{a}[/tex] [tex]c_{a}[/tex] ([tex]T_{f}[/tex] - [tex]T_{ai}[/tex] ) + [tex]m_{c}[/tex] [tex]c_{c}[/tex] ([tex]T_{f}[/tex] - [tex]T_{ci}[/tex] ) = 0

(0.250) (4186) ([tex]T_{f}[/tex] - 20) + (0.400) (900) ([tex]T_{f}[/tex] - 26) + (0.100) (386) ([tex]T_{f}[/tex] - 100) = 0

[tex]T_{f}[/tex] = 23.63 °C

To find the work needed to pump half of the water out of the aquarium, you need to calculate the volume of the aquarium and the mass of the water. The work done is equal to the force (weight of the water) multiplied by the distance (height of the aquarium). Therefore, the work needed is 29400 Nm or J (Joules).

To find the work needed to pump half of the water out of the aquarium, we first need to calculate the volume of the aquarium. The volume is given by length x width x height, which equals 3m x 2m x 1m = 6m³. Half of the water in the aquarium would be 6m³/2 = 3m³.

The density of water is given as 1000 kg/m³. So the mass of the water that needs to be pumped out is 3m³ x 1000 kg/m³ = 3000 kg.

The work done to pump the water out is given by the formula work = force x distance. In this case, the force required is the weight of the water, which is mass x gravity. The distance is the height of the aquarium, which is 1m.

Therefore, the work needed to pump half of the water out of the aquarium is 3000 kg x 9.8 m/s² x 1m = 29400 Nm or J (Joules).

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Answer:

electrical conductivity is given as

[tex]\sigma = 3.8 \times 10^{-15} [/tex]

Explanation:

For a current carrying conductor we will have relation between current density and electric field inside the conductor as

[tex]j = \sigma E[/tex]

here we have

current density = j

electric field = E

given that

[tex]j = 7.00 \times 10^{-13}[/tex]

[tex]E = 185 V/m[/tex]

now we will have

[tex]\sigma = \frac{j}{E}[/tex]

[tex]\sigma = \frac{7.00 \times 10^{-13}}{185}[/tex]

[tex]\sigma = 3.8 \times 10^{-15} [/tex]

Answer:203.60 N

Explanation:

Given

[tex]V_{inlet}[/tex]=3 m/s

[tex]d_{inlet}[/tex]=6 cm

[tex]d_{outlet}[/tex]=2 cm

Using continuity equation

[tex]A_{inlet}V_{inlet}=A_{outlet}V_{outlet}[/tex]

[tex]\frac{\pi \times d_{inlet}^2}{4}\times 3[/tex]=[tex]\frac{\pi \times d_{outlet}^2}{4}\times V_{outlet}[/tex]

[tex]V_{outlet}[/tex]=27 m/s

Horizontal Force required to hold elbow=Change in momentum of water

initial momentum=[tex]\rho \times A_{inlet}\times V_{inlet}=10^{3}\times \frac{\pi}{4}6^2\times 3^{2}\times 10^{-4}[/tex]

Final Momentum=[tex]\rho \times A_{outlet}\times V_{outlet}=10^{3}\times \frac{\pi}{4}2^2\times 27^{2}\times 10^{-4}[/tex]

Force required=[tex]10^{3}\times \frac{\pi }{4}\times 10^{-4}\left [ 27^{2}\times 2{2}-6{2}\times 3{2}\right ][/tex]=203.60 N

Answer:

72.75 kg

Explanation:

[tex]F[/tex] = force applied on a box = 750 N

[tex]m[/tex] = mass of the box

[tex]N[/tex] = Normal force on the box

[tex]\mu _{s}[/tex] = Coefficient of static friction = 0.66

From the force diagram, force equation along the vertical direction is given as

[tex]N = F Sin25 + mg[/tex]

[tex]N = 750 Sin25 + mg[/tex]                                                         eq-1

Static frictional force is given as

[tex]f_{s} = \mu _{s} N[/tex]

using eq-1

[tex]f_{s} = \mu _{s} (750 Sin25 + mg)[/tex]

For the box to move,

[tex]F Cos25 = f_{s}[/tex]

[tex]750 Cos25 = \mu _{s} (750 Sin25 + mg)[/tex]

[tex]750 Cos25 = (0.66) (750 Sin25 + m (9.8))[/tex]

m = 72.75 kg