What is radiogenic energy?

The flux through the paper when the paper is turned 25° with respect to field will be 22.67 \(Nm^{2} /C\).

We have a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25 \(Nm^{2} /C\).

Now we have to calculate the flux through the paper if it is turned 25° with respect to the field.

Electric flux (E) through a surface is defined as the dot product of Electric field vector and Normal vector to the surface. The formula to calculate the electric flux is -

Φ = \(\vec{E}.\vec{A}\) = E x A x cosθ

In the question it is given that when one face perpendicular to a uniform electric field the flux through it is 25 \(Nm^{2} /C\). In this case θ = \(0^{o}\).

Therefore -

E x A x cos \(0^{o}\) = 25

EA = 25 \(Nm^{2} /C\)

When the paper is turned \(25^{o}\) with respect to field, then the flux through it can be calculated as follows -

Φ = 25 x cos (\(25^{o}\)) = 22.67 \(Nm^{2} /C\).

Hence, the flux through the paper will be 22.67 \(Nm^{2} /C\).

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A very powerful vacuum cleaner has a hose 2.27 cm in diameter. The atmospheric pressure is 1.00 atm=101,300 Pa. then vacuume cleaner generate the force F = 65.1 N

A line that is straight and cuts through the middle of a body or figure. Particularly: the length of a diameter; a line segment passing through the center of a circle with its ends on the circumference.

Given,

diameter of the nose is d= 2.86 cm

radius of the nose r= 1.43 cm = 1.43× \(10^{-2}\) m

We know F=P.A

where P is the atmospheric pressure and A is the surface area of the circulsr nose F is the force exerted by vacuume cleaner on the air

∴ F=P ( \(\pi r^{2}\))

we know atmospheric pressure, p=1.013×\(10^{5}\) pa

F = 1.013 × \(10^{5}\) × 3.14 (1.43 × \(10^{-2}\)) ^2

⇒ 65.1 N

⇒ F = 65.1 N

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Explanation:

7.7 m/s

1st determine you subject slade to you have a mass of 15 kg being subjected to a force of 180N

so.

180N/15kg=180(kg m)/s^2/15kg=12 ms^2

now determine how long u pushed it

d=0.5 A T^2

sutable the known values getting

2.5 m = 0.5 12 ms^2 T^2

2.5m= 0.6 ms^2 T^2

0.41667 s^2=T^2

0.645497224s=T

the velocity multiply the time by the acceleration giving

0.645497224 s 12 ms^2 =7.745966692 ms.

after rounding the 2 significant u get 7.7 ms

Main answer-

PE=mgh

KE=E-PE

V = (2*g*h)^½

at point A

PE = 10*9.8*50=4900 J

KE = 0 J  ( because the velocity of the body is 0)

V=0

at point B

PE = 10*9.8*35=3430 J

KE = 4900-3430=1470 J

V= ( 2*9.8*15)^ ½

V=17.14 m/s

at point C

PE= 10*9.8*25=2450 J

KE=4900-2450=2450 J

V= (2*9.8*25)^ ½

V=22.135 m/s

at point D

PE=10*9.8*0=0 J

KE=4900-0=4900 J

V=(2*9.8*50)^ ½

V=31.30 m/s

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The net force on particle q₁ is 6.76 x 10⁻³ N.

The net electric force acting on a charged particle due to other charged particles is given by Coulomb's law:

F = k × (|q₁|×|q₂|)/r²

where F is the force, k is Coulomb's constant, q₁ and q₂ are the charges on the particles, and r is the distance between them.

The direction of the force is along the line connecting the two charges, and is attractive for opposite charges and repulsive for like charges. In this case, the net force on q₁ due to q₂ and q₃ is the vector sum of the individual forces. So we need to calculate the force on q₁ due to q₂ and the force on q1 due to q₃ separately, and then add them as vectors.

F1 = k × (|q₁||q₂|)/r₁², where r1 = 0.220 m

F1 = (8.99 x 10⁹ Nm²/C²) x (8.99 uC) x (5.16 uC) / (0.220 m)²

F1 = 1.04 x 10⁻³N (attractive force)

F2 = k × (|q₁||q₃|)/r₂², where r₂ = 0.330 m

F2 = (8.99 x 10⁹ Nm²/C²) x (8.99 uC) x (89.9 uC) / (0.330 m)²

F2 = 7.80 x 10⁻³ N (repulsive force)

The net force on q₁ is the vector sum of F₁ and F₂, taking into account their directions:

F_net = F₁ - F₂ = 1.04 x 10⁻³ N - 7.80 x 10⁻³ N

F_net = -6.76 x 10⁻³ N (attractive force)

So the net force on q₁ is 6.76 x 10⁻³ N, in the direction towards q₂.

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Answer:She can increase the number of wire loops

Explanation: When the number of loops of wire ( usually around the iron ) increases, the strength of the electromagnet is increased.

According to the equation

B=μ₀ N I

Where

μ₀ = permeability of the core

N= Number of turns of the coil

I= Current flowing through the coil

From the equation above , we can see that The strength of an electromagnet depends the amount of current, no of turns of coil and the permeability  core of coil. we can now say that Increasing the  number of wire loops  increases the strength of electromagnet.

Therefore, Lisa should increase the number of wire loops to increase the strength of the electromagnet

Answer:

C

Explanation:

Edge2021

Answer:

2.92 m

Explanation:

As we know, frequency × Wavelength = Speed of light

so here frequency of 102.7 MHz can be written as 102.7× 10⁶ Hz..

So Lambda (wavelength) = 3×10⁸/ 102.7 × 10⁶ which gives 2.92 metres or 2.92 × 10¹⁰ Å

Answer:

C. 199.9 s

Explanation:

3 minutes = 3×60 = 180 seconds.

the turtle moves in that time 180×10 = 1800 cm.

in other words the rabbit gave it that much head-start (it does not matter if that was at the begin of in the middle of the race).

the rabbit moves with 10×10cm/s = 100cm/s.

the rabbit needs therefore 1800/100 = 18 seconds for the

1800 cm.

at that time the turtle has added another 18×10 = 180 cm.

for which the rabbit needs 180/100 = 1.8 seconds.

during that time the turtle has added 1.8×10 = 18 cm.

and so on.

in formal mathematics this looks like this :

1800 + 10x = 100x

after x seconds of the rabbit running both will have run the same distance, and it is a tie.

1800 = 90x

x = 20 seconds

so, at that point, the rabbit was actively running for 20 seconds and raced 20×100 = 2000 cm

and the turtle was actively running for 180 + 20 = 200 seconds, and also covered 200×10 = 2000 cm.

but our question tells us that the turtle won by 10 cm.

so, the race was over a little bit before these 200 seconds (for a tie).

this means, the rabbit could not run the last 10 cm for the tie (because the race was over and the turtle had won).

the rabbit would have needed 10/100 seconds for these 10 cm.

as speed = distance/time

we need to divide distance by speed

distance/1 / distance/time

to get time.

so,

10cm/1 / 100cm/s = 10s/100 = 1/10 s

so, we need to deduct this 1/10 s from the 200 seconds of the turtle (and also from the 20 seconds for the rabbit).

the race lasted of course the whole time the turtle was running (while the rabbit was resting, officially still participating in the race with speed 0 for 3 minutes).

and so, the race was 199.9 s long.

The AM radio transmitter emits approximately 9.94 x \(10^{31}\) photons per second.

To calculate the number of photons emitted per second by an AM radio transmitter, we'll need to use the following terms and equations:
Power (P): This is the amount of energy transferred or converted per unit time. In this case, the transmitter radiates 500 kW (kilowatts).
Frequency (f): The number of cycles of a periodic wave that occur in a unit of time. In this case, the frequency is 760 kHz (kilohertz).
Planck's constant (h): A fundamental constant with a value of approximately 6.626 x \(10^{-34}\) Js (joule-seconds). This constant relates the energy of a photon to its frequency.
The energy (E) of a single photon can be calculated using the equation:
E = h x f
Next, we need to convert the power and frequency to the appropriate units:
P = 500 kW x 1,000 (to convert kW to W) = 500,000 W
f = 760 kHz x 1,000 (to convert kHz to Hz) = 760,000 Hz
Now, we can calculate the energy of a single photon:
E = (6.626 x 10^-34 Js) * (760,000 Hz) = 5.035 x \(10^{-28}\) J
To find the number of photons emitted per second (n), we will divide the total power (P) by the energy per photon (E):
n = P / E

n = (500,000 W) / (5.035 x \(10^{-28}\) J)

n ≈ 9.94 x \(10^{31}\) photons per second
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SICENCE

To measure the temperature of frying oil, you should attach a high-temperature immersion probe to your thermocouple or thermistor.


1. Choose a high-temperature immersion probe: This type of probe is designed to withstand high temperatures and is suitable for measuring the temperature of hot liquids like frying oil.
2. Ensure compatibility: Make sure the immersion probe is compatible with your thermocouple or thermistor. Consult the manufacturer's specifications for guidance.
3. Attach the probe: Connect the immersion probe to your thermocouple or thermistor according to the device's instructions.
4. Insert the probe into the frying oil: Carefully immerse the tip of the probe into the hot oil, ensuring it does not touch the bottom or sides of the pan.
5. Monitor the temperature: Observe the temperature reading on your thermocouple or thermistor to ensure the oil is at the desired temperature for frying.
By following these steps, you'll be able to accurately measure the temperature of frying oil using a thermocouple or thermistor with a high-temperature immersion probe.

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Beginning of double-shell fusion The core has double the mass. size and brightness grow while temperature drops.

a double-shell fire. He fuses to diamond in a shell it around carbon core when core helium fusing stops, while H fuses to He inside a shell all around helium layer. Because of a sequence of thermal pulses, the fusion rate in this double-shell burning stage never approaches equilibrium.

Whenever a star loses out of nitrogen, its core will begin to collapse once more until it temperature where carbon fusion can start. As the star burns up progressively heavier components, such as carbon, neon, oxygen, and silicon, this pattern will persist.

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Answer:

when the radius is halved, F becomes 4 times

Explanation:

The formula is given by

\(F = \frac{Gm1m2}{r^2}\)

When r is halved:

\(F = \frac{Gm1m2}{(1/2r)^2}\)

=> \(F = \frac{Gm1m2}{1/4 r^2}\)

=> \(F = 4(\frac{Gm1m2}{r^2} )\)

This means when the radius is halved, F becomes 4 times

Answer:

Explanation:

High concentrations of hydrogen ions yield a low pH (acidic substances), whereas low levels of hydrogen ions result in a high pH (basic substances). The overall concentration of hydrogen ions is inversely related to its pH and can be measured on the pH scale

a gamma-ray photon produces an electron and a positron, each with a kinetic energy of 375 keV.

wavelength of photon = 7.019 × 10⁻¹³ m

Total energy of photon = 1.77 MeV.

An object's kinetic energy is the energy it has as a result of motion. Applying force is necessary if we wish to accelerate an item. We must put forth effort in order to apply a force. Once the job is finished, energy is transferred to the item, which then moves at a new, constant speed.

The rest mass energy for electron and positron:

E₀ = m₀c²

E₀ = (9.1 × 10⁻³¹) × (3 × 10⁸)

E₀ = 8.198 × 10⁻¹⁴ J

E₀ = 512.4 × 10³ eV or 512.4 KeV

Total energy of photon:

= [2 × (375)] + [2 × ( 512.4 )]

= 1774.8 KeV

= 1.77 MeV.

Energy = hc/λ

or, λ = (6.626 × 10⁻³⁴) × (3 × 10⁸)

or, λ = 7.019 × 10⁻¹³ m

Thus, wavelength of photon = 7.019 × 10⁻¹³ m

Total energy of photon = 1.77 MeV.

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The study of galaxies found that their rotation rate measured in the halo is higher than would be expected, indicating the presence of dark matter.

Observations of galaxies have revealed that the rotation velocities of stars and gas in the outer regions of galaxies, known as the halo, do not follow the expected behavior based on the visible matter alone. According to Newtonian physics, the outer regions of galaxies should rotate at slower velocities as the gravitational force weakens with distance. However, measurements have consistently shown that the rotation velocities remain relatively constant or even increase with distance from the galactic center. This discrepancy can be explained by the presence of dark matter, an invisible and as-yet-undetected form of matter that provides the additional gravitational pull required to account for the observed high rotation velocities. Dark matter is believed to be a significant component of galaxies and plays a crucial role in their formation and dynamics.

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Hi there!

Recall the equation for electric potential energy for point charges:
\(\boxed{U = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r}}\)

U = Electric potential energy (J)

ε₀ = Permittivity of Free Space (8.85 × 10⁻¹² C/Vm)

q₁, q₂: Charges (C)
r = distance between charges (m)

Solving for r:
3 - (-3) = 6m

Now, plug in the values:

\(U = \frac{1}{4\pi \epsilon_0} \frac{(.00001)(.00001)}{6} = \boxed{1.498 J}\)

The area of the shaded region is 4621.5 cm2 - 1083.8 cm2 = 3537.7 cm.

Area is the space or size of a two-dimensional surface or shape. It is a measure of the size of a surface and is typically expressed in square units, such as square feet or square meters. Area is usually calculated by multiplying the length of a shape by its width. For example, the area of a rectangle with a length of 4 feet and a width of 3 feet is 12 square feet. Area can also be calculated for shapes with curved sides such as circles, triangles, and other polygons.

The shaded region is the area between the two circles. To calculate the area of the shaded region, we need to subtract the area of the inner circle from the area of the outer circle.

The area of the outer circle can be calculated using the formula A = πr2, where r is the radius of the circle. Since the two circles have the same center, the radius of the outer circle is twice the radius of the inner circle.

Therefore, the radius of the outer circle is 2 × (117.436/2π) = 38.212 cm.

The area of the outer circle is A = π × 38.212^2 = 4621.5 cm2.

The area of the inner circle can be calculated using the same formula: A = πr2. The radius of the inner circle is (117.436/2π) = 18.606 cm.

The area of the inner circle is A = π × 18.606^2 = 1083.8 cm2.

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Answer: 137.14 kg

Explanation: F=ma (Force = mass x acceleration)

So, 48N (newtons) = m×0.35m/s²

N = kg (m/s²)

      (48 kg m/s²)÷0.35m/s² = m

      137.14 kg

 note: answer is to the nearest hundredth because of the .035 m/s²

The neutron has no charge, and a mass of slightly over 1 amu. Some scientists propose the neutron is made up of a proton and electron-like particle. The electron is a very small particle located outside the nucleus.

Electron: -1

Neutron: 0

Proton: +1

Answer:

Explanation:

When a coil of wire is moved around a magnet, it induces electrons into the wire and creates electric current. Thus, the correct option is

moves through and cuts magnetic field lines

Answer:

The image is formed at 0.44 m in front of the mirror

magnification (M) = 0.44

Explanation:

Applying, mirror formular

1/f = (1/u)+(1/v).................... Equation 1

Where f = Focal length of the convex mirror, u = object distance, v = image distance.

Using the real is positive convection,

From the question,

Given: f = -1.6/2 = -0.8 m( The focal length of a convex mirror is vitual), u = 1.0 m

Substitute these values into equation 1

-1/0.8 = (1/v)+(1/1)

Solve for v

1/v = 1.25+1

1/v = 2.25

v = 1/2.25

v = 0.44 m

Hence the image is formed at 0.44 m in front of the mirror

Magnification (M) = v/u

m = 0.44/1

m = 0.44

Answer:

Work input W = -200.39 KJ

Explanation:

From the question, we are given;

m = 1.15 kg

Constant temperature T1 = T2 = 27 + 273 = 300k

Since the temperature is constant, we can say that the process is isothermal

P1 = 111 KPa

P2 = 0.84 MPa = 0.84 * 1000 KPa = 840 KPa

Now what we want to calculate is W1-2

Mathematically, for isothermal process;

W1-2 = mRTlnP1/P2

where R can be obtained from table and it is equal to 0.287 KJ/kg.k

Hence;

W1-2 = (1.15)(0.287)(300)(ln 111/840)

W1-2 = 99.015 * -2.023871690525 = -200.39 KJ

Kindly note that the value of the work is negative because work is done on the system and not by the system

The ice slab must have a minimum capacity of 50.0 L.

Yet water is special because its solid form (ice) is actually less thick than its liquid phase. As a result, walleye and other aquatic life may endure each winter in cold, unfrozen waters under the ice. The ice rises to the top because the water, which is heavier, pushes aside the lighter ice.

Buoyant force = Weight of water displaced = Density of water x Volume of water displaced x Gravity

Buoyant force = Weight of the woman = 50.0 kg x 9.81 m/s² = 490.5 N

Now, we can use the buoyant force equation to find the minimum volume of the ice slab required:

Buoyant force = Density of water x Volume of ice slab x Gravity

Volume of ice slab = Buoyant force / (Density of water x Gravity)

Volume of ice slab = 490.5 N / (1000 kg/m³ x 9.81 m/s²)

Volume of ice slab = 0.0500 m³ or 50.0 L (rounded to 3 significant figures)

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Answer:

Go out into the country. Look on a farm. Look in a barn. Listen for, "MOOOOO." You found a cow.

Explanation:

No, plants do not need soil for photosynthesis. Photosynthesis is the process by which plants use sunlight, water, and carbon dioxide to create food. The soil provides nutrients to plants, but it is not necessary for photosynthesis.

In fact, there are many ways to grow plants without soil. One way is to use hydroponics, which is a method of growing plants in water with nutrients added. Another way is to use aeroponics, which is a method of growing plants in a mist of water and nutrients.

Soil is not necessary for photosynthesis, but it does provide other benefits to plants. Soil helps to anchor plants and provides a source of nutrients. It also helps to regulate the temperature and moisture around the roots.

If you are growing plants in soil, it is important to make sure that the soil is fertile and well-drained. You should also fertilize the plants regularly to provide them with the nutrients they need.

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The assignment most likely to carry a works cited page is a research paper or an essay that requires the use of external sources.

A works cited page is essential in such assignments because it provides a list of the sources that have been referenced throughout the paper, acknowledging the original authors and their work. This practice is crucial in maintaining academic integrity and avoiding plagiarism. The works cited page should follow a specific citation style, such as APA, MLA, or Chicago, depending on the assignment guidelines provided by the instructor or institution.

In a research paper or essay, a works cited page helps readers verify the accuracy and credibility of the information presented, as well as locate the original sources for further study or understanding. Overall, a works cited page is an essential component of any assignment that relies on external sources to support arguments, ideas, or research findings.

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Answer:

2.65

Explanation:

Its simple just find the newtons