A man has 2 spheres A and B. He gently drops sphere A vertically down and throws the sphere B horizontally at the same time. Which of the following statements is correct?

A.sphere B reaches the ground earlier

B.not enough information

C.both spheres reach the ground at the same time

D.sphere A reaches the ground earlier​

Answer:

512.5 mJ

Explanation:

Let the two identical charges be q = +35 µC and distance between them be r₁ = 46 cm. A charge q' = +0.50 µC located mid-point between them is at r₂ = 46 cm/2 = 23 cm = 0.23 m.

The electric potential at this point due to the two charges q is thus

V = kq/r₂ + kq/r₂

= 2kq/r₂

= 2 × 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C/0.23 m

= 630/0.23  × 10³ V

= 2739.13 × 10³ V

= 2.739 MV

When the charge q' is moved 12 cm closer to either of the two charges, its distance from each charge is now r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm = 0.35 m and r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.

So, the new electric potential at this point is

V' = kq/r₃ + kq/r₄

= kq(1/r₃ + 1/r₄)

= 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C(1/0.35 m + 1/0.11 m)

= 315 × 10³(2.857 + 9.091) V

= 315 × 10³ (11.948) V

= 3763.62 × 10³ V

= 3.764 MV

Now, the work done in moving the charge q' to the point 12 cm from either charge is

W = q'(V' - V)

= 0.5 × 10⁻⁶ C(3.764 MV - 2.739 MV)

= 0.5 × 10⁻⁶ C(1.025 × 10⁶) V

= 0.5125 J

= 512.5 mJ

Work done will be:  \(512.5 \mu J\).

Given:

q = +35 µC

q' = +0.50 µC

r₁ = 46 cm

r₂ = 46 cm/2 = 23 cm = 0.23 m

The electric potential at this point due to the two charges q is thus

\(V = \frac{kq}{r_2}+\frac{kq}{r_2}\\\\ V= \frac{2kq}{r_2}\\\\V= \frac{2 * 9 * 10^9Nm^2/C^2 * 35 * 10^{-6} C}{0.23 m}\\\\V= \frac{630}{0.23*10^3}V\\\\V= 2739.13 * 10^3 V\\\\V= 2.739 \mu V\)

When the charge q' is moved 12 cm closer to either of the two charges, its distance from each charge is now;

r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm = 0.35 m and

r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.

Thus, the new electric potential at this point is

\(V' = \frac{kq}{r_3} + \frac{kq}{r_4}\\\\V= kq(\frac{1}{r_3}+\frac{1}{r_4})\\\\V= 9 * 10^9 Nm^2/C^2 * 35 * 10^{-6} C (\frac{1}{0.35m}+\frac{1}{0.11m})\\\\V= 315 * 10^3(2.857 + 9.091) V\\\\V= 315 * 10^3 (11.948) V\\\\V= 3763.62 * 10^3 V\\\\V= 3.764 \mu V\)

Now, the work done in moving the charge q' to the point 12 cm from either charge is:

\(W = q'(V' - V)\\\\W= 0.5 * 10^{-6} C(3.764 MV - 2.739 MV)\\\\W= 0.5 *10^{-6} C(1.025 * 10^6) V\\\\W= 0.5125 J\\\\W= 512.5 \mu J\)

Thus, the work done will be: \(512.5 \mu J\).

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For you to feel pressure on your skin, mechanoreceptor neurons are required.

The Pacinian corpuscles Pacinian corpuscles, like the ones shown in the bright field light microscopy image, are sensitive to pressure (touch) and high-frequency vibration.

Mechanoreceptors are sensitive to many physical alterations, including as touch, pressure, vibration, and stretch.

These sensors are capable of picking up feelings like pressure, vibration, and texture. Meissner's corpuscles, Pacinian corpuscles, Ruffini's corpuscles, and Merkel's disks are the four types of known mechanoreceptors whose sole purpose is to detect indentations and vibrations of the skin.

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If a student took seven science classes, his/her at score would be 1200.

A graph is a relationship in which data is presented on the axes of cartesian coordinates. The presentation of the graph must be such that it is easy to read off the graph.

Now we can see that the graph is a plot of SAT scores versus the number of science classes that students take. In this case, the slope would tell us the score per subject of a student in the SAT.

a) The mathematical model that re[resents this graph is;

y =100x + 500

b) The slope of the graph must be obtained by the use of two points thus;

m = y2 - y1/x2 - x1

m = 600 - 500/1 - 0

m = 100

c) Using the relation;

y = mx + c

m = 100

c = 500

y = 100(7) + 500

y = 1200

Thus if a student took seven science classes, his/her at score would be 1200.

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The values are: (a) α ≈ 0.47 Np/m, (b) β ≈ 26 rad/m, and (c) Z0 ≈ 320.70 Ω.

The given parameters for the planar transmission line can be used to calculate the values of α, β, and Z0.

(a) To calculate α, we can use the equation:

\(α = (ωε₀μ₀)^{(1/2)} (1 + jσ/ωε₀)^{(1/2)}\)

where ω is the angular frequency, ε₀ is the permittivity of free space, μ₀ is the permeability of free space, and σ is the conductivity.

Given:

\(ω = 2πf = 2π(750 MHz) = 1.5π × 10^9 rad/s

ε₀ = 8.854 × 10^{-12} F/m

μ₀ = 4π × 10^{-7} H/m

σ = -5.5 × 10^7 S/m\)

Substituting the values, we get:

\(α = (1.5π × 10^9 × 8.854 × 10^{-12} × 4π × 10^{-7})^{(1/2)} (1 + j(-5.5 × 10^7)/(1.5π × 10^9 × 8.854 × 10^{-12}))^{(1/2)}\)

Calculating the above expression gives:

α ≈ 0.47 Np/m

(b) To calculate β, we can use the equation:

\(β = (ωε₀μ₀)^{(1/2)} (1 + jσ/ωε₀)^{(-1/2)}\)

Substituting the given values, we get:

\(β = (1.5π × 10^9 × 8.854 × 10^{-12} × 4π × 10^{-7})^{(1/2)} (1 + j(-5.5 × 10^7)/(1.5π × 10^9 × 8.854 × 10^{-12}))^{(-1/2)}\)

Calculating the above expression gives:

β ≈ 26 rad/m

(c) To calculate Z0, we can use the equation:

Z0 = (η₀/2π) (β/α)

where η₀ is the intrinsic impedance of free space.

Given:

\(η₀ = (μ₀/ε₀)^{(1/2)} ≈ 377 Ω\)

Substituting the values, we get:

Z0 = (377/2π) (26/0.47) ≈ 320.70 Ω

Therefore, the values are:

(a) α ≈ 0.47 Np/m

(b) β ≈ 26 rad/m

(c) Z0 ≈ 320.70 Ω.

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Answer:

  zero

Explanation:

The gravitational force between two masses is jointly proportional to the masses and inversely proportional to the square of the distance between them. The constant of proportionality is the universal gravitational constant.

__

In this scenario, the force to the right is matched by an identical force to the left from an identical mass at an identical distance. The magnitudes of the forces are equal, and their directions are opposite. The forces cancel.

The net force on the 2.2 kg mass in the center is zero.

Answer:

if you stretch a spring with k = 2, with a force of 4N, the extension will be 2m. the work done by us here is 4x2=8J. in other words, the energy transferred to the spring is 8J. but, the stored energy in the spring equals 1/2x2x2^2=4J (which is half of the work done by us in stretching it).

Answer:

Dynamic stretches

Explanation:

Dynamic stretches are active movements where joints and muscles go through a full range of motion. When you do dynamic stretches you do not hold the stretch, instead you stretch with movement.

Answer: Proxima Centauri is the closet star about 40,208,000,000,000 km away. 

Explanation:

Answer:

The closest star is about 25,300,000,000,000 miles (39,900,000,000,000 kilometers) away

Explanation:

Answer:

Wieght = mass × Gravity

750 = M × 10. { M = mass , taking g = 10m/s² }

M = 75kg

now when elevator is moving down = Relative gravity = actual gravity - down acceleration = 10 - 6 = 4m/s²

So relative mass = Actual mass × relative gravity

75 × 4 = 300N

Note : If the elevator was moving upwards then Relative gravity = actual gravity + acceleration up

Answer:

the ebergy

Explanation:

energy does it have if a miving object

Answer:

The  value  is   \(W = -54.615 \ J\)

Explanation:

From the question we are told that

    The mass of the block is  \(m = 3.0 \ kg\)

     The force is \(F = 16 \ N\)

     The  angle is  \(\theta = 37^o\)

     The first speed of the block is  \(u = 2 \ m/s\)

     The second  speed of the block is  \(v = 3.8 \ m/s\)

     The displacement is  \(d = 5.5 \ m\)

Gnerally from kinematic equation we have that

      \(v^2 = u^2 + 2as\)

=>    \(3.8 ^2 = 2^2 + 2 * a* 5.5\)

=>    \(a = 0.9491 \ m/s^2\)

Generally the net force acting on the crate is mathematically represented as

        \(F_{net} = [ F cos (\theta ) - F_f ] = ma\)

Here  \(F_f\) is the frictional force acting on the crate

So    

          \([ 16 cos (37 ) - F_f ] = 3 * 0.9491\)    

=>       \(F_f = 9.93 \ N\)

Generally the work done by  friction  during the displacement is mathematically represented as

        \(W = F_f * d * cos ( 180 )\)

=>     \(W = 9.93 * 5.5 * cos ( 180 )\)

=>     \(W = -54.615 \ J\)

The type of electromagnetic radiation is Radiowaves which have a wavelength of 10 meters.

Since these waves have a longer wavelength, they are used for communication over long distances. Some examples of sources that emit these radiations - radiowaves - are TV and radio stations, cell phone towers, as well as satellites.

Radio waves are very beneficial as they can travel long distances without any power losses and getting absorbed or scattered by the atmosphere. Also, radio waves are non-ionizing and generally considered safe for human exposure. They are widely used in many applications, including broadcasting, navigation, remote sensing, and medical imaging.

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the frequency of the given wavelength of light, 682.0 nm, is 4.40 x 10^14 Hz.

we can use the equation c = λν, where c is the speed of light, λ is the wavelength, and ν is the frequency. We know the wavelength (682.0 nm) and the speed of light (3.00 x 10^8 m/s), so we can solve for the frequency:

c = λν
ν = c/λ
ν = (3.00 x 10^8 m/s) / (682.0 nm x 10^-9 m/nm)
ν = 4.40 x 10^14 Hz

Therefore, the frequency of the given wavelength of light is 4.40 x 10^14 Hz.

In conclusion, the frequency of a wavelength of 682.0 nm is 4.40 x 10^14 Hz.
Main Answer: The frequency of the 682.0 nm wavelength light is approximately 4.40 x 10^14 Hz.

Explanation:
To convert the wavelength (in nm) to frequency (in Hz), you can use the equation:

Frequency (v) = Speed of Light (c) / Wavelength (λ)

First, convert the wavelength from nanometers to meters:

Wavelength (λ) = 682.0 nm × (1 m / 1,000,000,000 nm) = 6.82 x 10^-7 m

Next, use the speed of light (c), which is approximately 3.00 x 10^8 m/s:

Frequency (v) = (3.00 x 10^8 m/s) / (6.82 x 10^-7 m)
Frequency (v) ≈ 4.40 x 10^14 Hz

The frequency of the given wavelength (682.0 nm) of light is approximately 4.40 x 10^14 Hz.

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Answer:

40sec

Explanation:

Data

Work = 440 J

Power= 11watt

time = ?

Power = work done/time

===> time = work done/power

= 440/11

= 40sec

The speed of the second bowling ball after collision is 1.55 m/s

Speed can be defined as the ratio of distance to time of a body.

To calculate the speed of the second bowling ball after collision, we use the formula below

Formula:

v = mu/m'................ Equation 1

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the second bowling ball is 1.55 m/s fast.

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Hi there!

Using the Impulse-Momentum theorem:

I = Δp = mΔv

Where:

I = Impulse (Ns)

m = mass of object (kg)

Δv = change in velocity (vf - vi, m/s)

Plug in the given values and solve:

60 = 5Δv

Δv = 12 m/s

The force of static friction is 10 newtons. The magnitude of the net force is 10 newtons - 10 newtons = 0 newtons.  It can be said that the crate is not accelerating and is in a state of equilibrium or constant velocity.

To calculate the magnitude of the force of friction between the crate and the floor, we need to use the equation of static friction: Fs = μs * N, where Fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force.

Since the crate is initially moving to the right, the force of static friction must equal the force applied in the opposite direction (10 newtons) to maintain equilibrium. Therefore, the force of static friction is 10 newtons.

The magnitude of the net force acting on the crate can be calculated using the equation: Net force = Applied force - Force of friction. In this case, the applied force is 10 newtons, and the force of friction is also 10 newtons. Therefore, the magnitude of the net force is 10 newtons - 10 newtons = 0 newtons.

Since the net force acting on the crate is zero (as calculated in the previous question), the crate is not experiencing a net force. According to Newton's second law of motion, F = m * a, where F is the net force, m is the mass, and a is the acceleration. Since the net force is zero, the crate does not have any acceleration. It can be said that the crate is not accelerating and is in a state of equilibrium or constant velocity.

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We can use the principle of Pascal's law to solve this problem. Pascal's law states that pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and the walls of the container.

First, we can calculate the pressure applied to the fluid by the smaller piston:

Pressure = Force / Area
Area = πr^2 = π(0.01m)^2 = 7.854 x 10^-4 m^2
Pressure = 120.0 N / 7.854 x 10^-4 m^2 = 152,913.4 Pa

Next, we can use Pascal's law to find the force on the larger piston:

Pressure = Force / Area
Area = πr^2 = π(0.125m)^2 = 0.0491 m^2
Pressure = 152,913.4 Pa
Force = Pressure x Area = 152,913.4 Pa x 0.0491 m^2 = 7500 N

Therefore, the force on the larger piston is 7500 N.

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Among the given options, the characteristic that is NOT true of a spiral galaxy is: Contains little gas and dust.

A spiral galaxy is a type of galaxy which is characterized by its long, spiral arms. These arms are found surrounding a center nucleus, creating a spiral shape. Spiral galaxies are further categorized into two main types depending on the size of the central bulge and how tightly wound their arms are, these are barred and unbarred galaxies. Furthermore, they are also one of the most common types of galaxies observed in the universe.

Characteristics of Spiral Galaxy

A few of the common characteristics of spiral galaxies are:

Ongoing star formation

Appears mostly blue

Contains a central bulge, disk, and diffuse extended halo

Contains both old and young star

Contains little gas and dust (This statement is NOT a characteristic of a Spiral Galaxy.)

They are one of the most common types of galaxies in the universe.

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Answer:

D. Malleable

Explanation:

Gold is metallic, with a yellow colour when in a mass, but when finely divided it may be black, ruby, or purple. It is the most malleable and ductile metal; 1 ounce (28 g) of gold can be beaten out to 300 square feet. It is a soft metal and is usually alloyed to give it more strength.

Answer:

v = 2.22 m/s

Explanation:

First we apply the second equation of motion to the vertical motion of the body:

s = Vi t + (1/2)gt²

where,

s = y = vertical distance covered = 200 m - 100 m = 100 m

Vi = V₀y = Vertical Component of Initial Velocity = 0 m/s  (because spider man jumps horizontally, thus his velocity has no vertical component initially)

t = Time Taken to Land on 100 m high building = ?

g = 9.8 m/s²

Therefore,

100 m = (0 m/s)t + (0.5)(9.8 m/s²)t²

t² = (100 m)/(4.9 m/s²)

t = √(20.4 s²)

t = 4.5 s

Now, we analyze the horizontal motion. Neglecting air friction, the horizontal motion is uniform with uniform velocity. Therefore,

s = vt

where,

s = x = horizontal distance covered = 10 m

v = V₀ₓ = Horizontal Component of Initial Velocity = Initial Velocity = ?

Therefore,

10 m = v(4.5 s)

v = 10 m/4.5 s

v = 2.22 m/s

Answer:

I hope this helps if not I am sorry good luck

Explanation:

The food of the foxes is the voles and if the their population went down of course the foxes would too if the voles went extinct what would the foxes eat surley they would die since their food source is shortened .

Answer:

Many ionic substances will dissolve completely in water to form an aqueous solution of cations and anions; this solution will conduct electricity.

Explanation:

Answer:

Many ionic substances will dissolve completely in water to form an aqueous solution of cations and anions; this solution will conduct electricity.

Explanation:

Overall, the difference between detached and semi-detached binary systems lies in the extent to which the stars interact with each other and exchange mass.

In a detached binary system, the two stars are far enough apart from each other that they do not significantly interact with each other. The stars have their own gravitational fields, which cause them to orbit around the common center of mass, but they do not exchange significant amounts of mass or significantly alter each other's evolution. Detached binary systems are the most common type of binary systems.

In contrast, in a semi-detached binary system, one of the stars is closer to the common center of mass than the other and is therefore more strongly affected by the gravitational pull of the other star. The closer star can transfer mass to the more distant star, often through an accretion disk, causing the more distant star to become more massive and the closer star to become less massive over time. Semi-detached binary systems can lead to various phenomena such as accretion disks, mass transfer, and outbursts of energy.

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When the aluminum and water are mixed together, heat transfer occurs between the two substances until thermal equilibrium is reached. The aluminum, initially at a higher temperature of 80.0 °C, will transfer heat to the water, which is initially at a lower temperature of 10.0 °C.

Since aluminum has a higher specific heat capacity than water (0.897 J/g°C for aluminum compared to 4.186 J/g°C for water), it can hold more heat energy per gram per degree Celsius. This means that the aluminum will transfer its heat energy to the water, causing its temperature to increase.

Based on the given information, we can expect that the final temperature of the water will be somewhere between 10.0 °C and 80.0 °C. However, without knowing the exact values of the specific heat capacities of aluminum and water, as well as the amount of heat transferred between them, it is not possible to predict the final temperature accurately. In real-world scenarios, the final temperature will depend on several factors, such as the exact masses of aluminum and water, their specific heat capacities, and the heat transfer mechanism (e.g., conduction). To determine the final temperature accurately, precise measurements and calculations involving heat transfer equations would be necessary.

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We have that the the minimum coefficient of static friction  is mathematically given as

\(\mu= 0.235\)

Generally the equation for the Force   is mathematically given as

\(F=n\mu\\\\Where\\\\n = mg\\\\Given\\\\Fc = m(vt2/R)\\\\Therefore\\\\\mu = vt2/gR\\\\\mu = 2.22/(9.8(2.1)) \\\\\)

\(\mu= 0.235\)

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The absolute maximum shear stress in the shaft are Tmax = 129.4 MPa  for the shaft that has an outer diameter of 100 mm and an inner diameter of 80 mm.

The shaft constitutes an important aspect of the hole shaft motor, that's utilized in electrically powered vehicles, along with trains. Hollow shafts also are appropriate for the development of jigs and furnishings in addition to computerized machines.

Here we have diameter of 80 mm and outer diameter of 100 mm .

\(T = π / 16 × T × (D^4 -d^4)/D\\15×1000 = π/16 × T × 0.1^4 - 0.08^4 /0.1\\Tmax = 129.4 × 10 ^6 N/m^2\\Tmax = 129.4 MPa.\)

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The wavelength of a wave moving at 450m/s with a frequency of 464Hz is 0.97m.

We know that wavelength of a wave is calculated by

\(\lambda=\frac{v}{f}\)  ...(i)

where λ ⇒ wavelength

v ⇒ velocity of a wave

f ⇒ frequency

Now, as per the question:

Velocity of wave, v = 450 m/s

Frequency, f = 464 Hz

Putting the values in equation (i),

λ = 450/464

λ = 0.97 m

Therefore, wavelength of a wave moving at 450m/s with a frequency of 464Hz is 0.97m.

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Answer:

Explanation:

Kindly see attached file for your reference

Step one:

given data

the horizontal component of the force= 12N

the vertical component of the force= 5N

The dashed arrow represents the hypotenuse of the triangle, hence the resultant of the force system.

By implication of this, we will use the Pythagoras theorem to solve for the resultant force

Step two:

\(F_R=\sqrt{F_H^2+F_V^2}\\\\F_R= \sqrt{12^2+5^2}\\\\F_R=\sqrt{144+25}\\\\F_R=\sqrt{169}\\\\F_R=13N\)

Answer:

3.07 kilometres per second