An automobile tire having a temperature of 6.7 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 25 lb/in2 . What would be the gauge pressure in the tire when its temperature rises to 33◦C? For simplicity, assume that the volume of the tire remains constant, that the air does not leak out and that the atmospheric pressure remains constant at 14.7 lb/in2 . Answer in units of lb/in2 .

Answer:

636.619772368 A

Explanation:

[tex]\tau[/tex] = Torque = [tex]1\times 10^{-3}\ N/m[/tex]

B = Magnetic field of Earth = [tex]5\times 10^{-5}\ T[/tex]

A = Area

d = Diameter = 20 cm

Current is given by

[tex]I=\dfrac{\tau}{BA}\\\Rightarrow I=\dfrac{1\times 10^{-3}}{5\times 10^{-5}\times \dfrac{\pi}{4}\times 0.2^2}\\\Rightarrow I=636.619772368\ A[/tex]

The current is 636.619772368 A

The amplitude of the induced emf can be calculated by first finding the magnetic field within the solenoid using the formula B = µonI, calculating the magnetic flux through the coil, and then applying Faraday's law to find the time derivative of the flux. This result is then multiplied by the number of turns in the coil.

The amplitude of the electromagnetic force (emf) induced in a coil located inside a solenoid can be calculated using Faraday's law. The first step in solving this problem involves obtaining the magnetic field B of the solenoid by using the formula B = µonI, where µo is the vacuum permeability, n represents the number of turns per unit length of the solenoid, and I is the amplitude of the current. Here, n can be obtained by converting the given number of turns per cm into turns per meter. This magnetic field B is then multiplied by the area of the coil to obtain the magnetic flux Φ through the coil.

Next, using Faraday's law, we take the time derivative of the obtained magnetic flux. Since the only variable in time here is the current I, we apply the derivative to it, with the rest of the quantities being treated as constant. Lastly, we multiply the result by the number of turns in the coil to determine the induced emf. Remember that the magnitude of the emf is the peak or 'amplitude' of the sinusoidal function associated with the time-varying current I.

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Answer:

a) final angular speed w2 = 414rpm

b) time to stop t = 312.5s

c) angular distance from start to stop s = s = 1171.88 rev

Explanation:

Average angular speed wa = (w1+w2)/2 = Angular distance/time = s/t

Given;

w1 = 450 rpm = 450/60 rev/s = 7.5 rev/s

s = 180 rev

t = 25s

a)

s/t = (w1+w2)/2

w2 = 2s/t - w1

Substituting the values;

w2 = 2×180/25 - 7.5 = 6.9 rev/s

w2 = 6.9×60rpm = 414rpm

b)

Angular deceleration a = (w1-w2)/t = (7.5-6.9)/25

a = 0.024 rev/s^2

time t for w2 = 0

t = (w1-w2)/a = (7.5-0)/0.024

t = 312.5 s

c)

Angular distance from start to stop s;

s = w1t - 0.5at^2

s = 7.5×312.5s - 0.5×0.024×312.5^2

s = 1171.88 rev

Answer: the nasal conchae also known as the turbinates

Explanation:

Answer:

15 miles

Explanation:

6 miles per hour

2 1/2 hours

6 x 2 = 12

6 x 1/2 = 3

12 + 3 = 15

The 40 kg skater moves approximately 6.19 meters towards the 65 kg skater when they pull themselves along a 10 m pole on ice, utilizing the conservation of momentum and considering the center of mass of the system.

To solve how far the 40 kg skater moves when two skaters pull themselves along a pole on ice, we use the principle of conservation of momentum. This principle states that in the absence of external forces, the total momentum of a system remains constant. Since both skaters start from rest, their initial total momentum is zero. As they pull towards each other, their momenta are equal in magnitude and opposite in direction, keeping the total momentum of the system at zero.

Let's denote the distance the 65 kg skater moves as d65 and the distance the 40 kg skater moves as d40. Because the pole has a length of 10 m, these two distances must add up to 10 m, so d65 + d40 = 10. To find out how far each skater moves, we look at their center of mass. The system's center of mass doesn't change because there are no external forces acting on the system.

The position of the center of mass xCM can be found using the formula xCM = (m1 * x1 + m2 * x2) / (m1 + m2), where m1 and m2 are the masses of the skaters, and x1 and x2 are their respective positions. Considering the system initially (before movement), the center of mass is at the midpoint of the pole as the skaters are at the ends. Therefore, xCM = 5 m.

Because the total mass of the system is 105 kg and the center of mass is 5 m from the starting point of either skater, we find that the 40 kg skater moves closer to the center of mass than the 65 kg skater to maintain the center of mass position. By directly calculating it, d40 = 10 * (65 / 105) = 6.19 m. Thus, the 40 kg skater moves approximately 6.19 meters towards the 65 kg skater.

Answer:

La energía es una cualidad o capacidad que tienen los cuerpos materiales y posee diversas funciones. Hay distintos tipos de energía, pero ésta permite que un cuerpo se mueva por ejemplo, cambie de temperatura o de estado.

Explanation:

La Energía de los cuerpos puede ser definida de distintas maneras según la ciencia en la cual estemos trabajando. Para la física, la energía es la capacidad para realizar un trabajo. Esa energía puede ser cinética, térmica, química, y cada una de ellas define distintas propiedades de los cuerpos, que hacen que puedan cambiar de posición, temperatura, estado.

Answer:

0.635 Ω

Explanation:

Using

E-E' = IR.................... Equation 1

Where E = total Emf of the of the batteries, E' = back emf of the motor, I = current of the motor, R = combined resistance of the battery and the motor.

Make R the subject of the equation

R = E-E'/I.............. Equation 2

Given: E = 6.3 V, E' = 2.1 V, I = 3.1 A.

Substitute into equation 2

R = (6.3-2.1)/3.1

R = 4.2/3.1

R = 1.355 Ω

Since the motor and the batteries are connected in series,

R = R'+r'....................... Equation 3

Where R' = Resistance of the motor, r' = resistance of the batteries.

make R' the subject of the equation

R' = R-r'...................... Equation 4

Given: R = 1.355 Ω, r' = 0.18×4 (four batteries) = 0.72 Ω

Substitute into equation 4

R' = 1.355-0.72

R' = 0.635 Ω

Hence the resistance of the motor = 0.635 Ω

To find the motor's resistance, subtract the back emf from the total emf to get the effective voltage, then divide by the current and subtract the total internal resistance of the batteries. The resistance of the motor is found to be 1.35 Ω.

The question involves a toy car motor that is powered by four batteries producing a total electromotive force (emf) of 6.3 V, with each battery having an internal resistance of 0.18 Ω. Upon running, the motor develops a 2.1 V back emf and draws a current of 3.1 A. To find the resistance of the motor, one can use the net emf equation and Ohm's law.

The total emf minus the back emf is the effective voltage across the motor. Thus, the effective voltage (Veff) is 6.3 V - 2.1 V = 4.2 V. The total internal resistance (Rint) in series is 4 batteries × 0.18 Ω = 0.72 Ω. Applying Ohm's law, V = IR where I is the current and R is the resistance, we solve for the resistance (Rmotor) of the motor:

Veff = I(Rmotor + Rint)
Rmotor = Veff/I - Rint
Rmotor = 4.2 V / 3.1 A - 0.72 Ω
Rmotor = 1.35 Ω

Therefore, the resistance of the motor is 1.35 Ω.

Answer:

there pieces of leftover comets :)

Explanation:

Answer:

Technician B is correct

Explanation:

Electrical Switches makes or brakes the flow of signal (current or voltage) in a circuit.

The Technician B that says the switch contacts are closed is correct because it is until when the circuit is continues that voltage flows across the switch and voltmeter will read the amount of voltage present in the circuit. This simply means closed circuit.

Technician A is not correct because when the switch contacts are open, voltage does not flow and no reading will be measured.

Answer:

Technician B is correct

Explanation:

When a circuit is open there is no energy that allows the flux of charge in the cable. Voltage determines the capability of doing a work over charges to generate a charge flux. An open circuit is not able to generate a current in the circuit and concequently, ther is no voltage in the circuit.

Hence, a measurement of 12V in a circuit means that the circuit is closed, store energy and can generate a flux of charge.

Technician B is correct.

Answer:

they are both equal

Explanation:

check the laws of reflection

Answer:

1.contact force

2. Non contact

Explanation:

1. Because bodies are in contact

2.Bodies are not in contact.

Answer:

1. contact

2. non contact

Explanation:

just did it

Answer:

A. The one with the circular cross section will stretch more.

Explanation:

According to the given data:

Two rods are made of brass and have the same length

Both rods having circular and square cross-section

Diameter of circular cross-section given is 2 a

therefore, Cross-section = [tex]A_c=\frac{\pi (2a)^2}{4}=\pi a^2[/tex]

If the length of square=2 a

then, Cross-section = [tex]A_{s}[/tex] = (2a)²=>4a²

Change in Length of rod = PL / AE

δL[tex]\alpha \frac{1}{A}[/tex]

Now, we are considering other factors same

the area of cross-section of square rod is more than Area of cross-section of circular rod

thus, the one with the circular cross section will stretch more

Answer:

The correct option is;

A. The one with the circular cross section will stretch more.

Explanation:

Here we have the cross section as being

1. Circular, with diameter, D = 2·a

2. Square cross section with each side  length = 2·a

The area of the circular rod is then

Area of circle = π·D²/4 which is equal to

π×(2·a)²/4 = π·4·a²/4 = π·a²

The area of the rod with square cross section is

Area of square = Side² which gives

Area of cross section = (2·a)² = 4·a²

Therefore, since π = 3.142, the cross sectional area of the circular rod is less than that of the one with a square cross section

That is,  π·a² = 3.142·a² < 4·a²

We note that the elongation or extension is directly proportional to the force applied as shown as follows

[tex]\frac{P}{A} = E\frac{\delta}{L}[/tex]

Where:

P/A = Force and

δ = Extension

The force is inversely proportional to the area, therefore a rod with less cross sectional area experiences more force and more elongation.

Answer:

Part(a): The expression for the velocity of the billiard ball is [tex]\bf{v_{2f} = \dfrac{2m_{1}v}{m_{1}+m_{2}}}[/tex]

Part(b): The value of the velocity of the billiard ball is [tex]\bf{5.57~m/s}[/tex].

Part(c): The expression for the velocity of the cue ball is [tex]\bf{v_{1f} = \dfrac{(m_{1} - m_{2})v}{m_{1}+m_{2}}}[/tex]

Part(d): The value of the velocity of the cue ball is [tex]\bf{1.93~m/s}[/tex].

Explanation:

Given:

The mass of the cue ball, [tex]m_{1} = 0.34~kg[/tex].

The mass of the billiard ball, [tex]m_{2} = 0.575~kg[/tex].

The initial velocity of the cue ball, [tex]u_{1} = 7.5~m/s[/tex]

The initial velocity of the billiard ball, [tex]u_{2} = 0[/tex]

(a)

Consider the final velocity of the cue ball be [tex]v_{1}[/tex] and the final velocity of the billiard ball be [tex]v_{2}[/tex].

From the conservation of linear momentum , we can write

[tex]~~~~&& m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\\&or,& m_{1}(u_{1} - v_{1}) = m_{2}(v_{2} - u_{2})~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]

From the conservation of energy, we can write

[tex]~~~&& \dfrac{1}{2}m_{1}u_{1}^{2} + \dfrac{1}{2}m_{2}u_{2}^{2} = \dfrac{1}{2}m_{1}v_{1}^{2} + \dfrac{1}{2}m_{2}v_{2}^{2}\\&or,& \dfrac{1}{2}m_{1}(u_{1}^{2} - v_{1}^{2}) = \dfrac{1}{2}m_{2}(v_{2}^{2} - u_{2}^{2})~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]

Dividing equation (1) by equation (2), we have

[tex]~~~&& \dfrac{m_{1}(u_{1} - v_{1}) }{m_{1}(u_{1}^{2} - v_{1}^{2})} = \dfrac{m_{2}(v_{2} - u_{2})}{m_{2}(v_{2}^{2} - u_{2}^{2})}\\&or,& u_{1} + v_{1} = u_{2} + v_{2}~~~~~~~~~~~~~~~~~~~~~~~~(3)[/tex]

Rearranging equation (3) for [tex]v_{1}[/tex], we have

[tex]v_{1} = u_{2} + v_{2} - u_{1}~~~~~~~~~~~~~~~~~~~~(4)[/tex]

Substitute equation (4) in equation (1), we can write

[tex]v_{2} = \dfrac{2m_{1}u_{1}}{m_{1}+m_{2}} + \dfrac{m_{2} - m_{1}}{m_{1} + m_{2}}u_{2}~~~~~~~~~~~~~~(5)[/tex]

Substituting [tex]0[/tex] for [tex]u_{2}[/tex], [tex]v[/tex] for [tex]u_{1}[/tex] and [tex]v_{2f}[/tex] for [tex]v_{2}[/tex] in equation (5), we have

[tex]v_{2f} = \dfrac{2m_{1}v}{m_{1}+m_{2}}~~~~~~~~~~~~~~~~~~~~~~~~(6)[/tex]

(b)

Substituting [tex]0.34~kg[/tex] for [tex]m_{1}[/tex], [tex]7.5~m/s[/tex] for [tex]v[/tex] and [tex]0.575~kg[/tex] for [tex]m_{2}[/tex] in equation (6), we have

[tex]v_{2f} &=& \dfrac{2(0.34~kg)(7.5~m/s)}{(0.34 + 0.575)~kg}\\~~~~~&=& 5.57~m/s[/tex]

(c)

Rearranging equation(3) for [tex]v_{2}[/tex], we have

[tex]v_{2} = u_{1} + v_{1} – u_{2}~~~~~~~~~~~~~~~~~~~~(7)[/tex]

Substitute equation (7) in equation (1), we can write

[tex]v_{1} = \dfrac{2m_{2}u_{2}}{m_{1}+m_{2}} + \dfrac{m_{1} - m_{2}}{m_{1} + m_{2}}u_{2}~~~~~~~~~~~~~~(8)[/tex]

Substituting [tex]0[/tex] for [tex]u_{2}[/tex], [tex]v[/tex] for [tex]u_{1}[/tex] and [tex]v_{1f}[/tex] for [tex]v_{1}[/tex] in equation (8), we have

[tex]v_{1f} = \dfrac{(m_{1} - m_{2})v}{m_{1}+m_{2}}~~~~~~~~~~~~~~~~~~~~~~~~(9)[/tex]

(d)

Substituting [tex]0.34~kg[/tex] for [tex]m_{1}[/tex], [tex]7.5~m/s[/tex] for [tex]v[/tex] and [tex]0.575~kg[/tex] for [tex]m_{2}[/tex] in equation (9), we have

[tex]v_{1f} &=& \dfrac{(0.34 - 0.575)~kg(7.5~m/s)}{(0.34 + 0.575)~kg}\\~~~~~&=& -1.93~m/s[/tex]

Negative sign indicates that the cue ball will bounce back.

The expressions for the horizontal component of the billiard ball's velocity after the collision and the cue ball's velocity after the collision are given. The velocities can be calculated using the provided equations and given values.

a) The expression for the horizontal component of the billiard ball's velocity, vr after the collision is given by:

vr = (m1 - m2) * (v1 / (m1 + m2))

b) Substituting the given values into the equation, we find that the horizontal component of the billiard ball's velocity after the collision is approximately 4.02 m/s.

c) The expression for the horizontal component of the cue ball's velocity, vr after the collision is given by:

vr = (2 * m2 * v1) / (m1 + m2)

d) Substituting the given values into the equation, we find that the horizontal component of the cue ball's final velocity is approximately 2.48 m/s.

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Answer:

w = I₂ / (I₁ -I₂) w₀ ,    L₂ = 2 L₁

Explanation:

This is an angular momentum exercise,

      L = I w

where bold indicates vectors

We must define the system as formed by the bicycle wheel, the platform, we create a reference system with the positive sign up

At the initial moment the wheel is turning and the platform is without rotation

The initial angle moment is

                 Lo = L₂ = I₂ w₀

L₁ is the angular momentum of the platform and L₂ is the angular momentum of the wheel.

In the Final moment, when the wheel was turned,

                [tex]L_{f}[/tex] = L₁ - L₂

                L_{f} = (I₁ - I₂) w

the negative sign of the angular momentum of the wheel is because it is going downwards since the two go with the same angular velocity

as all the force are internal, and there is no friction the angular momentum is conserved,

             L₀ =L_{f}

             I₂ w₀ = (I₁ -I₂) w

             w = I₂ / (I₁ -I₂) w₀

we can see that the system will complete more slowly

 we can also equalize the angular cognition equations

                 L₀ = Lf

                 L₁ = L₂-L₁

                 L₂ = 2 L₁

In this part we can see that the change in the angular momentum of the platform is twice the change in the angular momentum of the wheel.

This experiment with a rotating bicycle wheel demonstrates the conservation of angular momentum. Rotating the wheel changes the platform's rotational direction to maintain a balanced angular momentum. Returning the wheel to its initial position stops the platform's rotation, preserving the system's angular momentum at zero.

In this experiment, we examine the principles of conservation of angular momentum using a rotating bicycle wheel. When you and a platform are stationary, and a colleague spins the bicycle wheel, the system's initial angular momentum is zero.

Here’s a step-by-step explanation:

This experiment vividly shows how changing the direction of angular momentum through torque affects rotational motion while conserving the overall angular momentum of the system.

Answer:

Explanation:

Angular momentum is the product of inertial and angular frequency

L = I × ω

Where

L is angular momentum

I is inertia

ω is angular frequency

So, given that

Vinyl record has a massz

M = 0.115kg

Radius R = 0.0896m

Angular velocity of vinyl record

ω(initial) = 5.58 rad/s

Rotational inertial of vinyl record.

I(initial) = 4.84 × 10^-4 kgm²

Putty drop on the record

Mass of putty M' = 0.0484kg

Angular speed after putty drop ω'

Using conversation of angular momentum

Initial angular momentum is equal to final angular momentum

I(initial) × ω(initial) = I(final) × ω(final)

So, we need to find I(final)

Inertia log putty can be determine using MR² by assuming a thin loop

I(putty) = M'R² = 0.0484 × 0.0896

I(putty) = 3.89 × 10^-4 kgm²

I(final) = I(initail) + I(putty)

I(final) = 4.84 × 10^-4 + 3.89 × 10^-4

I(final) = 8.73 × 10^-4 kgm²

Therefore,

I(initial) × ω(initial) = I(final) × ω(final)

ω(final) = I(initial) × ω(initial) / I(final)

ω(final) = 4.84 × 10^-4 × 5.58 / 8.73 × 10^-4

ω(final) = 3.1 rad/s

Answer:

[tex]v_0=17.3m/s[/tex]

Explanation:

In this problem we have three important moments; the instant in which the ball is released (1), the instant in which the ball starts to fly freely (2) and the instant in which has its maximum height (3). From the conservation of mechanical energy, the total energy in each moment has to be the same. In (1), it is only elastic potential energy; in (2) and (3) are both gravitational potential energy and kinetic energy. Writing this and substituting by known values, we obtain:

[tex]E_1=E_2=E_3\\\\U_e_1=U_g_2+K_2=U_g_3+K_3\\\\\frac{1}{2}kd^2=mg(d\sin\theta)+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}m(v_0\cos\theta)^2[/tex]

Since we only care about the velocity [tex]v_0[/tex], we can keep only the second and third parts of the equation and solve:

[tex]mgd\sin\theta+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}mv_0^2\cos^2\theta\\\\\frac{1}{2}mv_0^2(1-\cos^2\theta)=mg(h-d\sin\theta)\\\\v_0=\sqrt{\frac{2g(h-d\sin\theta)}{1-\cos^2\theta}}\\\\v_0=\sqrt{\frac{2(9.8m/s^2)(4.4m-(0.21m)\sin32\°)}{1-\cos^232\°}}\\\\v_0=17.3m/s[/tex]

So, the speed of the ball just after the launch is 17.3m/s.

Final answer:

A car with tires of radius 0.280 m decelerating at 7.00 m/s² has an angular acceleration of 25 rad/s². To find the number of revolutions before stopping, kinematic equations for rotational motion are used with an initial angular velocity of 95.0 rad/s.

Explanation:

A car decelerating at 7.00 m/s² with tires of radius 0.280 m requires an understanding of the relationship between linear acceleration and angular acceleration. The formula linking these two is α = a/r, where α is the angular acceleration, a is the linear acceleration, and r is the radius of the tire. Using these values, the angular acceleration is calculated to be 25 rad/s².

For part (b), to find out how many revolutions the tires make before coming to rest, we need to use kinematic equations for rotational motion. Initially, we have an angular velocity (ω) of 95.0 rad/s and we want to find the total angle (θ) covered by the tires. Using the equation θ = (ω0²)/(2*α), we calculate the angle in radians and then convert this to revolutions by dividing by 2π. After finding the revolutions, the car's tires will have stopped rotating completely.

The voltage drop across each wire in the extension cord with a resistance of 0.04 ohms and a current of 15 A flowing through it is calculated using Ohm's Law (V = I x R) to be 0.6 volts.

To calculate the voltage drop across each wire in the extension cord, we can apply Ohm's Law, which states that voltage (V) is the product of current (I) and resistance (R). Given that each wire has a resistance of 0.04 Ω and a current of 15 A is flowing through it, we can calculate the voltage drop by multiplying these two figures:

V = I × R

V = 15 A × 0.04 Ω = 0.6 V

Therefore, the voltage drop across each wire in the extension cord is 0.6 volts.

Answer:

1) not so long (maybe an hour or two)

2) access to information through the internet will be most affected if my computer and mobile phone run out of battery power.

3) yes, one should prepare for power outage. This can be done by having a standby alternative source of power like the use of inverters that stores electrical energy in form of chemical energy, and small internal combustion engine powered electric generators.

4) solar panels can be used to draw power from incident sun rays, this power can be stored in an inverter for future use in case of a power outage.

5) energy from the sun is converted into direct current which is then supplied to an accumulator in the opposite direction to its flow of current. When the energy is needed, it can be used directly, or converted to an alternating current. This is achieved by connecting its terminal to the supply. Electric field is generated by flow of ions and electrons within the working chemical (e.g lithium).

Explanation:

Answer:

If there was a power outage, you could survive normally using older ways of life and with proper preparations; some ways to prepare is having canned foods that don't require heating or refrigeration, since you wouldn't have electronic means of preparing it, and supplies for starting camp fires. Some ways your life would be affected is lack of simple methods of day-to-day tasks that you have become accustomed to: no warm water, to refrigerators, or heating systems such as electronic kettles, microwaves or ovens. A backup system you and/or your family/community could install would be solar panels, to collect power during the day to then be used at night or when needed. Solar panels collect solar energy by catching the sun-rays in the cells of the panels and turning the energy into a direct current (DC), then the panels convert the direct current into usable alternating current (AC) energy with the help of inverter technology. That current is then distributed throughout the building accordingly.

Final answer:

The primary difference between gravitational and electric forces is that electric forces can be both attractive and repulsive, whereas gravitational forces are always attractive. Electric forces dominate on small scales, such as within atoms, while gravitational forces are significant on large scales, like between planets.

Explanation:

Main Differences Between Gravitational and Electric Forces:

A main difference between gravitational and electric forces is that electric forces can be either attractive or repulsive, while gravitational forces are always attractive. Gravitational force is proportional to the masses of the interacting objects and always pulls them together, which is important in interactions between large objects like planets and stars. Conversely, electric force is proportional to the charges of interacting objects and plays a crucial role at the microscopic level within atoms and molecules, often keeping them together. It is much stronger than gravity in most systems where both appear.

Another significant difference is that on a small scale, like that of electrons or protons, the electric force is dominant and significantly exceeds the gravitational force. In contrast, on a large scale, gravitational force typically predominates since most objects tend to be electrically neutral; thus, the net Coulomb forces cancel out.

To summarize, gravitational forces are crucial at cosmological scales, handling the structure of galaxies and the dynamics of celestial bodies, whereas electric forces are on a much smaller scale but are instrumental in the binding of atoms, the functioning of everyday materials, and biological processes.

Answer:

energy of activation is the correct answer.

Explanation:

The activation energy is the energy needed to initiate an exergonic reaction. It is required to kickstart the process, despite the reaction being one that eventually releases energy.

The energy required to initiate an exergonic reaction is called the activation energy. This is the minimum amount of energy required for a chemical reaction to occur. In an exergonic reaction, the activation energy is critical even though these reactions release energy overall, because it's needed to start the process. To envision this, think of pushing a boulder off a hill; the initial push (activation energy) is required for the action (the exergonic reaction) to begin, after which gravity (the reaction's natural tendency) takes over and the boulder rolls down (energy is released).

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Answer:

she used force to swing her ball all way back

Explanation:

Final answer:

Shannon swinging her racket to hit a tennis ball illustrates key physics concepts, including momentum and energy transfer, crucial for propelling the ball forward. The energy transferred from the racket to the ball depends on the racket's motion, enhancing the ball's speed when the racket moves toward it. Physics principles, such as Newtonian mechanics, are essential for understanding and improving tennis skills.

Explanation:

When Shannon swings her racket to hit a tennis ball, sending it flying in the same direction, several principles of physics come into play, particularly related to momentum and energy transfer. First, when the racket contacts the ball, momentum is transferred from the racket to the ball, propelling it forward. According to Newton's third law, for every action, there is an equal and opposite reaction. Therefore, as the racket exerts force on the ball, the ball exerts an equal and opposite force on the racket, though the effects are more noticeable on the ball due to its much smaller mass.

Energy transfer during the collision between the racket and the ball is significant. A moving racket transfers kinetic energy to the stationary ball, which then moves with a velocity dependent on the imparted energy. If the racket is moving towards the ball, the speed of the ball post-collision is greatly increased. This results from the addition of the racket's velocity to that imparted to the ball, showcasing an efficient energy transfer.

Understanding the dynamics of hitting a tennis ball, including the effects of the racket's speed, angle of swing, and point of impact, requires knowledge of Newtonian mechanics and is essential for improving one's tennis playing skills. Moreover, accounting for the arc of the ball and its spin are crucial for predicting how the ball will move, enhancing the player's ability to plan their shots strategically.

Answer:

Explanation:

Given that,

Assume number of turn is

N= 1

Radius of coil is.

r = 5cm = 0.05m

Then, Area of the surface is given as

A = πr² = π × 0.05²

A = 7.85 × 10^-3 m²

Resistance of

R = 0.20 Ω

The magnetic field is a function of time

B = 0.50exp(-20t) T

Magnitude of induce current at

t = 2s

We need to find the induced emf

This induced voltage, ε can be quantified by:

ε = −NdΦ/dt

Φ = BAcosθ, but θ = 90°, they are perpendicular

So, Φ = BA

ε = −NdΦ/dt = −N d(BA) / dt

A is a constant

ε = −NA dB/dt

Then, B = 0.50exp(-20t)

So, dB/dt = 0.5 × -20 exp(-20t)

dB/dt = -10exp(-20t)

So,

ε = −NA dB/dt

ε = −NA × -10exp(-20t)

ε = 10 × NA exp(-20t)

Now from ohms law, ε = iR

So, I = ε / R

I = 10 × NA exp(-20t) / R

Substituting the values given

I = 10×1× 7.85 ×10^-3×exp(-20×2)/0.2

I = 1.67 × 10^-18 A

Answer:

 The neutron core is completely destroyed

Explanation:

 A earth - supernova is an explosion resulting to the death of a star that occurs close enough to the earth but this does not completely destroy a star. Supernovae are the most violent explosions in the universe. But they do not explode like a bomb explodes, blowing away every bit of the original bomb. Rather, when a star explodes into a supernova, its core survives. The reason for this is that the explosion is caused by a gravitational rebound effect and not by a chemical reaction. Stars are so large that the gravitational forces holding them together are strong enough to keep the nuclear reactions from blowing them apart. It is the gravitational rebound that blows apart a star in a supernova.

Final answer:

The incorrect statement is that a supernova explosion results in the complete destruction of the neutron core; often, the core becomes a neutron star. Supernovae can be life-threatening to nearby planets, but the actual danger zone is closer than a few dozen light-years. They are also responsible for creating new elements heavier than iron.

Explanation:

A supernova explosion is a cataclysmic event at the end of a massive star's life cycle. During a supernova, a number of phenomena occur, including the ejection of elements the star fused during its lifetime, and the creation of new elements heavier than iron. One of the statements provided as a result of supernova explosions is incorrect: 'The neutron core is completely destroyed'. In many cases, the core does not get completely destroyed; instead, it often collapses into a densely packed neutron star. Another result is a potential threat to life on planets within a few dozen light-years due to a life-threatening dose of radiation. It is important to note, though, that planets would need to be quite close to the supernova, much closer than a few dozen light-years, to receive such a dose. Finally, supernovae also contribute to the formation of new elements, with the violent energies enabling the fusion of atoms into elements heavier than iron.

Answer:

The magnetic field is [tex]2 \times 10^{-4}[/tex] T

Explanation:

Given:

Velocity of electron [tex]v = 5 \times 10^{7} \frac{m}{s}[/tex]

Electric field [tex]E = 10^{4} \frac{V}{m}[/tex]

The force on electron in magnetic field is given by,

 [tex]F = qvB \sin \theta[/tex]                      ......(1)

The force on electron in electric field is given by,

 [tex]F = qE[/tex]                               ......(2)

Compare both equation,

   [tex]qE = qvB \sin \theta[/tex]

Here [tex]\sin \theta = 1[/tex]

  [tex]E = vB[/tex]

  [tex]B= \frac{E}{v}[/tex]

  [tex]B = \frac{10^{4} }{5 \times 10^{7} }[/tex]

  [tex]B = 2 \times 10^{-4}[/tex] T

Therefore, the magnetic field is [tex]2 \times 10^{-4}[/tex] T

Final answer:

The shortest time it takes for the child to catch up with the horse can be found by analyzing the angular motion of the merry-go-round and the child. Using the given values of angular speed and angular acceleration, we can set up an equation and solve for time using the quadratic formula.

Explanation:

To find the shortest time it takes for the child to catch up with the horse, we need to analyze the angular motion of the merry-go-round and the child.

Given that the child's angular speed is 0.233 rad/s and the merry-go-round starts rotating with an angular acceleration of 0.0136 rad/s^2, we can use the equation:
Δθ = ω_0 * t + (1/2) * α * t^2
where Δθ is the angle the child needs to catch up with the horse (π/2 radians in this case), ω_0 is the initial angular speed of the merry-go-round, α is the angular acceleration, and t is the time.

Plugging in the values, we have:
π/2 = 0.233 * t + (1/2) * 0.0136 * t^2

This equation is quadratic in form, so we can solve it using the quadratic formula. The positive root will give us the time it takes for the child to catch up with the horse.

Answer:

Decrease

Explanation:

If you crawl to the rim the rotational speed will decrease. The law of conservation of angular momentum supports this answer. And it states that :

"When the net external torque acting on a system about a given axis is. zero , the total angular momentum of the system about that axis remains constant."

Final answer:

When moving toward the edge of a spinning turntable, the rate of rotation decreases due to the principle of conservation of angular momentum, which necessitates a decrease in angular velocity to compensate for an increased moment of inertia.

Explanation:

When you crawl toward the edge of a large turntable at an amusement park while it is spinning, the rate of the rotation decreases. This phenomenon is explained by the principle of conservation of angular momentum, which states that if no external torque acts on a system, the total angular momentum of the system remains constant.

Angular momentum is given by the product of the moment of inertia (I) and the angular velocity (ω), represented by the equation L = Iω. As you move away from the axis of rotation, your moment of inertia increases because the moment of inertia is directly proportional to the square of the distance from the axis of rotation. In order to conserve angular momentum, if the moment of inertia increases, the angular velocity must decrease accordingly.

Answer:

a) [tex]f=0.64 Hz[/tex]

b) [tex]\lambda=62.5 cm[/tex]

c) [tex]y_{m}=5 cm[/tex]

d) [tex]k=0.1 cm^{-1}[/tex]

e) [tex]\omega=4 s^{-1}[/tex]

g) [tex]T=0.064 N[/tex]    

Explanation:

We know that the wave equation is of the form:

[tex]y(x,t) = y_{m}sin(kx \pm \omega t)[/tex]

Comparing with the equation of the sinusoidal wave [tex](y=(5.0cm)sin[1.0−(4.0s^{-1})t])[/tex] we will have:

[tex]\omega = 4 s^{-1}[/tex]

a) ω is the angular frequency and it can writes in terms of frequency as:

[tex]\omega = 2\pi f[/tex] , where f is the frequency.

[tex]f=\frac{\omega}{2\pi}[/tex]

[tex]f=0.64 Hz[/tex]

b) Let's recall that the speed of the wave is the product between the wave length and the frequency, so we have:

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

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

v is 40 cm/s

[tex]\lambda=\frac{40}{0.64}[/tex]

[tex]\lambda=62.5 cm[/tex]

If we compare each equation we can find y(m), k and ω:

c) [tex]y_{m}=5 cm[/tex]

d) [tex]kx=1[/tex] and we know that x = 10 cm, so:

[tex]k=\frac{1}{x}=\frac{1}{10}[/tex]

[tex]k=0.1 cm^{-1}[/tex]

e) [tex]\omega=4 s^{-1}[/tex]

f) The minus sign in front of the angular frequency in the equation is the correct choice, just by comparing.

g) We have to use the equation of the speed in terms of tension.

[tex]v=\sqrt{\frac{T}{\mu}}[/tex]

[tex]T=\mu*v^{2}[/tex]

[tex]T=4*40^{2}=6400 g*cm*s^{-2}[/tex]    

[tex]T=0.064 N[/tex]    

I hope it helps you!

(a) The frequency of the wave is 6.37 Hz and (b) the wavelength is 2π cm. (c) The values for y_m, k, and ω are 5.0 cm, (d) 1 cm^-1, and (e) 4.0 s^-1 respectively. (f) The correct choice of sign in front of ω is positive. (g) The tension in the string is 12533.5 g*cm²/s².

(a) To find the frequency of the wave, we can use the formula f = v/λ, where f is the frequency, v is the velocity, and λ is the wavelength. In this case, v = 40 cm/s and we need to find λ. Since y(x,t) = y_m sin(kx ± ωt), we can compare it to the given equation y=(5.0cm)sin[1.0−(4.0s−1)t] to find k and ω. From the comparison, we know k = 1 cm^-1 and ω = 4.0 s^-1. Therefore, the wavelength is given by λ = 2π/k, and plugging in the values, we get λ = 2π cm. The frequency can then be calculated as f = v/λ = 40 cm/s / 2π cm = 6.37 Hz.

(b) the wavelength is 2π cm.

(c) y_m is the amplitude of the wave, which is 5.0 cm.

(d) k is the wave number, which is 1 cm^-1.

(e) ω is the angular frequency, which is 4.0 s^-1.

(f) The correct choice of sign in front of ω depends on the direction of wave propagation. If the wave is traveling in the positive x-direction, the sign should be positive, so ωt is correct.

(g) To find the tension in the string, we can use the formula T = λ²μf², where T is the tension, λ is the wavelength, μ is the linear density, and f is the frequency. Plugging in the values, we get T = (2π cm)² * (4.0 g/cm) * (6.37 Hz)² = 12533.5 g*cm²/s².

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Answer:

an inhibitor of angiotensin II

Explanation:

Angiotensin, specifically angiotensin II binds to many receptors in the body to affect several systems. It can normally increase blood pressure by constricting the blood vessels but with the introduction of an inhibitor, it wouldn't bring about an increase in blood pressure.