Two equal forces are applied to a door at the doorknob. The first force is applied perpendicular to the door; the second force is applied at 30° to the plane of the door. Which force exerts the greater torque?

Answer:

Magnetic field shall be zero at exactly in between the wires.

Explanation:

We can find the magnetic field by biot Savart law as follows

[tex]\overrightarrow{dB}=\frac{\mu _{0}I}{4\pi }\int \frac{\overrightarrow{dl}\times \widehat{r}}{r^{2}}[/tex]

For current carrying wire in positive y direction we have

[tex]\overrightarrow{dB_{1}}=\frac{\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{1}}}{r_{1}^{2}}[/tex]

Similarly for wire carrying current in -y direction we have [tex]\overrightarrow{dB_{2}}=\frac{-\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{2}}}{r_{2}^{2}}[/tex]

Thus the net magnetic field at any point in space is given by

[tex]\overrightarrow{dB_{1}}+\overrightarrow{dB_{2}}[/tex]

[tex]\frac{\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{1}}}{r_{1}^{2}}+\frac{-\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{2}}}{r_{2}^{2}}=0\\\\\Rightarrow \overrightarrow{r_{1}}=\overrightarrow{r_{2}}[/tex]

For points with same position vectors from the 2 wires we have a net zero magnetic field. These points are exactly midway between the 2 wires

Answer:

The magnitude and direction of the non conservative force acting on the box is <F= 59.36 N - DUE EAST DIRECTION>.

Explanation:

m= 4 kg

Vi= 11 m/s

Vf= 1.5 m/s

d= 4m

d= Vi * t - a * t²/2

clearing a:

a= 2*(Vi * t - d)/ t²

Vf= Vi - a * t

replacing "a" and clearing t:

t= 2d/(Vf+Vi)

t= 0.64 s

found now the value of a:

a= Vi - Vf / t

a= 14.84 m/s ²

F= m * a

F= 59.36 N

Answer:

The pressure in the chamber are of p= 0.127 atm.

Explanation:

n= 2.5 moles

T= 310 K

V= 0.5 m³ = 500 L

R= 0.08205746 atm. L /mol . K

p= n*R*T/V

p= 0.127 atm

Answer:

57.94°

Explanation:

we know that the expression of flux

[tex]\Phi =E\times S\times COS\Theta[/tex]

where Ф= flux

           E= electric field

           S= surface area

        θ = angle between the direction of electric field and normal to the surface.

we have Given Ф= 78 [tex]\frac{Nm^{2}}{sec}[/tex]

                          E=[tex]1.44\times 10^{4}\frac{Nm}{C}[/tex]

                          S=[tex]\pi \times 0.057^{2}[/tex]

                         [tex]COS\Theta =\frac{\Phi }{S\times E}[/tex]

 =   [tex]\frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}[/tex]

 =0.5306

 θ=57.94°

A) The magnetic field doubles as well

The relationship between rms value of the electric field and rms value of the magnetic field for an electromagnetic wave is the following:

[tex]E_{rms}=cB_{rms}[/tex]

where

E_rms is the magnitude of the electric field

c is the speed of light

B_rms is the magnitude of the magnetic field

From the equation, we see that the electric field and the magnetic field are directly proportional: therefore, if the rms value of the electric field is doubled, then the rms value of the magnetic field will double as well.

B) The intensity will quadruple

The intensity of an electromagnetic wave is given by

[tex]I=\frac{1}{2}c\epsilon_0 E_0^2[/tex]

where

c is the speed of light

[tex]\epsilon_0[/tex] is the vacuum permittivity

[tex]E_0[/tex] is the peak intensity of the electric field

The rms value of the electric field is related to the peak value by

[tex]E_{rms}=\frac{E_0}{\sqrt{2}}[/tex]

So we can rewrite the equation for the intensity as

[tex]I=c\epsilon_0 E_{rms}^2[/tex]

we see that the intensity is proportional to the square of the rms value of the electric field: therefore, if the rms value of the electric field is doubled, the average intensity of the wave will quadruple.

Answer: A. force with which air rushes across the vocal folds

Explanation:

The human voice is produced in the larynx, whose essential part is the glottis. This is how the air coming from the lungs is forced during expiration through the glottis, making its two pairs of vocal folds to vibrate.

It should be noted that this process can be consciously controlled by the person who speaks (or sings), since the variation in the intensity of the sound of the voice depends on the strength of the breath.

The loudness of a person's voice mainly depends on the force of airflow across their vocal folds. This causes the vocal folds to vibrate and the sound to be produced. The volume of this sound is determined by the amplitude of the resulting sound pressure wave.

The loudness of a person's voice is primarily dependent on the force with which air rushes across the vocal folds. Sound is created when air is pushed up from the lungs through the throat, causing the vocal folds to vibrate. When air flow from the lungs increases, the amplitude of the sound pressure wave becomes greater, resulting in a louder voice. Changes in pitch are related to muscle tension on the vocal cords.

Vocal cord vibration and sound pressure wave amplitude are thus key factors in determining the loudness of a person's voice.

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Answer:

C) 7,500 m/s

Explanation:

The satellite's acceleration due to gravity equals its centripetal acceleration.

v² / r = GM / r²

Solving for velocity:

v² = GM / r

v = √(GM / r)

Given:

G = 6.67×10⁻¹¹ m³/kg/s²

M = 5.98×10²⁴ kg

r = 6.357×10⁶ m + 6.8×10⁵ m = 7.037×10⁶ m

Substituting the values:

v = √(6.67×10⁻¹¹ × 5.98×10²⁴ / 7.037×10⁶)

v = √(5.67×10⁷)

v = 7500 m/s

Answer:

Explanation:

Given that,

Number of mole

n = 2.1mole

Temperature

T = 278 K

Initial volume

V1 = 14.5L

Work done

W = 945J

R = 8.314 J/mol•K

Work done is given as at constant temperature is

W = -P1•V1 In(V2/V1)

Now, let know the pressure using is ideal gas law

PV = nRT

P = nRT/V

P = 2.1 × 8.314 ×278 / 14.5

P = 334.74 N/L

Then,

W = -P1•V1 In(V1/V2)

945 = -334.74×14.5 In(V1/V2)

-945/(334.74×14.5) = In(V1/V2)

In(V1/V2) = -0.1947

Take exponential of both sides

V1/V2 = exp(-0.1947)

14.5/V2 = 0.823

14.5 = 0.823V2

V2 = 14.5/0.823

V2 = 17.62 L

The third option is correct

Answer:

Part a)

[tex]L = 18 kg m^2/s[/tex]

Part b)

[tex]v = 0.6 m/s[/tex]

Part c)

[tex]R_{max} = 6.04 kg[/tex]

[tex]R_{min} = 5.96 kg[/tex]

Explanation:

As we know that Ferris wheel start from rest with angular acceleration

[tex]\alpha = 0.001 rad/s^2[/tex]

time taken = 2 min

so here we have its angular speed after t = 2min given as

[tex]\omega = \alpha t[/tex]

[tex]\omega = (0.001)(2\times 60)[/tex]

[tex]\omega = 0.12 rad/s[/tex]

Part a)

Angular momentum of the Penguine about the center of the wheel is given as

[tex]L = I\omega[/tex]

[tex]L = (6\times 5^2)(0.12)[/tex]

[tex]L = 18 kg m^2/s[/tex]

Part b)

tangential speed is given as

[tex]v = r\omega[/tex]

[tex]v = (5)(0.12)[/tex]

[tex]v = 0.6 m/s[/tex]

Part c)

Maximum reading of the scale at the lowest point is given as

[tex]R_{max} = \frac{m\omega^2 r + mg}{g}[/tex]

[tex]R_{max} = \frac{6(0.12^2)(5) + 6(9.81)}{9.81}[/tex]

[tex]R_{max} = 6.04 kg[/tex]

Minimum reading of the scale at the top point is given as

[tex]R_{min} = \frac{mg - m\omega^2 r}{g}[/tex]

[tex]R_{min} = \frac{6(9.81) - 6(0.12^2)(5)}{9.81}[/tex]

[tex]R_{min} = 5.96 kg[/tex]

Explanation:

   Force                     Description

1. [tex]F_g[/tex]         It is also known as the weight of an object. It is the force that is exerted on an object due to its mass

2. [tex]F_p[/tex]         It is force which is exerted by a push or a pull on an object. It is also known as applied force.

3. [tex]F_f[/tex]          It is known as resistive force. It opposes the motion of an object.

4. [tex]F_n[/tex]        It is the force which is at a right angle to the surface or                              perpendicular to the surface.

Answer:

Explanation:

Fp- Force exerted by a push or pull

Ff- force that comes from a resistances

Fg- also known as an objects weight

FN- support force at a right angle to a surface. JUST DID THIS QUESTION

Final answer:

To find the initial temperature, we can use the combined gas law. By plugging in the values of the initial and final volumes and the final temperature, we can solve for the initial temperature.

Explanation:

To solve this problem, we can use the combined gas law, which states that for an ideal gas at constant pressure, the ratio of the initial volume and the initial temperature is equal to the ratio of the final volume and the final temperature:

V1/T1 = V2/T2

Given that the initial volume is 2.70 L, the final volume is 1.75 L, and the final temperature is 17.00 °C, we can plug these values into the combined gas law to find the initial temperature:

2.70/T1 = 1.75/17.00

Solving for T1, we get T1 = 45.545 °C. Therefore, the initial temperature of the gas was approximately 45.545 °C.

6 ohms is the resistance of the unknown resistor

Electrical resistance, or resistance to electricity, is a force that opposes the passage of current. It acts as a gauge for the difficulty of current flow in this way. Ohms () are used to express resistance values.

When two resistors are connected in parallel, the equivalent resistance can be calculated using the following formula:

1/Req = 1/R1 + 1/R2

where Req is the equivalent resistance, R1 and R2 are the resistances of the two individual resistors.

In this case, we know the resistance of one of the resistors (R1 = 12 ohms) and we can find the current through it (I1) using Ohm's law:

I1 = V/R1 = 12 V / 12 ohms = 1 A

We also know the total current (It = 3 A) and we can find the current through the unknown resistor (I2) using Kirchhoff's current law:

It = I1 + I2

I2 = It - I1 = 3 A - 1 A = 2 A

Now we can use Ohm's law to find the resistance of the unknown resistor (R2):

R2 = V/I2 = 12 V / 2 A = 6 ohms

Therefore, the resistance of the unknown resistor is 6 ohms.

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Final answer:

The resistance of the unknown resistor is 4.0 ohms.

Explanation:

To find the resistance of the unknown resistor, we can use Ohm's Law, which states that V = IR, where V is the voltage, I is the current, and R is the resistance. In this case, the voltage across the unknown resistor is the same as the voltage across the 12-ohm resistor, which is 12V. The current through the unknown resistor is given as 3.0A.

Using Ohm's Law, we can rearrange the formula to solve for R:

R = V/I

R = 12V/3.0A

R = 4.0 ohms

Therefore, the resistance of the unknown resistor is 4.0 ohms.

Answer:

The towns are 32,635 ft apart.

Explanation:

From the image drawn below:

AB = x ft

BC = a ft

AC = (x + a) ft

Considering triangle PCB,

tan 50° = 25000 / a

Or,

a = 25000 / tan 50°

Since tan x = 1/ cot x

a = 25000×cot 50°------------------------------------1

Considering triangle PCA,

tan 25° = 25000 / (a + x)

Or,

a + x = 25000 / tan 25°

Since tan x = 1/ cot x

a + x = 25000×cot 25° -------------------------------2

Thus, finding x from equation 1 and 2, we get:

x = 25000 (cot 25° - cot 50°)

Using cot 25° = 2.1445 and cot 50° = 0.8394, we get:

x ≈ 32,635 ft

Thus, the distance between two towns is 32,635 ft.

Answer:

32635.17 ft

Explanation:

In the diagram, AB = 25000 ft

Let C and D be the town, where CD = d (Distance between two towns)

By triangle, ABC

tan 50 = AB / AC

AC = 25000 / tan 50 = 20977.5 ft

By triangle, ABD

tan 25 = AB/AD

AD = 25000 / tan 25 = 53612.67 ft

So, the distance between two towns

d = AD - AC = 53612.67 - 20977.5 = 32635.17 ft

Thus, the distance between two towns is 32635.17 ft.

The force by the pitcher's hand on the ball during the acceleration phase is found using Newton's Second Law of Motion and is calculated to be 116 Newton.

The force exerted by the pitcher's hand can be found using Newtons Second Law of Motion which states that the force acting on an object is equal to the mass of the object times its acceleration.

it can be expressed as

F = m * a

Here, the mass (m) is 0.145 kg, and the acceleration (a) can be found using the formula a = Δv/Δt, where Δv is the change in velocity (40 m/s) and Δt is the change in time (50 ms or 0.05 s).

So, a = 40/0.05 = 800 m/s², and then the force F = 0.145 * 800 = 116 N. Therefore, the force of the pitcher's hand on the ball during this acceleration phase is 116 Newton.

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The force of the pitcher's hand on the ball can be calculated using Newton's second law of motion.

The force of the pitcher's hand on the ball can be calculated using Newton's second law of motion, which states that force equals mass times acceleration. In this case, the mass of the ball is 0.145 kg and the acceleration is the change in velocity divided by the time interval. The change in velocity is 40 m/s (the final velocity) minus 0 m/s (the initial velocity), and the time interval is 50 ms (or 0.05 s). Therefore, the force can be calculated as:

Force = mass × acceleration = 0.145 kg × (40 m/s - 0 m/s) / 0.05 s = 0.116 N.

So, the force of the pitcher's hand on the ball during this acceleration phase is approximately 0.116 Newtons.

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Answer:

When the net external force is zero

Explanation:

We can answer to this question by referring to Newton's Second Law, which can be written in the following form

[tex]F=\frac{\Delta p}{\Delta t}[/tex]

where

F is the net external force acting on a system

[tex]\Delta p[/tex] is the change in momentum of the system

[tex]\Delta t[/tex] is the time interval

When the total momentum of a system is conserved (so, it does not change), its variation is zero:

[tex]\Delta p = 0[/tex]

In order to satisfy this condition, we see from the formula that we must also have

F = 0

so the net external force acting on the system must be zero.

Answer: 80J

Explanation:

According to the first principle of thermodynamics:  

"Energy is not created, nor destroyed, but it is conserved."  

Then this priciple (also called Law) relates the work and the transferred heat exchanged in a system through the internal energy [tex]U[/tex], which is neither created nor destroyed, it is only transformed. So, in this especific case of the compressed gas:

[tex]\Delta U=Q+W[/tex]  (1)

Where:

[tex]\Delta U[/tex] is the variation in the internal (thermal) energy of the system (the value we want to find)

[tex]Q=-100J[/tex] is the heat transferred out of the gas (that is why it is negative)

[tex]W[/tex] is the work is done on the gas (as the gas is compressed, the work done on the gas must be considered positive )

On the other hand, the work done on the gas is given by:

[tex]W=-P \Delta V[/tex]  (2)

Where:

[tex]P=450kPa=450(10)^{3}Pa[/tex] is the constant pressure of the gas

[tex]\Delta V=V_{f}-V_{i}[/tex] is the variation in volume of the gas

In this case the initial volume is [tex]V_{i}=600{cm}^{3}=600(10)^{-6}m^{3}[/tex] and the final volume is [tex]V_{f}=200{cm}^{3}=200(10)^{-6}m^{3}[/tex].

This means:

[tex]\Delta V=200(10)^{-6}m^{3}-600(10)^{-6}m^{3}=-400(10)^{-6}m^{3}[/tex]  (3)

Substituting (3) in (2):

[tex]W=-450(10)^{3}Pa(-400(10)^{-6}m^{3})[/tex]  (4)

[tex]W=180J[/tex]  (5)

Substituting (5) in (1):

[tex]\Delta U=-100J+180J[/tex]  (6)

Finally:

[tex]\Delta U=80J[/tex]  This is the change in thermal energy in the compression process.