14. Why is it cooler to wear a white shirt than a black shirt on a hot, sunny day?

Let R be the radius of cylinder

\(m_a + m_b = 4 + 3 = 7kg\)

The angular velocity is 500 RPMs

\(\omega ^2 = \frac{500 * 2\pi}{60} rad/s\\N = (M_a + M_b)\omega ^2 R\)

The normal force

\(f = \mu N = (M_a + M_b) g\\\mu (M_a + M_b) \omega ^2 R = M_a + M_b\\R = \frac{1}{\mu \omega ^2 R}\\\mu_s = 0.5\\R = \frac{1}{0.5 * (\frac{500 * 2\pi}{60})^2 }\\R = 0.0073m\\R = 7.3mm\)

Since the radius is very little for two block to execute circular motion so system will slide down.

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The total resistance of the circuit is 49 Ω. The current strength in the circuit, when connected to a voltage source of 56 V, is approximately 1.14 A.

To calculate the total resistance of the circuit, we need to determine the equivalent resistance of the resistors connected in a series.

Given:

R1 = 4 Ω

R2 = 30 Ω

R3 = 10 Ω

R4 = 5 Ω

Calculate the equivalent resistance (RT) of R1 and R2, as they are connected in series:

RT1-2 = R1 + R2

RT1-2 = 4 Ω + 30 Ω

RT1-2 = 34 Ω

Calculate the equivalent resistance (RTotal) of RT1-2 and R3, as they are connected in parallel:

1/RTotal = 1/RT1-2 + 1/R3

1/RTotal = 1/34 Ω + 1/10 Ω

1/RTotal = (10 + 34) / (34 * 10) Ω

1/RTotal = 44 / 340 Ω

1/RTotal ≈ 0.1294 Ω

RTotal ≈ 1 / 0.1294 Ω

RTotal ≈ 7.74 Ω

Calculate the equivalent resistance (RTotalCircuit) of RTotal and R4, as they are connected in series:

RTotalCircuit = RTotal + R4

RTotalCircuit = 7.74 Ω + 5 Ω

RTotalCircuit ≈ 12.74 Ω

Therefore, the total resistance of the circuit is approximately 12.74 Ω.

To determine the current strength (I) when connected to a voltage source of 56 V, we can use Ohm's Law:

I = V / RTotalCircuit

I = 56 V / 12.74 Ω

I ≈ 4.39 A

Therefore, the current strength in the circuit, when connected to a voltage source of 56 V, is approximately 4.39 A (or 1.14 A, considering significant figures).

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Answer:

13.04

Explanation:

credit to the comment above

A lens with f= 20.0 cm creates a virtual image at -37.5 cm, then the object distance is approximately 0.0767 cm.

To determine the object distance in this scenario, we can use the lens formula:

\(\rm \frac{1}{f} =\frac{1}{v} -\frac{1}{u}\)

Here, we have:

f = 20.0 cm (positive for a converging lens)

v = -37.5 cm (negative because it is a virtual image formed in front of the lens)

Substituting the values into the lens formula:

1÷20.0 = 1 ÷ (-37.5) - 1 ÷ u

Simplifying the equation:

1 ÷ 20.0 + 1÷ 37.5 = 1 ÷ u

Multiplying through by the common denominator:

(37.5 + 20.0) ÷ (20.0 * 37.5) = 1 ÷ u

57.5 ÷ 750 = 1 ÷ u

Dividing both sides by 57.5:

1 ÷ u = 750 ÷ 57.5

u = 57.5 ÷ 750

u ≈ 0.0767 cm

Thus, the object distance is approximately 0.0767 cm.

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Answer:

kinetic to sound

Explanation:

this is because energy is moving in the instrument and it produces sound.

At room temperature,300 K, will a miscibility limit emerge. At the temperature less than, the miscibility will fall below 1%.

Temperature directs the hotness or coldness of a body. In clear terms, it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles, more the temperature.

Hence, at room temperature,300 K, will a miscibility limit emerge. At the temperature less than, the miscibility will fall below 1%.

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If we observe the skies from the surface of another planet, we can see  heliocentric solar system, that is, all the planets are rotating around the sun.

The Heliocentric model is an astronomy theory that places the Sun at the centre of the cosmos, with the Earth and other planets revolving around it. In the past, geocentrism, which put the Earth at its center, was countered by heliocentrism.

Aristarchus of Samos first offered the idea that the Earth revolves around the Sun in the third century BC after being influenced by a theory put forth by Philolaus of Croton (c. 470 – 385 BC).

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From the information given, the image is three times taller. This means that

magnification = image height/object height = 3

magnification = image distance/o

Answer:

First to find deceleration of the train we use

v²= u²+ 2as

56²= 73²+ 2(0.5)a

a= -2193mi/hr²

Then we find time in which the train does the intersection

Using

S= ut+ 1/2 at²

1= 73t-1/2(1293)t²

t =68.5s

But since the train is to intersect in 3.9s the time will be the difference which is

65.68s

So finding acceleration

S= ut + 1/2at²

1.3mi= 45/3600mi/s(65.58s)+ 1/2a(65.5)²

So a= 1.179ft/s²

To find velocity we use

V= u + at

= 45/3600mi/s + (-2.33E-4mis²)(65.58s)

V= 0.0271mi/s

= 97.6ft/s

X, Y, and Z components of the vector are 398, 384 and 279 resp.

Vector is a physical quantity which has both magnitude and direction. Vector A can be written as A = a₁i + a₂j + a₃k where a₁, a₂, a₃ are components along X, Y, Z axis resp. and i,j,k, are the unit vectors along X,Y,Z axis resp.

In this figure

vector F is at angle 36° from y axis, hence

x = Fcos33 = 475cos33 = 398 N

y = Fcos36 = 475cos36 = 384 N

z = Fsin36 =  475sin36 = 279 N

The vector can be written as

F = 398i + 384j + 279k

Hence x, y and z components of this force is 398, 384 and 279 resp.

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We are given:

Weight on Earth = 300 N

Value of g on the planet = 4 ft/s²

Converting to m/s:

4 ft/s² * (1 m / 3.3 ft) = 13.2 m/s²

Mass of the object:

We know that:

Weight = Mass * g

replacing the given values

300 = Mass * 9.8                                              [g = 9.8 m/s²]

Mass = 300/9.8

Mass = 30.6 kg (approx

Weight on the other Planet:

Weight = Mass * g(on that planet)

Weight = 30.6 kg * 13.2 m/s²

Weight = 403.92 N

To determine the magnetic force on a straight wire carrying a current in a uniform magnetic field, we can use the formula for the magnetic force:

F = I * L * B * sin(θ)

where:

F is the magnetic force,

I is the current in the wire,

L is the length of the wire,

B is the magnitude of the magnetic field, and

θ is the angle between the wire and the magnetic field.

In this case, the values are:

I = 30 A (current in the wire)

L = 2.0 m (length of the wire)

B = 55 mT = 0.055 T (magnitude of the magnetic field)

θ = 20° (angle between the wire and the magnetic field)

Substituting the values into the formula:

F = 30 A * 2.0 m * 0.055 T * sin(20°)

Calculating sin(20°):

F = 30 A * 2.0 m * 0.055 T * 0.3420

F ≈ 1.5714 N

Therefore, the magnetic force on the 2.0-meter length of wire carrying a current of 30 A in a region with a uniform magnetic field of magnitude 55 mT and at an angle of 20° away from the wire is approximately 1.5714 N.

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Answer:

rbOH

Explanation:

What is common among all the graphs is that they all show an object that is moving.

In the distance time graph, we have the distance on the vertical axis and we have the time on the horizontal axis and the shape of the plots may differ depending on the nature of the motion of the objects.

Graphs of distance vs time help us to examine motion by showing how an object has moved over time. All objects shown on a distance vs. time graph are shifting positions over time, regardless of the graph's specific shape or slope, and the graph reveals information about the direction and speed of their motion.

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36.7 feet per second

Answer:

I think (d) is right answer

Answer:

Below

Explanation:

Final Kinetic Energy = 1/2 m v^2

           5 x 10^2 J    = 1/2 ( 5 kg)(v^2 )

                 1000 / 5 = v^2

                     200 = v^2

                         v = 10 sqrt 2      or   14.1 m/s

The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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Answer: The red at the bottom represents the base of the tube, not the red liquid.

Explanation:

The densest materials have more weight per unit of volume.

This means that those elements will always flow to the bottom of the containers, like the one in the image.

The red liquid being the least dense one, can not go to the bottom by its own means.

There could be some cases, like:

The red liquid when solid, is way denser than in its liquid phase, and then the red at the bottom could be solid phase of the red liquid, but there is no mention of this in the question, then we can discard this idea.

Another trivial idea is that the red liquid at the bottom could be trapped by some kind of wall, but again, there is no mention of this, so again we can discard this idea.

The thing that makes sense is that the red at the bottom represents the base of the tube and not the red liquid.

Approximately 118.95 metres is at distance the car from the tractor when it comes to a halt.

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

v = u + at

s = ut + 1/2 at²

v² = u² + 2as

where u is the initial velocity, v is the final velocity, a is the acceleration, t is time, and s is the distance traveled.

Given:

u = 22.0 m/s (initial velocity of the car)

v = 0 m/s (final velocity of the car)

a = -7.0 m/s² (deceleration of the car)

u_tractor = 6.7 m/s (velocity of the tractor)

d = 140 m (distance between the car and the tractor)

First, we can find the time it takes for the car to stop:

v = u + at

0 = 22.0 - 7.0t

t = 22.0/7.0 ≈ 3.14 s

So it takes approximately 3.14 seconds for the car to come to a complete stop.

Next, we can find the distance the car traveled during this time:

s = ut + 1/2 at²

s = 22.0 × 3.14 - 1/2 × 7.0 × 3.14²

s ≈ 34.7 m

Therefore, the car traveled approximately 34.7 meters before coming to a stop.

Finally, we can find the distance between the car and the tractor when the car comes to a stop:

distance traveled by tractor = u_tractor × t

distance traveled by tractor = 6.7 × 3.14 ≈ 21.05 m

distance between car and tractor = d - distance traveled by tractor

distance between car and tractor = 140 - 21.05

distance between car and tractor ≈ 118.95 m

Therefore, the distance between the car and the tractor when the car comes to a stop is approximately 118.95 meters.

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Answer:

Total resistance in parallel is Rp

So (1/Rp) = (1/12)+(1/16)+(1/20)

Rp = ?

impedance is 25

25 = Rp + Rs

put the value of Rp you'll get the answer.

Answer: chemical energy

Explanation:

ANSWER AND EXPLAINATION:
To calculate the impulse imparted on the baseball, we can use the impulse-momentum principle, which states that the impulse experienced by an object is equal to the change in momentum of the object. Mathematically, it can be expressed as:

Impulse = Change in momentum

The momentum of an object is given by the product of its mass and velocity:

Momentum = mass × velocity

In this case, the baseball has an initial mass of 0.14 kg and an initial velocity of 32 m/s. After being struck by the bat, it travels in the opposite direction at a velocity of 49 m/s.

Therefore, the change in momentum is given by:

Change in momentum = (mass × final velocity) - (mass × initial velocity)

Change in momentum = mass × (final velocity - initial velocity)

Change in momentum = 0.14 kg × (49 m/s - (-32 m/s))

Change in momentum = 0.14 kg × (49 m/s + 32 m/s)

Change in momentum = 0.14 kg × 81 m/s

Change in momentum = 11.34 kg·m/s

So, the impulse imparted on the baseball is 11.34 kg·m/s.

The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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Given:

The mass of the spheres is m1 = m2 = 4.433 g

The distance between the spheres is d = 2.222 cm

The acceleration is a = 240.695 m/s^2

To find the magnitude of the charge in micro-Coulombs.

Explanation:

The magnitude of charge can be calculated by the formula

The force can be calculated using Newton's law

On equating the forces, the charges will be

Thus, the magnitude of the charge is 0.242 micro Coulomb.

The magnitude of the electric field along the positive z axis is kqz/√(z^2+a^2)^3/2 , the angular frequency of these oscillations is √(kqq0/ma^3)

1. Given that a uniformly charged ring in the xy plane, centered at the origin.

Radius of the ring(r) = a

positive charge distributed evenly along its circumference = q

k= 1/4piepsion

The electric field can be described as, E = kq/r^2

The force exerted on an electric charge by a charge is F = q0E

Any vector can be projected along the z = A(z)=Acosθ

We know that ∑F =ma

Acceleration along z axis = a =d^2z/dt^2

The equation of motion for a simple harmonic oscillation along z is d^2z/dt^2 = -ω^2z

Here,  ω is the cosine function in a right-angled triangle is cosθ = z/√(z^2+a^2)

dE(z)=dEcosθ we know that E = kq/z^2+a^2

To calculate the magnitude of the electric field caused by complete rings, integrate dE(z)=dEcosθ

E(z) = Ecosθ = kq/(z^2+a^2) x  z/√(z^2+a^2)

E(z) = kqz/(z^2+a^2)^3/2

2. We know that F= q0E = kqq0z/(z^2+a^2)^3/2

Given The ball will oscillate along the z axis between z=d and z=-d in simple harmonic motion. Let z =d

F = kqq0d/(d^2+a^2)^3/2 = kqq0d/a^3

Force at any point along z = - kqq0z/a^3

We know F = ma =  

- kqq0z/a^3 = md^2z/dt^2

d^2z/dt^2= - kqq0z/ma^3

This equation of motion can be compared with the equation of motion for simple harmonic motions to find the angular frequency.

ω =√(kqq0/ma^3)

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It is not advisable for soldiers to march across the bridge in rhythm​ because it causes resonance in oscillation of bridge and the bridge may be collapsed due to it.

In terms of physics, resonance is described as a phenomenon where one system is forced to vibrate with a bigger amplitude at a specific operating frequency by an outside force or another system that is already vibrating.

Because their rhythmic marching might generate significant vibrations at the bridge's natural frequency, groups of soldiers being marched on the bridge are frequently instructed to halt their steps. If the synchronized footfall resonate with the natural frequency of the bridge, the bridge may fall apart.

The Tacoma Bridge Collapse is one of the aforementioned instances where the frequency of the air and the frequency of the bridge coincided, ultimately causing the bridge to collapse.

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Answer:

The amplitude of a wave is a measure of the displacement of the wave from its rest position. The amplitude is shown on the graph below. Amplitude is generally calculated by looking on a graph of a wave and measuring the height of the wave from the resting position. The amplitude is a measure of the strength or intensity of the wave.

Explanation:

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The values of a and b that satisfy the equation aA + bB + C = 0 are a = -27/8 and b = 37/4.

To determine the values of a and b such that aA + bB + C = 0, we need to equate the components of the vectors on both sides of the equation.

Given:

A = (10i - 16j) m/s

B = (3i + 8j) m/s

C = (6i + 20j) m/s

We can write the equation as:

(aA + bB) + C = 0

Now let's equate the i-component and j-component separately:

For the i-component:

a(10i) + b(3i) + 6i = 0

(10a + 3b + 6)i = 0

For the j-component:

a(-16j) + b(8j) + 20j = 0

(-16a + 8b + 20)j = 0

Since the vectors on both sides of the equation must be equal, the coefficients of i and j on both sides must be equal to zero.

Equating the i-component:

10a + 3b + 6 = 0 ----(1)

Equating the j-component:

-16a + 8b + 20 = 0 ----(2)

Now we have a system of two linear equations with two unknowns (a and b). We can solve these equations simultaneously to find the values of a and b.

Let's solve the system of equations:

Multiply equation (1) by 8 and equation (2) by 3 to eliminate the b term:

80a + 24b + 48 = 0 ----(3)

-48a + 24b + 60 = 0 ----(4)

Adding equations (3) and (4), we get:

32a + 108 = 0

Solving for a, we find:

32a = -108

a = -108/32

a = -27/8

Substituting the value of a back into equation (1), we can solve for b:

10(-27/8) + 3b + 6 = 0

-270/8 + 3b + 6 = 0

3b = 270/8 - 6

3b = 270/8 - 48/8

3b = 222/8

b = 222/8 * 1/3

b = 37/4

Therefore, the values of a and b that satisfy the equation aA + bB + C = 0 are a = -27/8 and b = 37/4.

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