When a sound wave moves past a point in air, are there changes in the density of air at this point?

Answer:

radio waves, microwaves, infrared, visible, UV, X-rays, gamma rays

Explanation:

Electromagnetic waves consist of oscillating electric and magnetic fields which vibrate in a direction perpendicular to the direction of motion of the wave (transverse wave). Electromagnetic waves travel in a vacuum with a speed of [tex]c=3.0\cdot 10^8 m/s[/tex] (speed of light), and they are classified into 7 different types according to their wavelength.

From longer to shorter wavelength, these types are:

Radio waves (wavelength > 30 cm)

Microwaves (30 cm - [tex]5 \mu m[/tex])

Infrared ([tex]5 \mu - 750 nm[/tex])

Visible light (750 nm - 380 nm)

UV radiation (380 nm - 8 nm)

X-rays (8 nm - 6 pm)

Gamam rays (< 6 pm)

Answer;

30.6 m

Explanation;

All objects accelerate at the constant rate in the Earth's gravitational field. The gravitational acceleration, g = 9.8 m/s².

Distance traveled by an object falling down under a constant acceleration will be given the formula;

s = ut² + 1/2(gt²); but u the initial velocity is o

thus;

S =1/2(gt²)

  = 0.5 × 9.81 × 2.5 ²

  = 30.65

  ≈ 30.6 m

The advancements in television and communication technologies during the 1950s and 1960s revolutionized entertainment, news consumption, and long-distance communication by the creation of a shared experience and offering better and cheaper services.

One of the major results of technological advancements like television and satellites revolutionizing life in the 1950s and 1960s was a transformation in modes of communication and entertainment. With the accessibility and popularity of television soaring during this period, there was a shift in the way people consumed news, entertainment and information. In the early 1950's, only around 9 percent of U.S households had a TV which rocketed to approximately 65 percent in just five years. This widespread distribution of television sets played an instrumental role in uniting people through shared experiences whether it was news broadcasts or popular TV shows. A sense of a national, and even global, community was formed through these shared experiences in real-time.

Another crucial impact was the innovations in communication technologies such as microwave transmission and satellites, allowing phone calls to be made in a wireless mode. More affordable and superior quality telephonic communication, especially for long-distance calls, was facilitated. Therefore, the technology of this era not only influenced entertainment and news consumption but also significantly improved communication.

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A) [tex]2.4\cdot 10^{-6} J[/tex]

The energy stored in a capacitor is given by:

[tex]E=\frac{1}{2}\frac{Q^2}{C}[/tex]

where

Q is the charge stored

C is the capacitance

The capacitance of a parallel-plate capacitor is

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

where

[tex]\epsilon_0 = 8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity

[tex]A=6.0 cm \cdot 6.0 cm=36.0 cm^2=36\cdot 10^{-4} m^2[/tex] is the area of each plate

[tex]d=1.5 mm=0.0015 m[/tex] is the distance between the plates

Substituting,

[tex]C=\frac{(8.85\cdot 10^{-12} F/m)(36\cdot 10^{-4} m^2)}{0.0015 m}=2.1\cdot 10^{-11} F[/tex]

The charge stored on the capacitor is

[tex]Q=10 nC=10\cdot 10^{-9}C[/tex]

So, the energy stored is

[tex]E=\frac{1}{2}\frac{(10\cdot 10^{-9}C)^2}{2.1\cdot 10^{-11} F}=2.4\cdot 10^{-6}J[/tex]

B) [tex]2.6\cdot 10^{-6}J[/tex]

This time, the separation between the plates is

d = 1.7 mm = 0.0017 m

So, the new capacitance is

[tex]C=\frac{(8.85\cdot 10^{-12} F/m)(36\cdot 10^{-4} m^2)}{0.0017 m}=1.9\cdot 10^{-11} F[/tex]

And so, the new energy stored is

[tex]E=\frac{1}{2}\frac{(10\cdot 10^{-9}C)^2}{1.9\cdot 10^{-11} F}=2.6\cdot 10^{-6}J[/tex]

C)

Energy must be conserved, so the difference between the initial energy of the capacitor and its final energy is just equal to the work done to increase the separation between the two plates from 1.5 mm to 1.7 mm (in fact, the two plates of the capacitor attract each other since they have opposite charge, so work must be done in order to increase their separation)

Final answer:

The energy stored in the capacitor initially is 2.355 × 10−9 J. After increasing the plate separation, the energy stored becomes 2.674 × 10−9 J. The difference in energy is accounted for by the work done to separate the plates, which adds to the electric potential energy.

Explanation:

Part A: Energy Stored in the Capacitor

The energy stored in a capacitor can be calculated using the formula:

U = ½ QV

where U is the stored energy, Q is the charge, and V is the potential difference across the capacitor plates. To find V, we need to use the capacitance C, which is given by:

C = ε0 (A/d)

where ε0 is the permittivity of free space (ε0 = 8.85 × 10−12 F/m), A is the area of one of the plates, and d is the separation between the plates. Substituting the values, we find:

C = 8.85 × 10−12 F/m · (0.06 m × 0.06 m) / 0.0015 m

C = 2.124 × 10−9 F

The potential difference V is found using:

V = Q/C

V = 10 × 10−9 C / 2.124 × 10−9 F

V = 4.71 V

Substituting back into the energy formula, we get:

U = ½ × 10 × 10−9 C × 4.71 V

U = 2.355 × 10−9 J

Part B: Energy Stored After Increasing Plate Separation

When the plate separation is increased, the capacitance C changes, which affects the stored energy, but the charge Q remains the same as the capacitor was disconnected from the battery. We recalculate C with the new distance:

C' = 8.85 × 10−12 F/m · (0.06 m × 0.06 m) / 0.0017 m

C' = 1.870 × 10−9 F

The stored energy now is:

U' = ½ × 10 × 10−9 C × (10 × 10−9 C / 1.870 × 10−9 F)

U' = 2.674 × 10−9 J

Part C: Accounting for Energy Difference

Since energy must be conserved, the work done to separate the plates must be accounted for. The increase in energy observed is the mechanical work done to pull the plates apart against the attractive force between them. This work is converted into additional electric potential energy stored in the capacitor as it has a larger separation and hence an increased potential difference.

1 mol of oxygen molecules = 2 * 16 = 32 grams.

x mol of oxygen = 16 grams

1/x = 32/16    Cross multiply

16 = 32x        Divide by 32

16/32 = x

x = 1/2 mol

1 mol of anything has 6.02 * 10^23 somethings (molecules in this case).

1/2 mol = 6.02 *10^23 / 2

1/2 mol = 3.01 * 10^23 molecules <<<< answer  

Answer:

KE = 0.16 J

Explanation:

As we know that total energy is always remains conserved

so here we can say that initial potential energy stored in the spring is equal to the kinetic energy of the object when it comes to relaxed state

So here we have

[tex]U = \frac{1}{2}kx^2[/tex]

here we know that

[tex]k = 125 N/m[/tex]

x = 0.05 m

now from above equation

[tex]U = \frac{1}{2}(125)(0.05)^2[/tex]

[tex]U = 0.16 J[/tex]

so total potential energy here is same as the final kinetic energy which is 0.16 J

Answer:

10 m/s

Explanation:

The speed of a wave is given by

[tex]v=\lambda f[/tex]

where

[tex]\lambda[/tex] is the wavelength

f is the frequency

For the wave in this problem,

- The frequency is given by the number of oscillations per second, so

[tex]f=\frac{2 cycles}{1 s}=2 Hz[/tex]

- The wavelength is

[tex]\lambda=5 m[/tex]

So, the wave speed is

[tex]v=(5 m)(2 Hz)=10 m/s[/tex]

The speed of the water wave is calculated using the formula speed = frequency × wavelength. With a frequency of 2 Hz and a wavelength of 5 meters, the wave speed is 10 m/s.

The student's question is about calculating the speed of a water wave based on the given wavelength and the frequency of an object oscillating on the water's surface. To find the wave speed, we can use the formula speed = frequency × wavelength. Given that the object completes 2 cycles per second (which is the frequency), and the wavelength is 5 meters, we simply multiply the two values.

Speed = Frequency × Wavelength = 2 Hz × 5 m = 10 m/s

Therefore, the speed of the water wave is 10 meters per second (m/s).

ok interesting what is you question

Answer:

If an object changes speed or  direction, its velocity also changes. Any change in  velocity results in acceleration.

If an object changes speed or direction , its velocity also changes. Any change in results velocity in acceleration.

What are the term velocity and direction means and related?

Direction and Velocity are the two missing words in the sentence.

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This is chemical energy the substance that makes it glow are chemicals.

Answer:

Chemical .

Explanation:

On Halloween, you take a glow stick, crack the capsule inside and shake it until it glows. This is an example of light energy being created from ___________________ energy

From the principle of the conservation of energy which states that energy can neither be created nor destroyed but can converted from one form to another

Phenyl oxalate ester is responsible for the luminescence in a glow stick.When it reacts with hydrogen peroxide, the liquid inside a glow stick to glow.

therefore we can say that chemical energy is been converted to light energy

the glow lights are by no means radioactive. it can be made to glow again by putting it in a freezer.

I shall draw this out as these tend to be answered best visually

To begin to lift the car, a hydraulic jack must produce more than 9,800N of force.

To keep the car in position once it is lifted, the jack must produce exactly 9,800N of force.

Answer:

To begin to lift the car, a hydraulic jack must produce more than 9,800N of force.

To keep the car in position once it is lifted, the jack must produce exactly 9,800N of force.

To calculate the electromagnetic energy contained in a given space just above the Earth's surface, multiply the intensity of the sunlight by the volume of the space. In this case, the energy is 8.40 × 10^3 W.

The electromagnetic energy contained in a given space can be calculated by multiplying the intensity of the sunlight by the volume of the space.

Given that the intensity of sunlight is 1.40 × 10^3 W/m² and the volume of the space is 6.00 m³, we can calculate the electromagnetic energy as follows:

Energy = Intensity × Volume

Energy = (1.40 × 10^3 W/m²) × (6.00 m³)

Energy = 8.40 × 10^3 W

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➷ A normal atom has the same amount of electrons and protons, making it neutral. An atom develops a positive charge when it loses an electron(s). Once it loses an electron(s), there would now be more protons that electrons.

Short answer: by losing an electron(s)

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Answer:

255.4 N/m

Explanation:

We can consider the system eyeball-attached to the musculature as a mass-spring system in simple harmonic motion, whose frequency of oscillation is given by

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

where in this case, we know:

f = 29 Hz is the frequency of oscillation

k is the spring constant, which is unknown

m = 7.7 g = 0.0077 kg is the mass of the eyeball

Solving the equation for k, we find the spring constant of the musculature attached to the eyeball:

[tex]k=(2\pi f)^2 m=(2 \pi (29 Hz))^2 (0.0077 kg)=255.4 N/m[/tex]

To find the effective spring constant of the musculature attached to the eyeball, use the formula k = (4π²m)/(frequency²)

To determine the effective spring constant of the musculature attached to the eyeball, we can use the formula for the resonant frequency of a mass-spring system:

frequency = 1 / (2π√(m/k))

where m is the mass of the eyeball and k is the spring constant. Rearranging the formula, we can solve for k:

k = (4π2m)/(frequency2)

Plugging in the values, we have m = 7.7 g and frequency = 29 Hz:

k = (4π2(7.7 g))/(29 Hz)2

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Answer:

13 Hz

Explanation:

The frequency of a wave is given by the equation:

[tex]f=\frac{v}{\lambda}[/tex]

where

f is the frequency

v is the speed of the wave

[tex]\lambda[/tex] is the wavelength

In this problem, we have:

v = 5.2 m/s is the speed of the wave

[tex]\lambda=0.40 m[/tex] is the wavelength (distance between two adjacent crests)

Substituting into the formula, we find the frequency of the wave:

[tex]f=\frac{5.2 m/s}{0.40 m}=13 Hz[/tex]

Answer:

D.13 Hz

Explanation:

A

p

e

x

A large elliptical galaxy

Answer:

A Super Massive black hole

Explanation:

Answer:

The Biot-Savart law

Explanation:

Answer:

The Biot-Savart law

Explanation:

Answer:

1. 5.12068 × 1011 N/C away from the proton

Explanation:

The electric field produced by a single point charge is given by:

[tex]E=k\frac{q}{r^2}[/tex]

where

k is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

[tex]q=1.6\cdot 10^{-19}C[/tex] is the charge of the proton

[tex]r=5.3\cdot 10^{-11} m[/tex] is the distance at which we want to calculate the field

[tex]k=8.99\cdot 10^9 Nm^2C^{-2}[/tex] is the Coulomb's constant

Substituting into the formula,

[tex]E=(8.99\cdot 10^9 Nm^2C^{-2})\frac{1.6\cdot 10^{-19}C}{(5.3\cdot 10^{-11}m)^2}=5.12068\cdot 10^{11} N/C[/tex]

And the direction of the electric field produced by a positive charge is away from the charge, so the correct answer is

1. 5.12068 × 1011 N/C away from the proton

The magnitude of the electric field set up by the proton at the position of the electron in a hydrogen atom is approximately 5.14 x 10¹¹ N/C. This is based on the formula for the electric field E = kQ/r². The direction of the electric field is away from the proton.

The electric field created by a charge is given by the formula E = kQ/r², where Q is the charge, r is the distance, and k is the Coulomb constant. In this case the magnitude of the charge of a proton (Q) is +1.602 x 10⁻¹⁹ C, the average distance (r) between a proton and an electron in a hydrogen atom is 5.3 x 10⁻¹¹ m, and the Coulomb constant (k) is 8.99 x 10⁹ N.m²/C².

Using these values in the formula gives E = (8.99 x 10⁹ N.m²/C²)(1.602 x 10⁻¹⁹ C)/(5.3 x 10⁻¹¹ m)² = 5.14 x 10¹¹ N/C. Therefore, the magnitude of the electric field set up by the proton at the position of the electron is approximately 5.14 x 10¹¹ N/C.

As for the direction of the electric field, electric field lines point in the direction that a positive test charge would move if placed in the field. Since the proton has a positive charge, the field lines (and hence the direction of the electric field) point away from the proton. So, the correct answer is ~5.14 x 10¹¹ N/C away from the proton.

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Answer:

[tex]R_3 < R_1 < R_2[/tex]

Explanation:

The resistance of a wire is given by:

[tex]R=\frac{\rho L}{A}[/tex]

where

[tex]\rho[/tex] is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

1) The first wire has length L and cross-sectional area A. So, its resistance is:

[tex]R_1=\frac{\rho L}{A}[/tex]

2) The second wire has length twice the first one: 2L, and same thickness, A. So its resistance is

[tex]R_2=\frac{2\rho L}{A}[/tex]

3) The third wire has length L (as the first one), but twice cross sectional area, 2A. So, its resistance is

[tex]R_3=\frac{\rho L}{2A}[/tex]

By comparing the three expressions, we find

[tex]R_3 < R_1 < R_2[/tex]

So, this is the ranking of the wire from most current (least resistance) to least current (most resistance).

(a) [tex]4.26\cdot 10^5 V/m[/tex]

In a velocity selector, the speed of the beam is related to the magnitude of the electric field and of the magnetic field by the formula:

[tex]v=\frac{E}{B}[/tex]

where

E is the magnitude of the electric field

B is the magnitude of the magnetic field

In this problem, we have

[tex]B=0.115 T[/tex] (magnetic field)

[tex]v=3.7\cdot 10^6 m/s[/tex] (speed of the particles)

Solving the equation for E, we find the electric field:

[tex]E=vB=(3.7\cdot 10^6 m/s)(0.115 T)=4.26\cdot 10^5 V/m[/tex]

(b) 3.2 kV

The relationship between electric field and potential difference between the two plates is:

[tex]V=Ed[/tex]

where, in this problem:

[tex]E=4.26\cdot 10^5 V/m[/tex] is the magnitude of the electric field

[tex]d=0.75 cm=0.0075 m[/tex] is the separation between the plates

Substituting into the equation, we find the potential difference:

[tex]V=(4.26\cdot 10^5 V/m)(0.0075 m)=3195 V=3.2 kV[/tex]

The answer is B

Bimetallic strip is used to create a bimetallic coil for a thermometer which reacts to the heat from a lighter, by uncoiling and then coiling back up when the lighter is removed.

A Bimetallic Coil is formed from two pieces of different metals stuck together lengthwise.

Option A

Explanation:

It is formed with two metal pieces that are stuck together in proper length. It is also called bimetallic strip which is mainly used for converting temperature into mechanical displacement. This coil or strip is consist of two different metal that are "steel and copper" or "steel and brass". With "riveting", welding and brazing the strip in the metals are joined length wise together. The displacement in the sideways of strips is more than smaller length ways expansion. The effect produced is used is basically used in mechanical and electrical devices.

Answer:

[tex]8.4\cdot 10^{-13} N[/tex]

Explanation:

The magnitude of the magnetic force on the proton is given by:

[tex]F=qvB sin \theta[/tex]

where:

[tex]q=1.6\cdot 10^{-19} C[/tex] is the proton charge

[tex]v=4.2\cdot 10^6 m/s[/tex] is the proton velocity

[tex]B=2.5 T[/tex] is the magnetic field

[tex]\theta=30^{\circ}[/tex] is the angle between the direction of v and B

Substituting into the formula, we find

[tex]F=(1.6\cdot 10^{-19}C)(4.2\cdot 10^6 m/s)(2.5 T) sin 30^{\circ}=8.4\cdot 10^{-13} N[/tex]

Answer:

2639 N

Explanation:

The impulse on the baseball is given by:

[tex]I=F \Delta t[/tex] (1)

where F is the force exerted on the ball and [tex]\Delta t[/tex] the contact time between the bat and the ball.

The impulse is also equal to the change in momentum of the ball:

[tex]I=\Delta p = m (v-u)[/tex] (2)

where m is the ball's mass, v is its final velocity, u is its initial velocity.

By using (1) and (2) simultaneously we can write an expression for F, the force exerted on the ball:

[tex]F=\frac{m(v-u)}{\Delta t}[/tex]

where:

m = 0.145 kg

u = 35.0 m/s

v = -56.0 m/s

[tex]\Delta t=5.00 \cdot 10^{-3}s[/tex]

Substituting,

[tex]F=\frac{(0.145 kg)((-56.0 m/s)-35.0 m/s)}{5.00\cdot 10^{-3} s}=-2639 N[/tex]

and the negative sign means that the force is simply in the opposite direction to the ball's initial direction.

option B open system

because in open system energy and mass can escape from the system or can be added to it.

Answer:

[tex]4.7\cdot 10^6 Pa[/tex]

Explanation:

The total force exerted by the man on the chair is equal to his weight:

[tex]F=W=mg[/tex]

where m=95.0 kg is the man's mass and g=9.8 m/s^2. Substituting,

[tex]F=(95.0 kg)(9.8 m/s^2)=931 N[/tex]

Since there are 4 legs, we can assume that the force is equally distributed over the 4 legs; so the force supported by each leg is

[tex]F=\frac{931 N}{4}=232.8 N[/tex]

The radius of each leg is [tex]r=0.400 cm=0.004 m^2[/tex], so the area of each leg is

[tex]A=\pi r^2 = \pi (0.004 m)^2=5\cdot 10^{-5} m^2[/tex]

And the pressure exerted on each leg is equal to the ratio between the force supported by each leg and the area:

[tex]p=\frac{F}{A}=\frac{232.8 N}{5\cdot 10^{-5} m^2}=4.7\cdot 10^6 Pa[/tex]

The pressure on each leg is 4655kNm-2.

The term pressure is defined as the ratio of force per unit area. Firts we have to find the force acting on the chair and the area covered by the force.

The force acting on the ground is the weight of the man and the chair.

F = W = mg = 95.0 kg * 9.8 ms-2 = 931 N.

Since this force is evenly distributed, the force on each leg = 931/4 = 232.75 N

The area of each leg = πr^2 = 3.142 * (0.4 * 10^-2)^2 = 5 * 10^-5 m^2

Pressure = force/area = 232.75 N/ 5 * 10^-5 m^2 = 4655kNm-2

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(a) 392 N/m

Hook's law states that:

[tex]F=k\Delta x[/tex] (1)

where

F is the force exerted on the spring

k is the spring constant

[tex]\Delta x[/tex] is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

[tex]F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N[/tex]

- The stretching of the spring is

[tex]\Delta x=15 cm-10 cm=5 cm=0.05 m[/tex]

Solving eq.(1) for k, we find the spring constant:

[tex]k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m[/tex]

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

[tex]F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N[/tex]

And so, the stretch of the spring is

[tex]\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm[/tex]

And since the initial lenght of the spring is

[tex]x_0 = 10 cm[/tex]

The final length will be

[tex]x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm[/tex]

(a) The spring constant of the spring is 392 N/m

(b) Length of the spring is 17.5 cm

[tex]\texttt{ }[/tex]

Hooke's Law states that the length of a spring is directly proportional to the force acting on the spring.

[tex]\boxed {F = k \times \Delta x}[/tex]

F = Force ( N )

k = Spring Constant ( N/m )

Δx = Extension ( m )

[tex]\texttt{ }[/tex]

The formula for finding Young's Modulus is as follows:

[tex]\boxed {E = \frac{F / A}{\Delta x / x_o}}[/tex]

E = Young's Modulus ( N/m² )

F = Force ( N )

A = Cross-Sectional Area ( m² )

Δx = Extension ( m )

x = Initial Length ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

initial length of spring = Lo = 10 cm

mass of object = m = 2.0 kg

extension of the spring = x = 15 - 10 = 5 cm = 0.05 m

mass of second object = m' = 3.0 kg

Asked:

a. spring constant of the spring = k = ?

b. length of spring = L = ?

Solution:

[tex]F = kx[/tex]

[tex]mg = kx[/tex]

[tex]k = mg \div x[/tex]

[tex]k = 2.0 ( 9.8 ) \div 0.05[/tex]

[tex]\boxed {k = 392 \texttt{ N/m}}[/tex]

[tex]\texttt{ }[/tex]

[tex]F' = kx'[/tex]

[tex]m' g = k x'[/tex]

[tex]x' = ( m' g ) \div k[/tex]

[tex]x' = ( 3.0 (9.8) ) \div 392[/tex]

[tex]x' = 0.075 \texttt{ m} = 7.5 \texttt{ cm}[/tex]

[tex]\texttt{ }[/tex]

[tex]L = Lo + x'[/tex]

[tex]L = 10 + 7.5[/tex]

[tex]\boxed {L = 17.5 \texttt{ cm}}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: College

Subject: Physics

Chapter: Elasticity

A wheelbarrow is a 2nd class lever. Your hands on the handles give the effort. The load inside the wheelbarrow is the resistance. The wheel is the fulcrum. When you lift up the handles, the load goes up in the same direction.

A wheelbarrow exemplifies a second-class lever, where the load is between the fulcrum (wheel's axle) and the input force, enabling the lifting of heavier loads through mechanical advantage.

A wheelbarrow is an example of a second-class lever. In this type of lever, the load is situated between the fulcrum and the effort force. A wheelbarrow enables the user to lift heavier loads more easily due to the arrangement of the forces involved.

A wheelbarrow is an example of a second-class lever. In this type of lever, the output force or load is located between the fulcrum (pivot point) and the input force. The fulcrum in the case of a wheelbarrow is the wheel's axle. This arrangement allows the wheelbarrow to lift heavier loads than one could manually, due to the mechanical advantage achieved by the lever's design. A key characteristic of second-class levers is that both the input and output forces are on the same side of the fulcrum, making tasks like lifting and carrying heavy materials more manageable. This principle underscores the effectiveness of simple machines in amplifying human effort to accomplish tasks more efficiently.

Answer:

The current halves

Explanation:

The relationship between voltage, current and resistance in a circuit is given by Ohm's law:

[tex]V=RI[/tex]

where

V is the voltage

R is the resistance

I is the current

We can rewrite the formula as

[tex]I=\frac{V}{R}[/tex]

we see that I is directly proportional to V and inversely proportional to R.  In this problem, V is held constant while R is doubled:

[tex]R'=2R[/tex]

so, the new current in the circuit will be

[tex]I'=\frac{V}{R'}=\frac{V}{2R}=\frac{1}{2}I[/tex]

So, the current halves.

Answer:

[tex]2.6 ^{\circ}C[/tex]

Explanation:

First of all, we need to calculate the penny's kinetic energy before hitting the ground. This is given by

[tex]K=\frac{1}{2}mv^2[/tex]

where m = 5.25 g = 0.00525 kg is the penny's mass and v = 3.27 m/s is its speed. Substituting,

[tex]K=\frac{1}{2}(0.00525 kg)(3.27 m/s)^2=0.028 J[/tex]

When the penny hits the ground, all this energy is converted into internal energy of the penny and the ground. If only half is converted into penny's internal energy, its increase in internal energy is

[tex]\Delta U= \frac{0.028 J}{2}=0.014 J[/tex]

And its formula is

[tex]\Delta U=m C_s \Delta T[/tex]

where m is the penny's mass, Cs its specific heat capacity (2.03 J/gC) and [tex]\Delta T[/tex] the increase in temperature. Solving for the last term,

[tex]\Delta T=\frac{\Delta U}{m C_s}=\frac{0.028 J}{(0.00525 kg)(2.03 J/gC)}=2.6 ^{\circ}C[/tex]