What are the correct methods of heat transfer

Answer:

They both intercept,focus and determine the intensity of an incoming radiation but at different degree.

Explanation:

Radio telescope is much larger than optical telescope because radio wavelengths are much longer than optical wavelengths. The longer the wavelengths implies that the radio waves has a lower energy than optical light waves. So, in order to collect enough radio photons to detect a signal, the radio dishes must be very large.

Answer:small barrel gun

Explanation:

Given

Muzzle velocity of bullet is greater in short barrel gun as compared to larger barrel gun

acceleration is given by change in velocity with respect to time

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

In case of short barrel bullet time taken by bullet to reach its muzzle velocity is less therefore acceleration of small barrel bullet is more compared to long barrel bullet.

Final answer:

The muzzle velocity of a bullet can be calculated using the formula v = u + at. Using the given acceleration of 6.20 x 10⁵ m/s² and time of 8.10 x 10⁻⁴ s, the muzzle velocity is 502 m/s.

Explanation:

To calculate the muzzle velocity of a bullet, which is the velocity of the bullet as it leaves the barrel of a gun, we can use the formula for uniform acceleration: v = u + at, where v is the final velocity, u is the initial velocity (which is 0 m/s in this case, as the bullet starts from rest), a is the acceleration, and t is the time. Here, we are given an acceleration of 6.20 × 10⁵ m/s² and a time of 8.10 × 10⁻⁴ s. Plugging these values into the formula gives us the muzzle velocity.

Using the formula, we have: v = 0 m/s + (6.20 × 10⁵ m/s²)(8.10 × 10⁻⁴ s), which simplifies to v = 502 m/s. Therefore, the muzzle velocity of the bullet is 502 meters per second. This value represents how fast the bullet is traveling when it exits the gun barrel.

Answer:

0.25m

Explanation:

Using the expression that relates the angular velocity(w) and the linear velocity (v).

v= wr where;

w is the angular speed= 128rad/s

v is the linear speed = 32m/s

r is the radius

r = v/w

r = 32/128

r = 0.25m

The radius of the blade is 0.25m

Answer: Tmelt = 19.89mins

Explanation: (complete question- 334×10^3j/kg. How much time, Tmelt passes before the ice starts to melt?)

For ice to melt, its temperature must be 0* C.

Let

Q1 = heat required to raise the ice temp from -16.1 °C. to 0* C.

Q2 = heat required to melt the ice

Q1 = MCp(delta T)

Q2 = m(Hf)

where

M = mass of the ice = 0.500 kg (given)

Cp = specific heat of ice = 2100 J/kg K (given)

delta T = temperature change = 0 - -16.1= 16.1 °C. 

Hf = heat of fusion of ice = 334 x 10^3 J/jg.

Substituting values,

Q1 = 0.500× 2100× 16.1 = 16905J

Q2 = 0.500 × 334×10^3 = 167,000J

Heat required to melt the ice = Q1 + Q2 = 16905 + 167000 = 183905 J

How much time Tmelts = Q1/ 850 = 16905/850 = 19.89mins.

Answer:

a) dissociates completely

Explanation: strong base is a base that is completely dissociated in an aqueous solution.

Answer:

6.2N force

Explanation:

According to Newton's second law of motion, force is equal to the product of the mass of a body and its acceleration. Mathematically,

Force = mass × acceleration

Given mass of bucket of water = 6.2kg

acceleration of the bucket = 1m/s²

Force exerted on the rope = 6.2 × 1

= 6.2N

Answer:

[tex]5.8\cdot 10^{-5}T[/tex]

Explanation:

The magnetic field produced by a current-carrying wire is:

[tex]B=\frac{\mu_0 I}{2\pi r}[/tex]

where

[tex]\mu_0=4\pi \cdot 10^{-7} H/m[/tex] is the vacuum permeability

I is the current in the wire

r is the distance from the wire

The direction of the field is given by the right-hand rule: the thum is placed in the same direction of the current, and the other fingers "wrapped" around the thumb gives the direction of the field.

First of all, here we are finding the magnetic field strength at a point 12.0cm from one wire and 13.6cm from the other: since the two wires are 6.50 cm apart, this means that the point at which we are calculating the field is located either on the left or on the right of both wires. So, by using the right-hand rule for both, we see that both fields go into the same direction (because the currents in the two wires have same direction).

Therefore, the net magnetic field will be the sum of the two magnetic fields.

For wire 1, we have:

[tex]I_1=18.5 A\\r_1=12.0 cm =0.12 m[/tex]

So the field is

[tex]B_1=\frac{(4\pi \cdot 10^{-7})(18.5)}{2\pi(0.12)}=3.1\cdot 10^{-5} T[/tex]

For wire 2, we have:

[tex]I_2=18.5 A\\r_2=13.6 cm=0.136 m[/tex]

So the  field is

[tex]B_2=\frac{(4\pi \cdot 10^{-7})(18.5)}{2\pi(0.136)}=2.7\cdot 10^{-5} T[/tex]

Therefore, the total magnetic field is

[tex]B=B_1+B_2=3.1\cdot 10^{-5}+2.7\cdot 10^{-5}=5.8\cdot 10^{-5}T[/tex]

Question:

Problem 14. Let f(x, y) = (x^2)y*(e^(x−1)) + 2xy^2 and F(x, y, z) = x^2 + 3yz + 4xy.

(a) (i) Find the gradient of f.

(ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing?

(b) (i) Find the gradient of F.

(ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.

Answer:

The answers to the question are

(a) (i)  the gradient of f =  ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) The direction in which f decreases most rapidly at the point (1, −1), ∇f(x, y) = -1·i -3·j is the y direction.

The rate is f decreasing is -3 .

(b) (i) The gradient of F is (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k is  ñ∙∇F =  4·x +⅟4 (8-3√3)y+ 9/4·z at (1, 1, −5)

4 +⅟4 (8-3√3)+ 9/4·(-5) = -6.549 .

Explanation:

f(x, y) = x²·y·eˣ⁻¹+2·x·y²

The gradient of f = grad f(x, y) = ∇f(x, y) = ∂f/∂x i+  ∂f/∂y j = = (∂x²·y·eˣ⁻¹+2·x·y²)/∂x i+  (∂x²·y·eˣ⁻¹+2·x·y²)/∂y j

= ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) at the point (1, -1) we have  

∇f(x, y) = -1·i -3·j  that is the direction in which f decreases most rapidly at the point (1, −1) is the y direction.  

The rate is f decreasing is -3

(b) F(x, y, z) = x² + 3·y·z + 4·x·y.

The gradient of F is given by grad F(x, y, z)  = ∇F(x, y, z) = = ∂f/∂x i+  ∂f/∂y j+∂f/∂z k = (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2·i + 3·j −√3·k

The magnitude of the vector 2·i +3·j -√3·k is √(2²+3²+(-√3)² ) = 4, the unit vector is therefore  

ñ = ⅟4(2·i +3·j -√3·k)  

The directional derivative is given by ñ∙∇F = ⅟4(2·i +3·j -√3·k)∙( (2·x+4·y)i + (3·z+4·x)j + 3·y·k)  

= ⅟4 (2((2·x+4·y))+3(3·z+4·x)- √3∙3·y) = 4·x +⅟4 (8-3√3)y+ 9/4·z at point (1, 1, −5) = -6.549

your question is missing the given conditions, here is the complete question;

Two machines, A and B, each working at a constant rate, can complete a certain task working together in 6 days. In how many days, working alone, can machine A complete the task?

(1) The average time A and B can complete the task working alone is 12.5 days.

(2) It would take machine A 5 more days to complete the task alone than it would take machine B to complete the task

Answer:

Machine A, working alone, can complete the task in 15 days.

Explanation:

As the average is given as 12.5 so,

A+B=2*12.5

A+ B=25

second condition says machine A takes 5 more days to complete the task alone than B so

A=B+5........Eq1

Now from the question statement we can get that

1/A+1/B=1/6, where A is time needed for A to do the task alone and B is the time needed for B to complete the task alone

thus, simplyfing above equation we get as

AB/A+B=6; \\ putting A+B=25 we get

AB/25=6

AB=6*25

AB=150; \\ putting Eq1 we get

(B+5)B=150

B^2+5B-150=0

simplifying the quadratic equation above we get B=10

putting B=10 in Eq1, we get

A+10=25

A=15

Thus, Machine A, working alone, can complete the task in 15 days.

Answer:

2.65 m

Explanation:

From work-kinetic energy principle,

workdone by friction + workdone by gravity on sled = kinetic energy change of sled

Let h be the  vertical height moved and d the distance moved along the incline. h = dsinθ   where θ is the angle of the incline = 20°.

The workdone by gravity on the sled is mghcos180 = -mgh = -mgdsinθ

The frictional force = -μmgcosθ where μ = 0.20 and the work done by friction = -μmgcosθd

The kinetic energy change = 1/2m(v₂² - v₁²) where v₁ = initial speed = 2.0 m/s and v₂ = final speed = 0 m/s (since the sled stops)

So, -mgdsinθ - μmgcosθd = 1/2m(v₂² - v₁²)

-gd(sinθ - μcosθ) = 1/2(v₂² - v₁²)

d = (v₂² - v₁²)/-2g(sinθ - μcosθ)

substituting the values for the variables,

d = (0 - 2²) /[- 2 × 9.8(sin20 - 0.2cos20)]

d = -4 /[- 2 × 9.8(sin20 - 0.2cos20)]

d = -4/ - 1.51 = 2.65 m

It has moved up the incline a distance of 2.65 m.

It is classified as Weakness in SWOT analysis.

Explanation:

SWOT analysis is performed to understand the characteristics of any start ups. It is the abbreviation of Strength, Weakness, Opportunities and Threats. It helps in self analyzing and to plan strategically for improvements. So in this different loopholes, advantages, benefits, profits and loss attained by any organized is classified in these four options like Strength, Weakness, Opportunities and Threats.

In the present case, the problem of slow service speed by Tito's company lead to reduction in customers. Also many of the customers are not returning back due to its slow speed in servicing any instrument. So this loophole is placed in the box of weakness in SWOT analysis. So it is classified as Weakness in SWOT analysis.

Answer:

a. v = 1.679 × 10⁹ m/s b. The answer of v = 1.679 × 10⁹ m/s is not reasonable

Explanation:

Here is the complete question

You read in a science magazine that on the Moon, the speed of a shell leaving the barrel of a modern tank is enough to put the shell in a circular orbit above the surface of the Moon (there is no atmosphere to slow the shell).

A.....What should be the speed for this to happen? Assume that the radius of the Moon is rM = 1.74×106m, and the mass of the Moon is mM = 7.35×1022kg.(Express your answer to two significant figures and include the appropriate units.)

B.....Is this number reasonable?

Solution

From the question, the centripetal force in orbit for the shell = gravitational force of moon on shell.

So, mv²/r = GMm/r²

So v = √(GM/r) where v = velocity of shell, G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of moon = 7.35 × 10²² kg and r = radius of moon = 1.74 × 10⁶ m.

v = √(GM/r) = v = √(6.67 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²² kg/1.74 × 10⁶ m)

v = √(49.0245/1.74 × 10¹⁷) m/s

v = √28.175 × 10¹⁷ m/s

v = √2.8175 × 10¹⁸ m/s

v = 1.679 × 10⁹ m/s

b. The answer of v = 1.679 × 10⁹ m/s is not reasonable because, it is over 1000000 km/s which is greater than the speed of light which is 300000 km/s

Answer:

Option E is correct.

none of the listed options are the same on the equator as at the celestial north pole; the region where Canada is.

Explanation:

Firstly, let's explain some terms.

The earth's rotation is not straight forward as a sphere just rotating. The shape of the earth (better described as geoid; spherical, but a bit flattened at the poles), some gravitational forces on the earth and the existence of seasons mean the earth's rotation is tilted at an angle of 23.5° relative to our orbital plane; the plane of the earth's orbit round the sun.

When the tilted earth is considered, the celestial north and south poles are the north and south poles on the tilted earth's rotational axis.

And the celestial equator is when the equator on the tilted earth, equidistant from the celestial north and south poles is projected into space. A portion of the celestial equator passes through the real equator.

Now, the options one by one,

A) the celestial poles are on the north and south points of the horizon

At the celestial north and south poles, the poles are observable at the north and south points of the horizon.

At the celestial north pole, it is usually located by locating the north star, Polaris, and at the celestial South poles, there are about 4 different methods of locating the poles on the horizon.

But these poles aren't noticeable on the horizon at the equator. The northern one can only be noticed at regions around the celestial north pole.

Hence, the statement A isn't correct.

B) the celestial equator is overhead and passes through the zenith.

Zenith describes the the point in the sky directly above one's head.

At some points on the equator, the celestial equator is truly overhead, but the celestial equator doesn't pass overhead at the north pole, So, this isn't something common to the equator and the north pole.

C) all stars rise and set (none remain in the sky all night long)

Stars remaining in the sky all night long only occurs at the poles (north pole especially). Depending on the seasons, there are times and/or places that have 24 hour sunlight or darkness.

But, this is usually not the case at the equator. At the equator, all stars rise and set all year round. Only small noticeable longer days, shorter nights or longet nights, shorter days are experienced at the equator.

Hence, the statement given isn't true for the north pole and the equator.

D) all-stars are above the horizon exactly half a day.

The paths of the stars are vertical and are cut exactly in half by the horizon at the equator. Each star is up half the time and down half the time.

Unlike at the poles, as explained in the previous statement, they enjoy more varied daytime and night times.

Again, the statement described isn't true for the north pole and the equator.

Answer:

The magnetic force on a free moving charge depends on the velocity of the charge and the magnetic field, direction of the force is given by the right hand rule. While gravitational depends on the mass and distance of the moving particle and electric forces depends on the magnitude of the charge and distance of separation.

Explanation:

The magnetic force on a free moving charge depends on the velocity of the charge and the magnetic field and direction of the force is given by the right hand rule. While gravitational depends on the mass and distance of the moving particle and electric forces depends on the magnitude of the charge and distance of separation.

The magnetic force is given by the charge times the vector product of velocity and magnetic field. While gravitational force is given by the square of the particle mass divided by the square its distance of separation. Also electric forces is given by the square of the charge magnitude divided by the square its distance separation.

Answer:

Acceleration =4.42m/s^2

Explanation:

As the mug is not sliding on the car roof ,there must be static friction between the car and the mug. The maximum limit of static friction that can act is given by:

Kmg

Where I is coefficient of static friction between car roof and mug

m = mass of mug

g= acceleration due to gravity

Acceleration of the car= Kmg

Acceleration = 1.1 × 0.41 × 9.8 =4.42m/s

Answer:

10.791 m/s^2

Explanation:

Solution:

- As the mug is not sliding on the car roof , there will be static friction between car and mug.

- The maximum limit of static friction that can act is given by = k*m*g

- Where, k = coefficient of static friction between the mug and the roof

              m = mass of mug

              g = acc. due to gravity

- So the maximum acc. of car ,

                                     (m*a = k*m*g)

                                     a = k*g

- Therefore max. a = 1.1*g = 1.1*9.81 = 10.791 m/s^2

Answer: True

Explanation:

Electromagnetic wave is defined as the wave which is associated with both electrical and magnetic component associated with them. They can travel in vacuum as well and travel with the speed of light.

Electromagnetic spectrum consists of electromagnetic waves called as radio waves, microwaves, infrared, Visible, ultraviolet , X rays and gamma rays in order of increasing frequency and decreasing wavelength.

Thus the statement that All light waves move through a vacuum with a constant speed is true.

Answer:

The frictional force will also act on the box. So, we can say that the 10 N force is an unbalanced force.

Explanation:

It is given that, a 10-Newton force is applied to a 2-kg block. The block slides across the floor at a constant speed of 5 m/s.

The block is sliding with a constant speed. This shows the acceleration of the block is zero.

When an object slides that means the force acting on it is unbalanced i.e. the object will move in a particular direction.

A)  The force of friction was 10-newtons.

The force of friction was 10-newtons. If the block is moving at constant speed, all the forces on the block must be balanced.

The potential difference across the capacitor is 7.5 V both before and after the Mylar is inserted. To calculate charge and electric field values, additional information about the system, such as the dielectric constant of Mylar or capacitance, is required. Without this information, only the potential difference can be definitively determined.

The subject concerns Physics, specifically the study of capacitors in the context of their behavior before and after a dielectric material is inserted. We are given the dimensions of the capacitor plates and the potential difference applied by a battery, and we need to deduce several properties of the capacitor both before and after the insertion of the Mylar dielectric sheet. The capacitance of the capacitor is not provided directly, but can be inferred using the provided dimensions and the permittivity of free space or air, as required.

The potential difference across a capacitor is determined by the voltage applied by the battery. Thus, the capacitor's potential difference before the Mylar is inserted is 7.5 V.

To find the electric field in the capacitor before the Mylar is inserted, we can use the formula E = V/d, where V is the potential difference and d is the separation between the plates. Thus, E = 7.5 V / 0.1 mm = 75,000 V/m.

The charge on a capacitor is given by Q = CV, where C is the capacitance and V is the potential difference. However, without the capacitance, we cannot calculate the charge directly. More information, like the dielectric constant of the Mylar or the capacitance of the capacitor with air between the plates, would be needed.

Once the Mylar is inserted, if we assume the dielectric is not discharged, the potential difference across the capacitor remains at 7.5 V, since the battery is still connected.

If the Mylar has a dielectric constant, we can calculate the new electric field using the formula E' = E / K, where K is the dielectric constant of the Mylar. But again, without K, we cannot calculate E' directly.

The charge on the capacitor after the Mylar is inserted will be the same as before, provided the capacitor is still connected to the battery, since potential difference and charge are related by Q = CV, and V remains unchanged at 7.5 V.

A) Potential difference before the Mylar is inserted: [tex]\( 7.5 \, \text{V} \).[/tex] B) Electric field before the Mylar is inserted: [tex]\( 7.5 \times 10^4 \, \text{V/m} \)[/tex]. C) Charge before the Mylar is inserted: [tex]\( 1.66 \times 10^{-11} \, \text{C} \)[/tex]. D) Potential difference after the Mylar is inserted: [tex]\( 7.5 \, \text{V} \)[/tex]. E) Electric field after the Mylar is inserted: [tex]\( 7.5 \times 10^4 \, \text{V/m} \).[/tex] F) Charge after the Mylar is inserted: [tex]\( 5.14 \times 10^{-11} \, \text{C} \).[/tex]

To solve this problem, we need to use the concepts of capacitance, electric field, and the effect of a dielectric on a capacitor.

- Electrode area [tex]\( A = 5.0 \, \text{mm} \times 5.0 \, \text{mm} = 25.0 \, \text{mm}^2 = 25.0 \times 10^{-6} \, \text{m}^2 \)[/tex]

- Separation distance [tex]\( d = 0.10 \, \text{mm} = 0.10 \times 10^{-3} \, \text{m} \)[/tex]

- Voltage [tex]\( V = 7.5 \, \text{V} \)[/tex]

- Dielectric constant of Mylar [tex]\( \kappa \approx 3.1 \)[/tex]

A) Potential difference before the Mylar is inserted

The potential difference across the capacitor before the Mylar is inserted is simply the voltage of the battery since the battery is connected directly to the capacitor.

[tex]\[ V_{\text{before}} = 7.5 \, \text{V} \][/tex]

B) Electric field before the Mylar is inserted

The electric field [tex]\( E \)[/tex] in a parallel-plate capacitor is given by:

[tex]\[ E = \frac{V}{d} \][/tex]

Substituting the given values:

[tex]\[ E_{\text{before}} = \frac{7.5 \, \text{V}}{0.10 \times 10^{-3} \, \text{m}} \][/tex]

[tex]\[ E_{\text{before}} = 7.5 \times 10^{4} \, \text{V/m} \][/tex]

C) Charge on the capacitor before the Mylar is inserted

The capacitance [tex]\( C \)[/tex] of a parallel-plate capacitor without a dielectric is given by:

[tex]\[ C = \frac{\epsilon_0 A}{d} \][/tex]

where [tex]\( \epsilon_0 \)[/tex] is the permittivity of free space [tex](\( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \)).[/tex]

[tex]\[ C_{\text{before}} = \frac{(8.85 \times 10^{-12} \, \text{F/m})(25.0 \times 10^{-6} \, \text{m}^2)}{0.10 \times 10^{-3} \, \text{m}} \][/tex]

[tex]\[ C_{\text{before}} \approx 2.21 \times 10^{-12} \, \text{F} \][/tex]

The charge [tex]\( Q \)[/tex] on the capacitor is given by:

[tex]\[ Q = CV \][/tex]

[tex]\[ Q_{\text{before}} = (2.21 \times 10^{-12} \, \text{F})(7.5 \, \text{V}) \][/tex]

[tex]\[ Q_{\text{before}} \approx 1.66 \times 10^{-11} \, \text{C} \][/tex]

D) Potential difference after the Mylar is inserted

When the Mylar is inserted, the potential difference across the capacitor remains the same because the battery is still connected.

[tex]\[ V_{\text{after}} = 7.5 \, \text{V} \][/tex]

E) Electric field after the Mylar is inserted

The electric field [tex]\( E \)[/tex] in a capacitor with a dielectric is given by the same formula:

[tex]\[ E = \frac{V}{d} \][/tex]

Since the voltage and the separation remain the same, the electric field remains the same:

[tex]\[ E_{\text{after}} = \frac{7.5 \, \text{V}}{0.10 \times 10^{-3} \, \text{m}} \][/tex]

[tex]\[ E_{\text{after}} = 7.5 \times 10^{4} \, \text{V/m} \][/tex]

F) Charge on the capacitor after the Mylar is inserted

The capacitance with the dielectric [tex]\( C' \)[/tex] is given by:

[tex]\[ C' = \kappa \cdot C \][/tex]

[tex]\[ C_{\text{after}} = 3.1 \times 2.21 \times 10^{-12} \, \text{F} \][/tex]

[tex]\[ C_{\text{after}} \approx 6.85 \times 10^{-12} \, \text{F} \][/tex]

The charge [tex]\( Q \)[/tex] on the capacitor is given by:

[tex]\[ Q = CV \][/tex]

[tex]\[ Q_{\text{after}} = (6.85 \times 10^{-12} \, \text{F})(7.5 \, \text{V}) \][/tex]

[tex]\[ Q_{\text{after}} \approx 5.14 \times 10^{-11} \, \text{C} \][/tex]

Answer:

The electrostatic potential within the conductor will be negative and constant throughout the conductor.

Explanation:

Whenever a conductor is charged the total charge is spread throughout the entire surface uniformly. Inside the conductor the net electric field is always zero. Also we know if [tex]\overrightarrow{E}[/tex] e the electric field and V is the potential, then

[tex]\overrightarrow{E}= - \overrightarrow{\nabla}V[/tex]

Now as in this case electric field within the conductor is zero, so potential (V) has to be constant.

Final answer:

The electric potential within a charged conductor is constant and uniform, though not necessarily zero, while outside the conductor, the potential is nonzero. This is because the electric field inside a conductor in electrostatic equilibrium is zero, leading to a uniform potential throughout the conductor's inside and surface.

Explanation:

When considering an arbitrarily shaped piece of conductor with a net negative charge in space, we can assert that the electric potential within the conductor is constant throughout its volume. By the principle that in electrostatic equilibrium, the electric field inside a conductor is zero, the potential has the same value in all points inside the conductor. However, this potential value is not necessarily zero; it is referenced to a point far away from the conductor, where the potential is assumed to be zero. Outside of the conductor, the potential is nonzero and governed by the distribution of charge on the conductor's surface and the distance from it.

Considering a perfect conductor, when a test charge is moved from one point to another within or on the surface of the conductor, no work is done, because the electric field is zero. Hence, the electric potential difference between any two points is zero, and all points in and on the conductor share the same potential value with respect to infinity or ground.

Despite the conductor’s potential not being zero in general, its uniformity throughout highlights a fundamental concept in electrostatics: that upon reaching equilibrium, charge distributes on the conductor's surface to ensure the electric field within the conductor is nonexistent, thus establishing a uniform potential.

We don't fall through the floor because of the electrical forces, specifically the electrostatic repulsion between negatively charged electrons surrounding atoms in both our bodies and the floor.

Although solid matter is mostly empty space, we don't fall through the floor because of electrical forces. Atoms consist mostly of empty space, with the volume of protons, neutrons, and electrons comprising less than 1% of an atom's total volume. The reason we perceive matter as solid and do not pass through the floor is that the electrons surrounding atoms are negatively charged, and the same charges repel each other. This repulsive electrostatic force is stronger at the scale we experience than gravitational force, which is comparatively weak.

Answer:

(a) 9.402 meters above the roof

(b) Velocity is 31.193 m/s just before hitting the ground

(c) Horizontal distance traveled is 104.2 meters

Explanation:

Let's first find the horizontal and vertical velocity components of the rock when it is thrown. These are:

[tex]V_x=31.9 * Cos(25.2)[/tex]

[tex]V_y=31.9*Sin(25.2)[/tex]

So we have,

Horizontal velocity = 28.864 m/s

Vertical velocity = 13.582 m/s

Now let's solve the three parts with this information:

(a) To find the maximum height, we need to use the fact that vertical velocity at this point will be zero, as the rock is just about to start falling downward. So we have:

[tex](V_f)^2 - (V_i)^2 = 2*a*s[/tex]

where [tex]V_f[/tex] is the final velocity = 0 m/s

[tex]V_i[/tex] is the initial velocity = 13.582 m/s

and a is the acceleration = -9.81 m/s^2

Solving, we get:

[tex]0^2-13.582^2=2*(-9.81)*s[/tex]

s = 9.402 m (distance above roof)

(b) Using the maximum height from (a), we can solve the following equation:

[tex](V_f)^2 - (V_i)^2 = 2*a*s[/tex]

where [tex]V_f[/tex] is the final velocity we need to find,

[tex]V_i[/tex] is the initial velocity = 0 m/s (from maximum height)

a is the acceleration = 9.81 m/s^2

and s = 14.9 + 9.402 = 24.302 m

Solving, we get:

[tex](V_f)^2 - (0)^2 = 2*(9.81)*(24.302)[/tex]

[tex]V_f = 21.836[/tex] m/s

As this is just the vertical velocity, to find the total velocity we have:

V = [tex]\sqrt{(V_x)^2+(V_y)^2}[/tex]

V = [tex]\sqrt{28.864^2 + 21.836^2}[/tex]

V = 36.193 m/s (Total velocity just before it hits the ground)

(c) To solve for this, we need to know the total time for this projectile motion, we can calculate this as follows:

[tex]s=u*t+\frac{1}{2} (a*t^2)[/tex]

here, s = -14.9 m

u = 13.582 m/s (initial vertical velocity)

a = - 9.81 m/s^2 (acceleration due to gravity)

Solving, we get:

[tex]-14.9 = 13.582t+0.5(-9.81t^2)[/tex]

and get the answers:

t1 = 3.61 s

t2 = -0.84 s

Since t2 isn't possible, our total time is t1 = 3.61 seconds.

Using this and our horizontal velocity, we can find the total distance traveled:

Distance = 28.864 * 3.61

Distance = 104.2 m (horizontal)

Answer: 1.9394m/s

Explanation: m1 (mass of 3 billiard ball) = 0.16kg

U1(initial velocity of the 3 billiard ball) = 4.0m/s

M2 (mass of the cue ball) = 0.17kg

U1 (initial velocity of the cue ball) = 0m/s

Final velocities of both the cue ball and 3 billiard ball after collision = ?

According to the principle of conservation of linear momentum, total momenta before collision = total momenta after collision:

M1U1 + M2U2 = (M1+M2)V

0.16*4.0 + 0.17* 0 = (0.16 + 0.17)V

0.64 + 0 = 0.33V

V = 0.64/0.33 = 1.9394m/s

Answer: 1.25

Explanation:

The ratio of the amplitude of the sound can be calculated using the wave intensity by squaring the amplitude of the two wave particles.

Let the sound level be S2 and S1 and intensities be I1 and I2

Difference in sound level(S) = S2 - S1 = 2dB

Using the relationship between difference in sound level and Intensity

S2 - S1 = 10log(I2 ÷ I1)

Note : Intensity is the square of the amplitude of sound

Therefore,

I1 = A1^2 and I2 = A2^2

Substituting the values into the equation:

S2 - S1 = 10log (I2 ÷ I1)

2dB = 10log (A2^2 ÷ A1^2)

2dB = 10log(A2 ÷ A1)^2

2 dB= 20log(A2 ÷ A1)

2÷20 = log (A2 ÷ A1)

0.1 = log (A2 ÷ A1)

(A2 ÷ A1) = 10^0.1

A2/A1 = 1.25

Answer:

Jupiter

Explanation:

This is because it has the largest mass. Jupiter  is massive and has the highest mass in the solar system.

Gravitational pull is dependent on the mass of body based on the newton's law of gravitation.

I hope this was helpful, please mark as brainliest

Answer:i had this on usatest prep it is jupiter

Explanation:

Answer:

4.38 × 10⁻³ A

0.395mΩ

452 µΩ

Explanation:

(a)the current in each strand is = the total current/ number of strand

0.876/ 200 = 4.38 × 10⁻³ A

(b) potential difference = IR

the total resistance = 2.26 µΩ × 0.876

= 0.395mΩ

c) resistance of the cable =  2.26 µΩ  × 200

=452 µΩ

Answer: Tympanum

Explanation: A Tympanum could be described as the semicircular space above the doorway or window or entrance of a building. The space is usually designed using elaborate sculptural designs, Rocky costumes and ornaments. The tympanum could also be triangular-shaped and attributed to classical Greece and Roman architecture. It is usually located between the portals archivolt and the lintel. The tympanum is commonly observed in the churches and temples of ancient Rome and Greece.

The semicircular space above a doorway in a Romanesque church portal is called a tympanum and it was often decorated with intricate carvings relating to religious themes.

The prominent semicircular space above a doorway in a Romanesque church portal is referred to as a tympanum. These were often covered in elaborate carvings that represented religious imagery or important biblical stories. The Romanesque style of architecture, prevalent in Europe from the 9th to 12th centuries, was known for its massive quality, thick walls, and robust pillars. This style also romanticized elaborate decorative details, among them the beautifully carved tympanum.

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Answer:

B

Explanation:

Because they have a full valence shell they don't need to make bonds to stabilize because they are already stable. These will be your Noble Gasses (ie Neon)

Answer:

A

Explanation:

- The complete question is:

" A crate is lifted vertically 1.5 m and then held at rest. The crate has weight 100 N (i.e., it is  supported by an upward force of 100 N).Now suppose the crate is lifted so rapidly that air resistance was significant during the raising. How much work was done by the lifting force as the box was raised 1.5 m?"

Options:

1. More than 150 J

2. A bit less than 150 J because the air partially supported the crate.

3. No work was done.

4. Still 150 J

5. None of these

Solution:

- As the box is raised the work is done against gravity and air resistance opposes the motion of box. Hence, both the force of gravity ( Weight and air resistance act downward).

- The amount of work done is the sum of both work done against gravity and against air resistance.

-                         W = W_g + W_r

                         W > W_g

                         W > 100*1.5

                         W > 150 J

- Hence, the work done is greater than 150 J i.e work is also done against ai r resistance.  

Answer:

3seconds

Explanation:

Time of flight of an object is the time taken by the object to spend in the air before landing. Mathematically it is represented as;

T = 2u/g

Where g is the acceleration due to gravity = 10m/s²

If the vertical velocity component (u) is given as 15m/s, the time of flight will be;

T = 2(15)/10

T = 30/10

T = 3seconds.

Therefore the total time by the ball in the air is 3seconds

Explanation:

Below is an attachment containing the solution.

The ratio of the work delivered by the turbine to the work consumed by the pump in the steam power plant is 1.

To determine the ratio of the work delivered by the turbine to the work consumed by the pump in the steam power plant, we need to consider the energy conversions that occur in the cycle. The work done by the turbine is equal to the area under the pressure-volume curve in the turbine phase, while the work consumed by the pump is equal to the area under the pressure-volume curve in the pump phase. Since the process is reversible and heat losses are negligible, the amount of work done in the turbine is equal to the amount of work consumed by the pump. Therefore, the ratio of the work delivered by the turbine to the work consumed by the pump is 1.

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