Help me answer this. Here, the distance to school is representing the remaining distance left to lucy while traveling to school.

V = 23V

R1 = 15 Ω

R2= 15 Ω

Both resistors in series

Rt = total resistance = R1 + R2 = 15 + 15 = 30Ω

Ohm's law:

V = I x R

I = V/R = 23V/30Ω= 0.766 Ampere

Current = 0.766 A

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Answer:

motion of an object—even if the object is maintaining a constant speed—still count as acceleration

Explanation:

a higher energy than an electron near the nucleus

In an electron cloud, an electron that is farther away from the nucleus has a higher energy than an electron near the nucleus. Thus, the correct option is C.

An electron cloud is the region which is surrounding an atomic nucleus in which there is a very high probability of an electron being located in an atom. The probability of finding an electron is greater than the more dense regions of the electron cloud in the atom.

One way in which the elements differ from each other is the structure of the electron cloud in each of the element's atoms in an electron cloud and the electrons which is farther from the nucleus has a higher energy than an electron near the nucleus of the atom.

Therefore, the correct option is C.

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Answer:

d= -2

Explanation:

t=5s

S=100 meters

V= 10 m/ s

d=?

deceleration = (final velocity - initial velocity) / time. d = (vf - vi)/t.

The formula for acceleration can be used, recognizing that the final result must have a negative signe.

d= (0-10)/5

d= -2

A fundamental particle with no electric charge and a spin of one-half is called an electron neutrino (v).

The subatomic particle known as a neutrino resembles an electron in many ways, but it differs in that it lacks an electrical charge and has a very small mass that may even be zero. One of the cosmological particles with the highest density is the neutrino. However, they are exceedingly difficult to find because they barely interact with matter.

Small, almost massless particles known as neutrinos move at speeds close to light. They are phenomenally common in the universe and can move through lead as readily as we do through air. They are the product of intense astronomical events like exploding stars and gamma ray bursts.

A flat recoil energy spectrum is the result of monoenergetic electron neutrinos scattering from electrons.

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Answer:

\( \fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}\)

Explanation:

Given:

Force between pair of charges= 900 newtons

The distance between the charges = 0.01 meters

Strength of Charge first q1 = 2e-10 Coulomb

To find:

Strength of Charge second q2 = ____ Coulomb?

Solution:

We know that,

Force between two charges separate by distance r is given by the equation,

\(|F| = K_e \frac{q1 \cdot \: q2}{ {r}^{2} } \\ 900 =K_e \frac{(2e - 10)\cdot \: q2}{ {0.01}^{2} } \\ 900 \times {10}^{ - 4} = K_e {(2e - 10)\cdot \: q2} \\ q2 = \frac{9 \times {10}^{ - 2} }{(2e - 10) K_e} \\ \\ \fbox{We \: know \: that \: e = 2.71 } \\ substituting \: the \: value \: \\ q2 = \frac{9 \times {10}^{ - 2} }{(2 \times 2.71 - 10)K_e} \\ q2 = \frac{0.09}{ - 4.58 K_e} \\ q2 = \frac{-0.0196}{K_e}\: coulomb\)

\( \fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}\)

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Answer:

Other

Explanation:

charge quantity - Q

Big Q represents the source charge which creates the electric field. Little q represents the test charge which is used to measure the strength of the electric field at a given location surrounding the source charge. Give considerable attention to the charge quantity - Q or q - being used in each equation.

The object distance of the lens is 10 cm and the image distance of the lens is 30 cm.

The object and image distance formed by the lens is calculated by applying the following lens formula.

v + u = 40 ------- (1)

v/u = 3 ------------ (2)

v = 3u

Substitute v into equation (1);

3u + u = 40

4u = 40

u = 40/4

u = 10 cm

The image distance = 3u

= 3 x 10 cm

= 30 cm

Thus, the object distance is 10 cm and the image distance is 30 cm.

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Comparing my diet to the food pyramid, I can see that my diet is similar in some ways and different in others.

For breakfast, I had a bowl of oatmeal with almonds, blueberries, and a drizzle of honey. For lunch, I had a turkey sandwich with lettuce, tomato, and avocado on whole wheat bread. I also had a side of carrot sticks and hummus. For dinner, I had grilled salmon with roasted sweet potatoes and broccoli. For snacks throughout the day, I had an apple, a handful of grapes, and a small serving of dark chocolate.

Comparing my diet to the food pyramid, I can see that my diet is similar in some ways and different in others. For example, my breakfast of oatmeal with blueberries is a good source of whole grains and fruits, which are both recommended in the food pyramid. Similarly, my lunch of turkey sandwich on whole wheat bread with vegetables provides a good balance of protein, whole grains, and vegetables. My snack of fruit, such as apples and grapes, is also recommended in the food pyramid as a healthy option for snacking.

However, there are also some differences in my diet compared to the food pyramid. For instance, my diet did not include any dairy products or legumes, which are both recommended in the food pyramid. Additionally, my dinner did not include any grains, which are also recommended in the food pyramid. Instead, I had sweet potatoes, which are a good source of complex carbohydrates.

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The kinematics of the uniform motion and the addition of vectors allow finding the results are:

Vectors are quantities that have modulus and direction, so they must be added using vector algebra.

A simple method to perform this addition in the algebraic method which has several parts:

 Indicate that Barry's velocity is constant, let's find using the uniform motion thatthe distance traveled in ad case

              v = \(\frac{\Delta d}{t}\)

              Δd = v t

Where  v is the average velocity, Δd the displacement and t the time

We look for the first distance traveled at speed v₁ = 600 m / s for a time

          t₁ = 2 min = 120 s

          Δd₁ = v₁ t₁

          Δd₁ = 600 120

          Δd₁ = 72 10³ m

Now we look for the second distance traveled for the velocity v₂ = 400 m/s    

  time t₂ = 1 min = 60 s

          Δd₂ = v₂ t₂

          Δd₂ = 400 60

          Δd₂ = 24 103 m

In the attached we can see a diagram of the different Barry trajectories and the coordinate system for the decomposition,

We must be careful all the angles must be measured counterclockwise from the positive side of the axis ax (East)

Let's use trigonometry for each distance

Route 1

          cos (180 -35) = \(\frac{x_1}{\Delta d_1}\)

          sin 145 = \(\frac{y_1}{\Delta d1}\)

          x₁ = Δd₁ cos 125

          y₁ = Δd₁ sin 125

          x₁ = 72 103 are 145 = -58.98 103 m

          y₁ = 72 103 sin 155 = 41.30 10³ m

Route 2

          cos (90+ 30) = \(\frac{x_2}{\Delta d_2}\)

          sin (120) = \(\frac{y_2}{\Delta d_2}\)

          x₂ = Δd₂ cos 120

          y₂ = Δd₂ sin 120

          x₂ = 24 103 cos 120 = -12 10³ m

           y₂ = 24 103 sin 120 = 20,78 10³ m

The component of the resultant vector are

              Rₓ = x₁ + x₂

              R_y = y₁ + y₂

              Rx = - (58.98 + 12) 10³ = -70.98 10³ m

              Ry = (41.30 + 20.78) 10³ m = 62.08 10³ m

We construct the resulting vector

Let's use the Pythagoras' Theorem for the module

             R = \(\sqrt{R_x^2 +R_y^2}\)

             R = \(\sqrt{70.98^2 + 62.08^2}\)   10³

             R = 94.30 10³ m

We use trigonometry for the angle

             tan θ ’= \(\frac{R_y}{R_x}\)

             θ '= tan⁻¹ \(\frac{R_y}{R_x}\)

             θ '= tan⁻¹ \(\frac{62.08}{70.98}\)

             θ ’= 41.2º

Since the offset in the x axis is negative and the displacement in the y axis is positive, this vector is in the second quadrant, to be written with respect to the positive side of the x axis in a counterclockwise direction

            θ = 180 - θ'

            θ = 180 -41.2

            θ = 138.8º

Finally, let's calculate the speed for the way back, since the total of the trajectory must be 5 min and on the outward trip I spend 3 min, for the return there is a time of t₃ = 2 min = 120 s.

The average speed of the trip should be

             v = \(\frac{\Delta R}{t_3}\)  

             v = \(\frac{94.30}{120} \ 10^3\)

              v = 785.9 m / s

in the opposite direction, that is, the angle must be

               41.2º to the South of the East

In conclusion, using the kinematics of the uniform motion and the addition of vectors, results are:

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Radio waves are safe for diagnosing illnesses through magnetic resonance imaging due to their low energy.

Radio waves are useful for space research due to their long wavelengths.

Radio waves are waves which form a portion of the electromagnetic spectrum  that occur at lower frequencies than microwaves.

The frequencies of radio waves are as low as 3 Hz and as high as 1 gigahertz.

The wavelengths of radio waves occur within the range of thousands of meters to 30 cm.

The property radio waves that makes them safe for diagnosing illnesses through magnetic resonance imaging is their low energy as they are not ionizing.

The property that makes radio waves useful for space research is their long wavelengths which extends to thousands of meters.

In conclusion, radio waves are safe for magnetic resonance imaging and are applied in space research.

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Answer:

Which property best makes radio waves safe for diagnosing illnesses through magnetic resonance imaging?

✔ low energy of radio waves

Which property best makes radio waves useful for space research?

✔ long wavelengths of radio waves

Answer:

reflect and absorb light

Explanation:

Opaque objects reflect some amount of light and absorb the rest. They do not transmit light.

Transparent objects transmit most light falling on them. They reflect a very small amount of light and absorb the rest.

Translucent objects undergo all three processes, i.e., they reflect, absorb and transmit the light that falls on them

In magnetic elements, each atom’s magnetic domains are aligned which creates magnetic force. In nonmagnetic elements, each atom’s domains are pointing in the different directions so, resulting element has nonmagnetic force.

Attraction or repulsion that arises between electrically charged particles because of their motion is called magnetic force and it is the basic force responsible for such effects as action of electric motors and also attraction of magnets for iron.

Magnetism is caused by motion of electric charges. Every substance is made up of tiny units called atoms and each atom has electrons, particles that carry electric charges. Spinning like tops, electrons circles the nucleus, or core, of an atom.

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Answer:

Direction and magnitude

The block moving at instant the spring is compressed 15cm is  √[(25m - 45)/m]

When there is no external force acting on the system, the total mechanical energy remains conserved.

K.E i = K.E f + SPE

1/2 mu² =1/2 mv² + 1/2 kx²

Substitute u = 5m/s, k =2000N.m and x =0.15m, we get the final velocity

1/2 mu² =1/2 mv² + 1/2 kx²

v =√[(25m - 45)/m]

Thus, the block moving at the instant the spring has been compressed 15 cm is √[(25m - 45)/m]

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The wing is made of aluminum, therefore: α = 25 x 10-6 /oC Length: l = 29 m, Temperature: T1 = 29  C and Change in length: Δl = -7 cm (-0.07 m)

α = 25 x 10-6 /oC,Δl = αlΔT = αl(T2 - T1)

The equation for what you don't know: Δl = αl(T2 - T1); divide both sides by αl Δl/ αl = (αl/ αl )(T2 - T1) = T2 - T1; add T1 to both sides. Δl/αl + T1 = T2 - T1 + T1 = T2 and Δl/ αl + T1 = T2

Δl/ αl + T1 = T2

= (-0.07 m)/( 25 x 10-6 /oC x 29 m) + 29 C = -96.551 oC + 29 C

= -67.551 oC = -67.6 C.

Therefore, The wing is made of aluminum, therefore: α = 25 x 10-6 /oC Length: l = 29 m, Temperature: T1 = 29 C and Change in length: Δl = -7 cm (-0.07 m).

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Answer:   The volume of the block will be \(100cm^3\)

Explanation:

Density is defined as the mass contained per unit volume.

\(Density=\frac{mass}{volume}\)

Given : Mass of cube = 100 grams

Density of cube = \(1g/cm^3\)

Putting in the values we get:

\(Volume=\frac{mass}{density}\)

\(Volume=\frac{100g}{1g/cm^3}=100cm^3\)

 Thus volume of the block will be \(100cm^3\)

To create a resonant circuit with a 1H inductor at 100Hz, a capacitance of 0.0159 microfarads (uF) should be used.

1. Determine the frequency of the resonant circuit, which is given as 100Hz.

2. The formula for calculating the resonant frequency of an LC circuit is: f = 1 / (2π√(LC)), where f is the frequency, L is the inductance, and C is the capacitance.

3. Rearrange the formula to solve for C: C = 1 / (4π²f²L).

4. Substitute the given values into the formula: C = 1 / (4π² * 100² * 1).

5. Calculate the capacitance using a calculator or mathematical software: C ≈ 1.5915 × 10^(-5) farads.

6. Convert the capacitance to microfarads for convenience: 1 farad = 1,000,000 microfarads, so C ≈ 15.915 uF.

7. Round the capacitance to an appropriate number of significant figures, resulting in a capacitance of 0.0159 uF (or 15.9 nF).

Therefore, to produce a resonant circuit with a 1H inductor at 100Hz, a capacitance of approximately 0.0159 microfarads (uF) should be used.

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The effort distance of a lever should be increased to lift a heavier load because it provides a mechanical advantage, allowing for easier lifting of the load.

The effort distance of a lever should be increased to lift a heavier load because it allows for a mechanical advantage that compensates for the increased weight.

In a lever system, the effort distance is the distance between the point of application of the input force (effort) and the fulcrum, while the load distance is the distance between the point of application of the output force (load) and the fulcrum. The mechanical advantage of a lever is determined by the ratio of the load distance to the effort distance.

By increasing the effort distance, the mechanical advantage of the lever system is increased. This means that for the same input force (effort), a greater output force (load) can be achieved. When dealing with a heavier load, a higher mechanical advantage is required to overcome the increased resistance.

By increasing the effort distance, the lever system can effectively multiply the applied force, making it easier to lift the heavier load. This allows for the redistribution of force and facilitates the efficient use of human effort in various applications, such as in construction, engineering, and even everyday tools like scissors and pliers.

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The RMS value of the current,phase angle and power dissipate will be 1191.51 ohm -89.01°,.283 watts respectively.

A capacitor is a device that can store electrical energy. It is a two-conductor configuration separated by an insulating medium that carries charges of equal size and opposite sign.

The r.m.s. current is found as;

\(\rm X_L = 2 \pi f L \\\\ \rm X_L = 2 \times 3.14 \times 60 \times 35 \times 10^{-3} \\\\ X_L = 13.19\ ohm\)

\(\rm X_C = \frac{1}{2\pi fC} \\\\X_C = \frac{1}{2 \times 3.14 \times 60 \times 20 \times 10^{-6}}\\\\ X_C=13.69 \ ohm\)

\(\rm Z = \sqrt{R^2+(X_L-X_C)^2} \\\\ \rm Z = \sqrt{2^2+(13.19-13.26)^2} \\\\ Z= \sqrt{14284.25} \\\\ Z = 119.51 \ ohm\)

The RMS value of the current is;

\(\rm I_{rms}=\frac{V_{rms}}{Z} \\\\ \rm I_{rms}=\frac{V_{45}}{119.51} \\\\ \rm I_{rms}=0.376\ ohm\)

The phase angle and indicate if the current lags or leads the total voltage is;

\(\rm tan \theta =tan^{-1} \frac{X_L-X_C}{R} \\\\ \rm tan \theta =tan^{-1} \frac{13.19132.69}{119.51} \\\\ \theta = -89.01^0\)

The power dissipate in this circuit is found as;

\(\rm P = IV cos \theta \\\\ \rm P = 0.376 \times 45 cos (-89.04) \\\\ P=0.376 \times 45 cos (-8904^0)\\\\ P=0.283 \ watt\)

Hence, the power dissipate in this circuit is found is 0.283 watt.

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Stochastic impacts of radiation allude to those that happen arbitrarily and are not reliant upon the portion got. These impacts are related to the likelihood of events and incorporate disease and hereditary changes. Non-stochastic impacts, then again, have a limit, and their seriousness increments with expanding portions.

Models incorporate radiation consumption and intense radiation conditions. Understanding the qualification among stochastic and non-stochastic impacts of radiation is significant in fields like radiation security, atomic medication, and radiobiology.

It assists in setting radiation with dosing limits, creating well-being rules, and carrying out suitable radiation safeguarding measures. By separating these impacts, experts can evaluate and deal with the dangers related to openness to ionizing radiation all the more successfully.

This information guides choices in regard to radiation wellbeing conventions, word-related openness limits, and the improvement of radiation therapy systems in medication.

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Answer:

Explanation:

Part A:

Using the formula for electric potential due to a charged spherical shell:

V1 = kq1/R1 and V2 = kq2/R2

where k is the Coulomb constant.

For r < R1, the electric potential due to both shells is:

V = V1 + V2 = kq1/R1 + kq2/R2

For R1 < r < R2, the electric potential due to the outer shell is:

V = V2 = kq2/R2

For r > R2, the electric potential due to both shells is:

V = V1 + V2 = kq1/r + kq2/R2

Substituting the given values, we get:

V = (9.0x10^9 Nm^2/C^2)(5.00x10^-6 C)/0.025 m + (9.0x10^9 Nm^2/C^2)(-6.00x10^-6 C)/0.15 m

V = -4.32x10^5 V

Therefore, the electric potential due to the two shells at a distance of 2.50 cm from their common center is -4.32x10^5 V.

Part B:

For r < R1, the electric potential due to both shells is:

V = V1 + V2 = kq1/R1 + kq2/R2

For R1 < r < R2, the electric potential due to both shells is:

V = V1 + V2 = kq1/r + kq2/R2

For r > R2, the electric potential due to both shells is:

V = V1 + V2 = kq1/r + kq2/r

Substituting the given values, we get:

V = (9.0x10^9 Nm^2/C^2)(5.00x10^-6 C)/0.1 m + (9.0x10^9 Nm^2/C^2)(-6.00x10^-6 C)/0.1 m

V = 4.80x10^5 V

Therefore, the electric potential due to the two shells at a distance of 10.0 cm from their common center is 4.80x10^5 V.

Jonathan needs to maintain a separation of 0.543 mm between the plates to get the desired charge, and a dielectric constant of 92.6 to achieve a separation of 5 mm with a dielectric.

(a) Using the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage, we can solve for the capacitance: C = Q/V =\((8.15 x 10^-9 C) / (50 V) = 1.63 x 10^-10 F.\)

Then, using the formula for capacitance of parallel plate capacitors: C = ε0A/d, where ε0 is the permittivity of free space, A is the area of the plates, and d is the separation distance, we can solve for the separation distance: d =\(_{3}OA/C = (8.85 x 10^-12 F/m) x (0.01 m^2) / (1.63 x 10^-10 F) = 0.543 mm.\)

(b) To find the dielectric constant, we can use the formula for capacitance of a parallel plate capacitor with a dielectric: C = εrε0A/d, where εr is the relative permittivity or dielectric constant of the material. Solving for εr, we get: εr = Cd / ε0A = \((1.63 x 10^-10 F)\) x (0.005 m) / \((8.85 x 10^-12 F/m)\) x \((0.01 m^2)\) = 92.6.

Therefore, Jonathan should use a dielectric with a relative permittivity of 92.6 to achieve a separation of 5 mm.

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Answer:

friction

Explanation:

the road is the pathway for the vehicle so the tures create friction to stay sturdie

a. the landing ramp should be placed at 276.298 ft horizontally from the launch point.

(b) the angle of the landing ramp is 30°.

(a)

Launch speed = 55.0 mph * (1.467 ft/s)

= 80.685 ft/s

Horizontal distance = Launch speed * Time of flight

Vertical velocity = Launch speed * sin(30.0°)

Time to reach maximum height = Vertical velocity / g

Vertical distance = (1/2) *g * t²

and Time = √(2 * Vertical distance / g)

Total time of flight = 2 * Time to reach maximum height + Time for descent

Horizontal distance = Launch speed * Total time of flight

The Vertical velocity = 80.685 ft/s * sin(30.0°)

= 40.3425 ft/s

Time to reach maximum height = 40.3425 ft/s / 32.2 ft/s²

Time to reach maximum height  = 1.253 s

Time of descent = √(2 * 25.0 ft / 32.2 ft/s²)

Time of descent  = 0.913 s

Total time of flight = 2 * 1.253 s + 0.913 s

Total time of flight = 3.419 s

In conclusion, the horizontal distance = 80.685 ft/s * 3.419 s

horizontal distance = 276.298 ft

b.

the angle of the landing ramp is  30.0 be the same as the launch angle.

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Answer:   the acceleration of the masses is given by = 0, which means the angular acceleration of the pulley is zero. This implies that the masses m and 2m move with constant velocity,  they are in equilibrium.

In order to answer the question of where the arrows of the magnetic domains point, one must consider the properties of the magnets' poles: A) The arrows line up and point upward.

Iron or cobalt is a common component of permanent magnets. An alloy composed of aluminum is called alnico. Involves a systematic process minerals are capable of making powerful electromagnets. They are widely utilized in industrial and consumer electronics applications.

The word "lithos magnes" from classical Greek is where the name first appeared. According to Pliny's explanation in his "Naturalis Historia," the term is derived from the myth of the Greek shepherd Magnes on Mount Ida, whose iron staff and the nails in his shoes were drawn to the magnetite stones.

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Answer:

POINTS LEFT

Explanation:

just did it

Answer:

The more the mass, the more the gravity. The more the distance the less the gravity.

Explanation:

I just learned about it on E_d_g_e_n_u_i_t_y