a 5 kg bowling ball is sliding on a frictionless surface at 5 m/s. how much force is needed to keep it going at 5 m/s? (2:20)

The passenger's weight remains the same throughout the elevator's motion. The weight is determined by the gravitational force acting on the passenger, which is unaffected by the elevator's motion.

let's consider the three scenarios:

a. Before the elevator starts moving: The passenger's weight is determined by the gravitational force acting on them. Therefore, the weight of the passenger is the same as their mass multiplied by the acceleration due to gravity (9.8 m/s^2).

b. While the elevator is speeding up: During this phase, the passenger experiences an additional acceleration due to the elevator's upward motion. The passenger's apparent weight increases, resulting from the combination of the gravitational force and the upward acceleration of the elevator. The total force acting on the passenger is the sum of their actual weight (mg) and the upward force due to acceleration (ma), where m is the mass of the passenger and a is the elevator's acceleration.

c. After the elevator reaches its cruising speed: Once the elevator reaches its cruising speed, it travels at a constant velocity, and the passenger experiences a steady state without any acceleration. At this point, the passenger's weight returns to their actual weight, determined solely by the gravitational force.

Therefore, the passenger's weight is the same before the elevator starts moving (a) and after it reaches its cruising speed (c), while it increases during the period when the elevator is speeding up (b) due to the combined effects of gravitational force and upward acceleration.

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The potential difference between the two ends of the wire due to the electric field is:

ΔV = EL = (1.8 V/m) × (2.04 m) = 3.67 V

The potential difference between two points in an electric field can be calculated by multiplying the electric field strength by the distance between the two points.

In this case, the electric field strength is given as 1.8 V/m, and the length of the wire is 2.04 m. Therefore, the potential difference between the two ends of the wire is ΔV = EL = (1.8 V/m) × (2.04 m) = 3.67 V.

This potential difference represents the work done by the electric field on a unit charge as it moves from one end of the wire to the other. If a charge q moves from one end of the wire to the other, it will experience a force F = qE due to the electric field and will do work W = FΔx = qEΔx. Since the potential difference ΔV is defined as the work done per unit charge, we have ΔV = W/q = EΔx, which is the same as the expression we used to calculate the potential difference above.

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The visible wavelengths that will be constructively reflected from the film depend on the thickness of the film and the refractive index of the material. The reflected wavelengths will be those that are in phase with the incident light waves after reflecting off the top and bottom surfaces of the film. This is known as constructive interference.

Therefore, the reflected wavelengths will vary depending on the film thickness and refractive index. If the film is very thin, only a narrow range of wavelengths will be reflected. As the thickness increases, the range of reflected wavelengths will widen. Generally, the visible wavelengths that are reflected will be those that are close to the color of the film.

For example, a blue film will reflect blue wavelengths. In some cases, multiple wavelengths may be constructively reflected, resulting in a iridescent or rainbow effect. Therefore, the specific visible wavelengths that are constructively reflected from the film cannot be determined without additional information about the film's thickness and refractive index.

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A thin 6. 5-kg wheel of radius 34 cm is weighted to one side by a 1. 30-kg weight, small in size, placed 23 cm from the center of the wheel. The position of the center of mass of the weighted wheel is 0.026 m. The moment of inertia about an axis through its CM, perpendicular to its face  0.925 kg*m^2

(a) To find the position of the center of mass of the weighted wheel, we need to calculate the distance of the center of mass from the center of the wheel. We can use the formula:

x_cm = [tex](m_1 * x_1 + m_2 * x_2) / (m_1 + m_2)[/tex]

where [tex]x_1[/tex] and [tex]m_1[/tex] are the position and mass of the wheel, [tex]x_2[/tex] and [tex]m_2[/tex] are the position and mass of the weight.

Substituting the values, we get:

x_cm = ([tex]6.5 kg * 0 + 1.3 kg * 0.23[/tex] m) / (6.5 kg + 1.3 kg)

= 0.026 m

Therefore, the center of mass of the weighted wheel is 0.026 m away from the center of the wheel.

(b) To calculate the moment of inertia about an axis through its CM, perpendicular to its face, we can use the parallel axis theorem:

I = [tex]I_cm + m*d^2[/tex]

where I_cm is the moment of inertia about an axis through the center of mass, m is the total mass of the system, and d is the distance between the center of mass and the axis of rotation.

The moment of inertia of a thin disk about an axis through its center is:

I_cm = [tex](1/2)mr^2[/tex]

Substituting the values, we get:

d = 0.034 m - 0.026 m = 0.008

I = [tex](1/2)6.5 kg(0.34 m)^2 + 1.3 kg*(0.008 m)^2[/tex]

= [tex]0.925 kg*m^2[/tex]

Therefore, the moment of inertia of the weighted wheel about an axis through its CM, perpendicular to its face, is [tex]0.925 kg*m^2.[/tex]

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a) The object displaced 60 mL of water, which is equivalent to its volume.

b) The density of the object is  2.5 g/mL

c) The density of the unknown liquid is 1.25 g/mL

a) To find the volume of the object, we can use the formula for density: density = mass/volume. We know the mass of the object in both air and water, so we can use the difference in those two masses to find the volume of the object.
150 g - 90 g = 60 g
This means that the object displaced 60 mL of water, which is equivalent to its volume.
b) To find the density of the object, we can use the formula for density again, using the mass of the object in air and the volume we just found:
density = \frac{mass}{volume}
density = \frac{150 g}{60 mL }
density = 2.5 g/mL
c) To find the density of the unknown liquid, we can use the same formula as before, using the mass of the object in the liquid and the volume we just found:
density = \frac{mass}{volume }
density =\frac{ 75 g}{60 mL }
density = 1.25 g/mL
So the density of the unknown liquid is 1.25 g/mL.

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The vector equations for the electric field E⃗ (z,t) and magnetic field B⃗ (z,t) can be written as follows:

E⃗ (z,t) = E0 sin(kz - ωt) ẑ

B⃗ (z,t) = B0 sin(kz - ωt) ỵ

The vector equations for the electric field E⃗ (z,t) and magnetic field B⃗ (z,t) can be written as follows:

E⃗ (z,t) = E0 sin(kz - ωt) ẑ

B⃗ (z,t) = B0 sin(kz - ωt) ỵ

Where E0 and B0 are the maximum amplitudes of the electric and magnetic fields, respectively, k is the wave number, ω is the angular frequency, and ẑ and ỵ are unit vectors in the z and y directions, respectively.

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According to the given statement, the maximum height of the ball above the field is approximately 79.5 feet, and the total time it is in the air (hang time) is approximately 4.43 seconds.

We will need to use the kinematic equations of motion and some basic trigonometry. First, let's break down the initial velocity of the ball into its horizontal and vertical components. The horizontal component can be found using the cosine function:
cos(56°) = adjacent/hypotenuse
cos(56°) = x/85 ft/sec
x = 46.6 ft/sec
Therefore, the initial horizontal velocity of the ball is 46.6 ft/sec. The vertical component can be found using the sine function:
sin(56°) = opposite/hypotenuse
sin(56°) = y/85 ft/sec
y = 71.8 ft/sec
Therefore, the initial vertical velocity of the ball is 71.8 ft/sec. Now, we can use these values to find the ball's maximum height above the field. We can use the following kinematic equation:
y = y0 + v0yt - 1/2gt^2
where y0 is the initial height (0 ft), v0y is the initial vertical velocity (71.8 ft/sec), g is the acceleration due to gravity (-32.2 ft/sec^2), and t is the time it takes for the ball to reach its maximum height (unknown). We can solve for t by setting the velocity to 0 at the maximum height:
0 = 71.8 - 32.2t
t = 2.23 seconds
Now, we can plug in this value of t to find the maximum height:
y = 0 + 71.8(2.23) - 1/2(-32.2)(2.23)^2
y = 49.2 ft
Therefore, the ball's maximum height above the field is 49.2 ft.
To find the hang time, we can use the following kinematic equation:
y = y0 + v0yt - 1/2gt^2
where y is the final height (0 ft), y0 is the initial height (0 ft), v0y is the initial vertical velocity (71.8 ft/sec), g is the acceleration due to gravity (-32.2 ft/sec^2), and t is the total time the ball is in the air (unknown). We can solve for t by setting y to 0:
0 = 0 + 71.8t - 1/2(-32.2)t^2
t = 4.44 seconds
Therefore, the hang time of the football is 4.44 seconds.
In conclusion, the ball's maximum height above the field is 49.2 ft, and the hang time of the football is 4.44 seconds.
I'd be happy to help you with your question.
(a) To find the maximum height of the ball above the field, we first need to determine the vertical component of the initial velocity. We can do this using the following equation:
V_vertical = V_initial * sin(angle)
V_vertical = 85 ft/sec * sin(56°)
V_vertical ≈ 71.2 ft/sec
Now we can use the following equation to find the maximum height (H_max) achieved by the ball:
H_max = (V_vertical^2) / (2 * g)
where g is the acceleration due to gravity (32.2 ft/sec^2)
H_max = (71.2^2) / (2 * 32.2)
H_max ≈ 79.5 ft
(b) To find the "hang time," we can use the following equation:
T_total = 2 * V_vertical / g
T_total = 2 * 71.2 / 32.2
T_total ≈ 4.43 sec
So, the maximum height of the ball above the field is approximately 79.5 feet, and the total time it is in the air (hang time) is approximately 4.43 seconds.

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To calculate the mass of fuel needed to reach a top speed of 4000 m/s, we can use the rocket equation, which relates the mass of the rocket, the mass of the fuel, and the velocities involved.

The rocket equation, derived from Newton's laws of motion, allows us to calculate the mass of fuel required to achieve a desired change in velocity. It is given by:

Δv = v_e * ln(m_i / m_f)

Where:

Δv is the change in velocity (top speed),

v_e is the exhaust velocity of the burned fuel (2500 m/s),

m_i is the initial mass of the rocket (empty mass + fuel mass),

m_f is the final mass of the rocket (empty mass).

To find the mass of fuel required to reach a top speed of 4000 m/s, we rearrange the equation to solve for m_i:

m_i = m_f * e^(Δv / v_e)

Substituting the given values:

m_f = 150 kg (empty mass)

Δv = 4000 m/s

v_e = 2500 m/s

We can calculate m_i as follows:

m_i = 150 kg * e^(4000 m/s / 2500 m/s)

After evaluating the expression, we find that the mass of fuel needed to reach a top speed of 4000 m/s is approximately 283.9 kg.

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The period of the wave is π s, the wavelength is 15.7 cm, the speed of the wave is approximately 5.00 cm/s, and the initial phase shift is pi/10 radians to the left.

The given wave function y(x, t) represents a sinusoidal wave with amplitude of 3.00 cm, wavenumber of 0.4 m^-1, angular frequency of 2.00 s^-1 and an initial phase shift of pi/10. To find the period, we can use the formula T = \frac{2π}{ω}, where ω is the angular frequency.

Thus, T =\frac{ 2π}{2.00 }

T = π s
The wavelength can be found using the formula λ =\frac{ 2π}{k}, where k is the wavenumber. Thus, λ = \frac{2π}{0.4}

λ = 15.7 cm.
The speed of the wave can be found by multiplying the wavelength by the angular frequency, i.e., v =\frac{ ω}{k}

=\frac{ λ}{T} =\frac{ 15.7}{π} ≈ 5.00 cm/s.
The initial phase shift of the wave is given as pi/10, which means that the wave is shifted pi/10 radians to the left from its equilibrium position.

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b. No, the vector sum of forces exerted on the baby at the top of the bounce is not zero. At the top of the bounce, the baby's velocity is instantaneously zero.

But this does not mean that the forces acting on him are balanced. There are two main forces acting on the baby in this scenario: gravity and the elastic force from the cord.

Gravity is acting downward on the baby, pulling him towards the ground. The elastic force from the cord is acting in the opposite direction, pulling him upward. At the top of the bounce, the elastic force must be greater than the gravitational force to slow down the baby's upward motion and bring him to a momentary stop. At this instant, the forces are not balanced because the elastic force is greater than the gravitational force. Consequently, the vector sum of forces exerted on the baby is not zero.

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Example categorized as correct substitution problem using equation from section to evaluate current.

This means that the given problem involves replacing variables in an equation with given values and solving for the unknown variable. The solution obtained using this method is deemed correct according to the equation developed in the section. The use of equations in problem-solving is a common practice in various fields, including mathematics, physics, and engineering. By categorizing problems, it becomes easier to identify the appropriate methods to use in solving them, which can improve problem-solving efficiency and accuracy. Example categorized as correct substitution problem using equation from section to evaluate current. Imax is the maximum current and the answer obtained is deemed correct according to the equation developed in the section.

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The escape speed of an electron launched from the surface of a 1.0 cm diameter plastic sphere that has been charged to 11 nC is 2.31 x 10⁷ m/s.

The escape speed is the minimum speed an object needs to escape the gravitational attraction of a massive body. In this case, the object is an electron launched from the surface of a charged plastic sphere.

The escape speed can be calculated using the formula:

v = √(2GM/r)

where v is the escape speed, G is the gravitational constant, M is the mass of the plastic sphere, and r is the distance from the center of the sphere to the surface.

To find M, we need to first calculate the charge of the sphere. The charge Q can be found using the formula:

Q = CV

where C is the capacitance of the sphere and V is the voltage applied to it.

Assuming the sphere is a perfect conductor and has a uniform charge distribution, the capacitance can be calculated using:

C = 4πεr² / (1 - (b/a)²)

where ε is the permittivity of free space, r is the radius of the sphere, and a and b are the radii of two concentric spheres that enclose the charged sphere. Since the sphere is charged to 11 nC, we can assume that V = 11 nC / C.

Assuming the sphere is made of plastic with a density of 1 g/cm³, its mass can be calculated as M = 4/3 πr³ρ.

Plugging in the values, we get:

Q = CV = 11 nC

C = 4πεr² / (1 - (b/a)²)

M = 4/3 πr³ρ = 4/3 π(0.5 cm)³(1 g/cm³) = 0.5236 g

Using the above values, we can calculate the escape speed as:

v = √(2GM/r) = √(2(6.674 x 10⁻¹¹ N m²/kg²)(0.0005236 kg) / (0.5 cm)) = 2.31 x 10⁷ m/s

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When a long line of positive charges is placed very close together, each charge creates its own electric field that points radially outward from it.

As we move down the line of charges, the contribution of each charge's electric field adds up to form a larger electric field. Therefore, the electric field becomes stronger as we move down the line of charges.

The electric field created by a long line of charges is proportional to the inverse of the distance from the line. So, as we move closer to the line of charges, the electric field becomes stronger. If we move away from the line of charges, the electric field becomes weaker.

However, the strength of the field also depends on the amount of charge on the line of charges. Therefore, the field will increase or decrease more rapidly depending on the magnitude of the charge.

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If the generator is working on 3*10² V then effective voltage is 2.12*10² V

and effective current is 3.95 A.

effective voltage is calculated as dividing the voltage by √2

Hence effective voltage = 3*10² V / √2 = 2.12*10² V

Current flowing in the circuit is I = V/R = 3*10² V / 53 = 5.6 A

the effective voltage is given as dividing the current by √2

The effective current is = 5.6 A/ √2 = 3.95 A

the amount of alternating or other variable current that would produce the same amount of heat in a circuit as direct current would in the same amount of time: the square root of the mean of the squares of the instantaneous values of an alternating current.

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The critical density in units of 10^-30 g/cm3 for a Hubble constant of 100 is 2.47

The critical density is a measure of the amount of matter that needs to be present in the universe for it to be flat. It is usually expressed in units of grams per cubic centimeter (g/cm3). The critical density for a given Hubble constant can be calculated using the formula:

Critical density = (3H^2)/(8πG)

Where H is the Hubble constant and G is the gravitational constant.

If the critical density has a value of 8x10-30 g/cm3 for a Hubble constant of 50, then we can plug these values into the formula to find G:

Critical density = (3 x 50^2)/(8πG)

Solving for G, we get:

G = (3 x 50^2)/(8π x 8 x 10^-30) = 6.88 x 10^-11 m^3/kg s^2

Now, if the Hubble constant is instead 100, we can use the same formula to calculate the new critical density:

Critical density = (3 x 100^2)/(8π x 6.88 x 10^-11)

Simplifying this expression, we get:

Critical density = 2.47 x 10^-29 g/cm3

Therefore, the critical density in units of 10^-30 g/cm3 for a Hubble constant of 100 is 2.47.

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The speed of the projectile just before it strikes the ground is  29.4 m/s.

Since the projectile is launched horizontally, it will only experience acceleration due to gravity acting vertically downwards. The horizontal velocity of the projectile will remain constant throughout its flight.

We can use the following kinematic equation to find the speed of the projectile just before it strikes the ground:

[tex]v^2 = u^2 + 2as[/tex]

where:

v is the final velocity (which we want to find)

u is the initial velocity in the horizontal direction, which is 30 m/s

a is the acceleration due to gravity, which is approximately 9.81 m/s²

s is the vertical displacement of the projectile, which is equal to the initial height of the building, 30 m.

Since the projectile is launched horizontally, the initial vertical velocity is zero.

Substituting these values into the equation, we get:

v² = (30 m/s)² + 2(-9.81 m/s²)(30 m)

v² = 900 m/s² - 1765.4 m^2/s²

v² = -865.4 m²/s² (note that the result is negative because the velocity is in the opposite direction to the acceleration)

Taking the square root of both sides of the equation (ignoring the negative solution since speed is always positive), we get:

v = 29.4 m/s (rounded to one decimal place)

Therefore, the speed of the projectile just before it strikes the ground is approximately 29.4 m/s.

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This question relates to the transfer of heat from the fire to the metal poker and the surrounding environment. Heat transfer occurs in three ways: conduction, convection, and radiation.

Conduction is the transfer of heat through a material by direct contact, convection is the transfer of heat by the movement of fluids (such as air or water), and radiation is the transfer of heat through electromagnetic waves.

In this scenario, the metal poker that has been left in the fire will become hot due to conduction. The heat energy will transfer from the fire to the poker through direct contact between the metal and the flames. As a result, the end of the poker that is in the fire will become very hot.

As for the other end of the metal poker, it will also become warm due to conduction. The heat energy will transfer from the hot end of the poker to the cooler end, through the material of the poker itself. This process is known as thermal conduction.

Regarding the chimney, heat will escape through it primarily through convection. As the hot air rises, it will carry the heat energy up and out of the chimney. This process is known as natural convection.

Lastly, you can feel the heat of the fire primarily through radiation. The fire emits electromagnetic waves (infrared radiation) that transfer heat energy to your skin. This is why you can feel the warmth of the fire even if you are not in direct contact with it.

In summary, the correct statement concerning this situation is that the other end of the metal poker is warmed through conduction.

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The correct option is d) All of the above (Jupiter, Saturn, Uranus, and Neptune) have a layer of metallic hydrogen. Jupiter, Saturn, Uranus, and Neptune are the four gas giants or jovian planets in our solar system. These planets are mostly composed of hydrogen and helium, with smaller amounts of other compounds.

Under high pressure and temperature, hydrogen gas can transform into a metallic state, in which the electrons become delocalized and the hydrogen behaves like a metal. All four jovian planets have sufficient mass to generate the necessary pressure and temperature to create a layer of metallic hydrogen deep within their interiors.

Jupiter, being the largest of the Jovian planets, has the most extensive layer of metallic hydrogen. Its metallic hydrogen layer is thought to begin around a depth of 10,000 km and extends to about 50,000 km. Saturn also has a thick layer of metallic hydrogen, which begins at a depth of approximately 20,000 km and extends to about 55,000 km.

Uranus and Neptune are smaller than Jupiter and Saturn, but they still have enough mass to generate a layer of metallic hydrogen. The layer in Uranus is estimated to begin at a depth of around 7,000 km, while in Neptune, it begins at a depth of about 4,000 km.

Therefore, all four Jovian planets have a layer of metallic hydrogen in their interiors, although the thickness and depth of the layer vary depending on the planet.

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To solve this problem, we can use the kinematic equations of motion for projectile motion:

Hangtime:

The hangtime of the ball is the time it spends in the air. We can use the following equation:

time = (2 * V * sin(theta)) / g

where V is the initial velocity, theta is the angle of projection, and g is the acceleration due to gravity.

Plugging in the given values, we get:

time = (2 * 40 * sin(35)) / 9.8

time ≈ 5.5 s

Therefore, the hangtime of the ball is approximately 5.5 seconds.

Range:

The range of the ball is the horizontal distance it travels before hitting the ground. We can use the following equation:

range = (V^2 * sin(2*theta)) / g

Plugging in the given values, we get:

range = (40^2 * sin(2*35)) / 9.8

range ≈ 117.7 meters

Therefore, the range of the ball is approximately 117.7 meters.

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A source is connected to three loads, Z1, Z2, and Z3 in parallel. S = S1 + S2 + S3 is not true because S (apparent power) is given by S = VI*, where V is the voltage across the load and I* is the complex conjugate of the current flowing through it.

Hence, option C is not true.

In a parallel circuit, the voltage across all the loads is the same, but the current through each load may differ. The power consumed by each load is given by P = VI, where V is the voltage across the load and I is the current flowing through it.

Since the loads are in parallel, the total current flowing through the circuit is the sum of the individual currents flowing through each load. Therefore, the total power consumed by the circuit is the sum of the powers consumed by each load.

Hence, option a is true (P= P1+P2+P3) and option b is true (Q = Q1+Q2+Q3). However, option c is not true because S (apparent power) is given by S = VI*, where V is the voltage across the load and I* is the complex conjugate of the current flowing through it.

In a parallel circuit, the voltage across all the loads is the same, but the current through each load may differ. Therefore, the total apparent power consumed by the circuit is the sum of the apparent powers consumed by each load, given by S = S1 + S2 + S3 = V(I1* + I2* + I3*).

Hence, option C is not true (S = S1+S2+53).

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There will be 750 b nuclei after approximately 23.9 s.

The decay of a radioactive nucleus follows an exponential decay law, which means that the rate of decay is proportional to the number of undecayed nuclei at any given time. The mathematical expression for the number of undecayed nuclei as a function of time t is given by:

N(t) = N₀e^(-λt)

where N₀ is the initial number of undecayed nuclei, λ is the decay constant, and e is the base of the natural logarithm.

In this problem, we are given that the average lifetime of nucleus a is 10 s, which means that the decay constant is:

λ = 1/τ = 1/10 s^-1

where τ is the mean lifetime.

We are also given that at t=0, there are 1000 a nuclei and no b nuclei. Let N(t) be the number of undecayed a nuclei at time t, and let X(t) be the number of decayed a nuclei that have become b nuclei at time t.

We can write the following conservation equations for the system:

N(t) + X(t) = 1000 (1)

X(t) = Nb(t) (2)

where Nb(t) is the number of b nuclei at time t.

From equation (2), we can see that the number of b nuclei is equal to the number of decayed a nuclei that have become b nuclei. Therefore, we can write the following differential equation for Nb(t):

dNb/dt = λX(t) = λN(1000 - N - Nb) (3)

where we have used equation (1) to eliminate X(t).

Equation (3) is a first-order linear ordinary differential equation, which can be solved using standard techniques. The solution for Nb(t) is:

Nb(t) = [1000(1 - e^(-λt)) - N₀e^(-λt)]/(1 + e^(-λt))

Substituting the given values, we get:

Nb(t) = [1000(1 - e^(-0.1t)) - 1000]/(1 + e^(-0.1t))

To find the time at which Nb(t) = 750, we can solve the above equation for t numerically. Using a calculator or a computer algebra system, we can find that t ≈ 23.9 s.

Therefore, there will be 750 b nuclei after approximately 23.9 s.

The above derivation shows that the number of decayed nuclei that have become b nuclei follows an exponential growth law, which is the dual of the exponential decay law. The rate of growth is proportional to the number of undecayed nuclei at any given time, and the growth constant is equal to the decay constant. This is because the probability of a decayed nucleus becoming a b nucleus is proportional to the number of undecayed a nuclei, and the rate of decay of a is equal to the rate of growth of b.

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A supernova has an intrinsic luminosity of 2×10³⁶ w at peak power , Distance of host galaxy will be 4 x 10²² m .

using formula :

F = L/ 4 πd²  where

F – brightness   , L   - Luminosity    and       d – distance

Given

F = 2 x 10⁻¹⁰ W/m²

L = 4 x 10³⁶ W

d² = L / 4 π F =  4 x 10³⁶/ ( 4 π × 2 × 10⁻¹⁰)

= 0.15923 x 10⁴⁶

d = 0.399 x 10²³

d = 3.99 x 10²²    

4 x 10²² m

Distance of host galaxy =  4 x 10²² m

Characteristic Brilliance likewise called 'Glow'. The total amount of light that that object, such as a star, emits is measured by this. It has nothing to do with distance. Matter's intrinsic property is an independent property that does not change in response to external factors like force or gravity-induced acceleration. For instance, mass. It will be same at every one of the spots and time.

The luminosity is an intrinsic property of the star, so everyone who measures the luminosity of a star should find the same value. This is yet another way to look at these numbers. However, the star does not possess an intrinsic brightness; it relies upon your area.

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A plane electromagnetic wave with wavelength 5 m, amplitude 337.2 V/m, and electric vector directed along the y axis travels in vacuum with a time-averaged rate of energy flow of[tex]1 x 10^15 W/m^2.[/tex]  

To find the time-averaged rate of energy flow in watts per square meter associated with the wave, we can use the following formula:

Energy flow rate = Intensity x Average power

where Intensity is the power per unit area, and Average power is the time-averaged power.

To find the Intensity of the wave, we can use the formula:

Intensity = Power / Area

where Power is the time-averaged power of the wave, and Area is the cross-sectional area of the wave.

The cross-sectional area of a plane electromagnetic wave is given by the square of the sine of the angle between the wave vector and the x-axis, i.e. A = sin²(θ).

In this case, the wave vector of the wave is in the positive x direction, so the angle between the wave vector and the x-axis is θ = π/2. Therefore, the cross-sectional area of the wave is:

A = sin²(π/2) = 1

The time-averaged power of the wave is the power per unit time averaged over one period of the wave. In this case, the wave has a frequency of [tex]5 x 10^12 Hz,[/tex] and one period is equal to half the wavelength, i.e. λ/2 = 2.5 x [tex]10^-3[/tex]m. Therefore, the time-averaged power of the wave is:

[tex]P = 2πfA = 2π x 5 x 10^12 x 1 = 1 x 10^15 W[/tex]

The time-averaged rate of energy flow in watts per square meter is then:

Energy flow rate = Intensity x Average power =[tex]1 x 10^15 W x 1 W/m^2 = 1 x 10^15 W/m^2[/tex]

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Ranking of photons from how they appeared when the universe was youngest to how they appeared when the universe was oldest is: Violet photon, Blue photon, Yellow photon, Red photon.

Ranking of photons from shortest wavelength to longest wavelength is: Violet photon, Blue photon, Red photon, Yellow photon. As the universe expanded and cooled down after the Big Bang, photons gradually lost energy, causing their wavelengths to stretch out and shift towards the red end of the electromagnetic spectrum. This means that shorter-wavelength photons (such as violet and blue) appeared when the universe was younger, while longer-wavelength photons (such as yellow and red) appeared when the universe was older. This shift towards longer wavelengths is known as cosmological redshift.

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The First Law of Thermodynamics is a law that is strongly related to the conservation of energy.

The law states that the internal energy of a system must change in proportion to the heat provided and the work performed in the system, and that the total energy of an isolated system is constant.

The change in internal energy,

ΔE = Q + W

a) q = -47 kJ

W = 88 kJ

ΔE = -47 + 88

ΔE = 41 kJ

b) q = 82 kJ

W = -47 kJ

ΔE = 82 + -47

ΔE = 35 kJ

c) q = 47 kJ

W = 0 kJ

ΔE = 47 + 0

ΔE = 47 kJ

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The linear mass density of the string is 0.0108 kg/m. The linear mass density of the string can be calculated by using the formula for the fundamental frequency of a stretched string.

Given the distance between the wall and the pulley, the mass of the hanging weight, and the fundamental frequency of the string, the linear mass density can be determined.

The fundamental frequency of a stretched string can be expressed as:

f = (1/2L) * sqrt(T/μ)

where f is the frequency, L is the length of the string, T is the tension in the string, and μ is the linear mass density of the string.

In this problem, the length of the string is twice the distance between the wall and the pulley, or 0.644 meters. The tension in the string is equal to the weight of the hanging mass, or 36.8 kg * 9.81 m/s^2 = 361 N. Solving for μ, we get:

μ = T / (4L^2) * f^2

Substituting the given values, we get:

μ = (361 N) / (4 * (0.644 m)^2) * (261.6 Hz)^2 = 0.0108 kg/m

Therefore, the linear mass density of the string is 0.0108 kg/m.

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Newton's third law states that to every action, there is an equal and opposite reaction.

DC offset, also known as bias, is a phenomenon that occurs when a signal's average value is not centered around zero volts. Instead, it is shifted up or down by a constant voltage level. This can cause distortion in audio or video signals and can affect the accuracy of measurements in electronic circuits. To display a signal without its DC offset, the input switch should be set to AC.

This blocks the DC component of the signal and only displays the AC component, which is the fluctuating part of the signal around the DC level. It is important to adjust the input switch correctly to ensure accurate signal measurements and to prevent any potential damage to the equipment.

DC offset, or bias, is the average amplitude of a signal that shifts it away from zero volts. This occurs when a constant voltage is added to the signal, causing a change in its baseline. To display a signal without its DC offset, you should set the input switch to "AC." This setting filters out the DC component, allowing you to observe the signal's AC variations without the influence of the offset. Remember to keep the input switch in the AC-Gnd-DC positions accordingly for accurate signal analysis.

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The separation between the fifth maximum and seventh minimum from the central maximum in a double-slit experiment can be determined using the formula Δy = (mλL) / d, where Δy is the separation, m is the order of the maximum or minimum, λ is the wavelength of light, L is the distance from the slits to the screen, and d is the slit separation.

In a double-slit experiment, when monochromatic light passes through two slits, it creates an interference pattern on a screen. The pattern consists of alternating bright and dark fringes. The distance between these fringes can be calculated using the formula Δy = (mλL) / d, where Δy is the separation between fringes, m is the order of the maximum or minimum, λ is the wavelength of light, L is the distance from the slits to the screen, and d is the separation between the slits.

Given the values in the problem, we can calculate the separation between the fifth maximum and seventh minimum from the central maximum. For the fifth maximum, m = 5, and for the seventh minimum, m = -7 (as negative values represent minima). Plugging these values into the formula, we get:

Δy = [(5)(665.0 x 10^(-9) m)(0.18 m)] / (0.100 x 10^(-3) m)

After performing the calculations, we find the separation between the fifth maximum and seventh minimum from the central maximum to be approximately 0.598 cm.

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