A small object has a mass of 3.0 × 10-3 kg and a charge of -32C. It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of 2.4 × 103 m/s2 in the direction of the +x axis. Determine the electric field, including sign, relative to the +x axis.

Answer:

T = 2010 N

Explanation:

m = mass of the uniform beam = 150 kg

Force of gravity acting on the beam at its center is given as

W = mg

W = 150 x 9.8

W = 1470 N

T = Tension force in the wire

θ = angle made by the wire with the horizontal =  47° deg

L = length of the beam

From the figure,

AC = L

BC = L/2

From the figure, using equilibrium of torque about point C

T (AC) Sin47 = W (BC)

T L Sin47 = W (L/2)

T Sin47 = W/2

T Sin47 = 1470

T = 2010 N

The tension in the cable supporting a 150 kg beam can be calculated by equating the weight of the beam to the vertical component of the tension in the cable. Solve 150 kg * 9.8 m/s² = T * sin(47°) for the tension T to find the force exerted by the cable.

The question concerns the calculation of the tension in the cable supporting a uniform beam. We begin by recognizing that this is a static situation with a beam of mass 150 kg in equilibrium. Thus, the total of the forces in the vertical direction must equal zero.

The forces acting on the beam are its weight (downwards) and the vertical component of the tension in the cable (upwards). The weight of the beam can be calculated by multiplying its mass by gravity g (approximately 9.81 m/s²), resulting in a downward force of 1471.5 N. The vertical component of the tension can be calculated using the sine function: T_vertical = T * sin(θ), where T is the total tension in the cable.

By setting the weight equal to the vertical component of the tension, we can solve for T: 150 kg * 9.8 m/s² = T * sin(47°). Solve this equation for the tension T to find the magnitude of the force exerted by the cable. This tension, along with its vertical and horizontal components, help maintain the beam in its horizontal position.

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Answer:

The centripetal acceleration is 30 m/s²

Explanation:

Given that,

Radius = 0.30 m

Acceleration = 2.0 rad/s²

Time = 5.0

We need to calculate the centripetal acceleration

Using formula of angular velocity

[tex]\omega=a\times t[/tex]

Where,

a = acceleration

t = time

Put the value into the formula

[tex]\omega=2.0\times5.0[/tex]

[tex]\omega=10\ rad/s[/tex]

Using formula of centripetal acceleration

[tex]a_{c}=r\omega^2[/tex]

[tex]a_{c}=0.30\times10^2[/tex]

[tex]a_{c}=30\ m/s^2[/tex]

Hence, The centripetal acceleration is 30 m/s²

The centripetal acceleration of a point on the outer edge of the tire after 5.0 seconds is [tex]\( 30 \, \text{m/s}^2 \)[/tex].

The centripetal acceleration of a point on the outer edge of the tire after 5.0 seconds is given by the formula [tex]\( a_c = \omega^2 r \)[/tex], where [tex]\( \omega \)[/tex] is the angular velocity and [tex]\( r \)[/tex] is the radius of the tire.

First, we need to find the angular velocity at 5.0 seconds. Since the tire starts from rest and accelerates at a constant rate.

We can use the formula [tex]\( \omega = \omega_0 + \alpha t \)[/tex], where [tex]\( \omega_0 \)[/tex] is the initial angular velocity (0 rad/s in this case, since the tire starts from rest), [tex]\( \alpha \)[/tex] is the angular acceleration ([tex]2.0 rad/s^2[/tex]), and [tex]\( t \)[/tex] is the time (5.0 s).

Plugging in the values, we get:

[tex]\( \omega = 0 + (2.0 \, \text{rad/s}^2)(5.0 \, \text{s}) = 10.0 \, \text{rad/s} \)[/tex]

Now that we have the angular velocity, we can calculate the centripetal acceleration:

[tex]\( a_c = \omega^2 r = (10.0 \, \text{rad/s})^2(0.30 \, \text{m}) \)[/tex]

[tex]\( a_c = 100 \, \text{s}^{-2} \times 0.30 \, \text{m} \)[/tex]

[tex]\( a_c = 30 \, \text{m/s}^2 \)[/tex]

Answer:

160 km

Explanation:

g = 9.8 m/s^2, g' = 9.3 m/s^2

Let h be the height from ocean level.

Use the formula for acceleration due to gravity at height.

g' = g (1 - 2h/R)

The radius of earth is 6400 km

9.3 = 9.8 ( 1 - 2 H / R)

0.05 = 2 h/R

h = 0.05 × 6400 / 2 = 160 km

Answer:

The current source is 1 mA.

Explanation:

Given that,

Voltage = 10 V

Resistor = 10 k ohm

We need to calculate the current

Using formula of transformation theorem

The source transformation must be constrained is given by

[tex]v_{s}=iR_{s}[/tex]

Where, V = voltage

I = current

R = resistor

Put the value into the formula

[tex]10=i\times10\times10^{3}[/tex]

[tex]i =\dfrac{10}{10\times10^{3}}[/tex]

[tex]i = 0.001\ A[/tex]

[tex]i=1\ mA[/tex]

Hence, The current source is 1 mA.

By using Ohm's Law (I = V/R), it can be determined that for a voltage source Vs = 10 V in series with a resistor of 10 kOhm, the current source would be 1mA after source transformation.

The question asks for the current value in a circuit when the source transformation theorem is applied. This involves using Ohm's law, which in basic form states that the current through a resistor equals the voltage across the resistor divided by the resistance. It's represented by the equation I = V/R.

In our case, the voltage (V) is 10 volts, and the resistance (R) is 10 kilo-Ohms or 10,000 Ohms. Applying Ohm's law gives I = V/R, so I = 10V/10,000Ω = 0.001 A. However, because the question asks for the current in milliamperes (mA), we need to convert from amperes to milliamperes, knowing that 1A = 1000mA. Therefore, 0.001A = 1mA.

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Answer:

Voltage of the secondary coil is 720 volts.

Explanation:

Number of turns in primary coil, N₁ = 60

Number of turns in secondary coil, N₂ = 360

Input rms voltage of primary coil, V₁ = 120 V

We have to find the output rms voltage of the secondary coil. The relationship between the number of coil and voltage is given by :

[tex]\dfrac{N_1}{N_2}=\dfrac{V_1}{V_2}[/tex]

[tex]\dfrac{60}{360}=\dfrac{120}{V_2}[/tex]

V₂ = output rms voltage of the secondary coil.

After solving above equation, we get :

V₂ = 720 V

Hence, the output rms voltage of the secondary coil is 720 volts.

Answer:

a) 8.61 m/s, b) 5.73 m

Explanation:

a) During the collision, momentum is conserved.

mv = (m + M) V

(12.5 g) (86.4 m/s) = (12.5 g + 113 g) V

V = 8.61 m/s

b) After the collision, energy is conserved.

Kinetic energy = Work done by friction

1/2 (m + M) V² = F d

1/2 (m + M) V² = N μk d

1/2 (m + M) V² = (m + M) g μk d

1/2 V² = g μk d

d = V² / (2g μk)

d = (8.61 m/s)² / (2 × 9.8 m/s² × 0.659)

d = 5.73 m

Notice we used the kinetic coefficient of friction.  That's the friction when an object is moving.  The static coefficient of friction is the friction on a stationary object.  Since the bullet/block combination is sliding across the surface, we use the kinetic coefficient.

Answer:

No

Explanation:

When the person slides down, the change in gravitational potential energy is converted into kinetic energy, according to

[tex]\Delta U = \Delta K\\mg\Delta h = \frac{1}{2}mv^2[/tex]

where

m is the mass of the person

g is the acceleration of gravity

v is the final speed

[tex]\Delta h[/tex] is the change in heigth of the person

Here we have assumed that the initial speed is zero.

Re-arranging the equation,

[tex]v = \sqrt{2g \Delta h}[/tex]

and we see that this quantity does not depend on the mass of the person, so every person will have the same speed at the bottom of the slide, equal to:

[tex]v=\sqrt{2(9.8 m/s^2)(3.2 m-1.2 m)}=6.3 m/s[/tex]

(a) 26.1 Nm

The magnitude of the torque exerted by a force acting perpendicularly to a surface is given by:

[tex]\tau = Fr[/tex]

where

F is the magnitude of the force

r is the distance from the pivot

In this situation,

F = 47.5 N is the force applied

[tex]r=\frac{1.10 m}{2}=0.55 m[/tex] is the distance from the hinges (the force is applied at the center of the door)

So, the magnitude of the torque is

[tex]\tau = (47.5 N)(0.55 m)=26.1 Nm[/tex]

(b) 52.3 Nm

In this case, the force is applied at the edge of the door farthest from the hinges. This means that the distance from the hinges is

r = 1.10 m

So, the magnitude of the torque is

[tex]\tau =(47.5 N)(1.10 m)=52.3 Nm[/tex]

The magnitude of torque applied about the axis of the door when the force is applied at the center is 26.125 N.m

The magnitude of the torque applied at the edge farthest from the center  is 52.25 N.m.

The given parameters;

The torque applied by the janitor is the product of applied force and perpendicular distance.

τ = F.r

The torque applied about the axis of the door when the force is applied at the center.

[tex]\tau = F \times \frac{r}{2} \\\\\tau = 47.5 \times \frac{1.1}{2} \\\\\tau = 26.125 \ N.m[/tex]

The magnitude of the torque applied at the edge farthest from the center ;

[tex]\tau = F\times r\\\\\tau = 47.5 \times 1.1\\\\\tau = 52.25 \ N.m[/tex]

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Answer:

a) 51.8 m, b) 27.4 m/s, c) 142 m

Explanation:

Given:

v₀ = 42.0 m/s

θ = 60.0°

t = 5.50 s

Find:

h, v, and H

a) y = y₀ + v₀ᵧ t + ½ gt²

0 = h + (42.0 sin 60.0) (5.50) + ½ (-9.8) (5.50)²

h = 51.8 m

b) vᵧ = gt + v₀ᵧ

vᵧ = (-9.8)(5.5) + (42.0 sin 60.0)

vᵧ = -17.5 m/s

vₓ = 42.0 cos 60.0

vₓ = 21.0 m/s

v² = vₓ² + vᵧ²

v = 27.4 m/s

c) vᵧ² = v₀ᵧ² + 2g(y - y₀)

0² = (42.0)² + 2(-9.8)(H - 51.8)

H = 142 m

(a) The height of the cliff is 51.82 m

(b) The final velocity of the stone before it strikes A is 27.36 m/s

(c) The maximum height reached by the stone is 67.50 m

The given parameters;

(a) The height of the cliff is calculated as;

[tex]h_y = v_0_yt - \frac{1}{2} gt^2\\\\h_y = (v_0\times sin(\theta)) t \ - \ \frac{1}{2} gt^2\\\\h_y = (42\times sin(60))\times 5.5 \ - \ (0.5\times 9.8\times 5.5^2)\\\\h_y = 200.05 - 148.23\\\\h_y = 51.82 \ m[/tex]

(b) The final velocity of the stone before it strikes A is calculated as;

The vertical component of the final velocity

[tex]v_f_y = v_0_y - gt\\\\v_f_y = (v_0 \times sine (\theta)) - gt\\\\v_f_y = (42\times sin(60) - (9.8\times 5.5)\\\\v_f_y = -17.53 \ m/s[/tex]

The horizontal component of the final velocity

[tex]v_x_f = v_0\times cos(\theta)\\\\v_x_f = 42 \times cos(60)\\\\v_x_f = 21 \ m/s[/tex]

The resultant of the velocity at point A;

[tex]v_f = \sqrt{v_x_f^2 + v_y_f^2} \\\\v_f = \sqrt{(21)^2 + (-17.53)^2} \\\\v_f = 27.36 \ m/s[/tex]

(c) The maximum height reached by the projectile;

[tex]H = \frac{v_o^2 sin^2(\theta)}{2g} \\\\H = \frac{[42 \times sin(60)]^2}{2\times 9.8} \\\\H = 67.50 \ m[/tex]

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Answer:

Explanation:

We know that the formula for resistance in terms of length and area is given by

R = p l/ a

Where p be the resistivity and a be the area of crossection

a = pi × d^2 / 4

Where d be the diameter

a = 3.14 × (1.628 × 10^-3)^2 / 4

a = 2 × 10^-6 m^2

R = 1.68 × 10^-8 × 24 / (2 × 10^-6)

R = 20.16 × 10^-2 ohm

By use of Ohm's law

V = IR

V = 13 × 20.16 × 10^-2

V = 2.62 Volt

Answer:

Induced emf, [tex]\epsilon=2.2\ mV[/tex]

Explanation:

It is given that,

Rate of change of magnetic field, [tex]\dfrac{dB}{dt}=0.23\ T/s[/tex]

A UHF antenna is oriented at an angle of 47° to a magnetic field, θ = 47°

Diameter of the antenna, d = 13.4 cm

Radius, r = 6.7 m = 0.067 m

We need to find the induced emf of the antenna. It is given by :

[tex]\epsilon=-\dfrac{d\phi}{dt}[/tex]

Where

[tex]\phi[/tex] = magnetic flux, [tex]\phi=BA\ cos\theta[/tex]

So, [tex]\epsilon=\dfrac{d(BA\ cos\theta)}{dt}[/tex]

B = magnetic field

[tex]\epsilon=A\dfrac{d(B)}{dt}\ cos\theta[/tex]

[tex]\epsilon=\pi r^2\times \dfrac{dB}{dt}\times cos(47)[/tex]

[tex]\epsilon=\pi (0.067\ m)^2\times 0.23\ T-s\times cos(47)[/tex]

[tex]\epsilon=0.0022\ V[/tex]

[tex]\epsilon=2.2\ mV[/tex]

So, the induced emf of the antenna is 2.2 mV. Hence, this is the required solution.

Final answer:

The type of friction acting between the tires and the road when a car's tires roll freely without any slippage is static friction, as the tire's contact point with the road is momentarily at rest.

Explanation:

When a car's tires roll freely without any slippage, the type of friction acting between the tires and the road is static friction. This is because the bottom of the tire is at rest with respect to the ground for a moment in time, ensuring there's no relative movement between the contact surfaces. Static friction is what allows the car to move forward as it prevents the tires from slipping on the surface of the road. Once slipping occurs, for instance, if the tires are spinning without moving the car forward, it becomes kinetic friction. However, in a scenario where rolling without slipping occurs, it's due to the presence of static friction, which is necessary for proper motion and control of the vehicle.

To find the normal component of acceleration, differentiate the velocity and acceleration functions.

In order to find the normal component of the acceleration, we need to first find the velocity and acceleration vectors as functions of time. Using the given position function, we can differentiate it to find the velocity function:

v(t) = dr/dt = -3(4sin(3t)i + 4cos(3t)ĵ)

Next, we can differentiate the velocity function to find the acceleration function:

a(t) = dv/dt = -3(3cos(3t)i - 3sin(3t)ĵ)

Since the tangential component of the acceleration is constant, we can set the coefficient of the tangential component equal to a constant value:

-3(3sin(3t)) = Constant

Solving this equation will give us the value of the constant. Once we know the value of the constant, we can substitute it back into the acceleration function to find the normal component of the acceleration as a function of time.

I₃ = 0.17 A into the junction.

The key to solve this problem is using Kirchhoff's Current Law which statements  that the algebraic sum of the currents that enter and leave a particular junction must be 0 (I₁+I₂+...+In = 0). Be careful, the currents that leaves a point is considered positive current, and the one that enters a point is considered negative.

In other words,  the sum of the currents that enter to the joint is equal to the sum of the currents that come out of the joint.

Three wire meet at junction. Wire 1 has a current of 0.40 A into the junction, the current of the wire 2 is 0.57 A out of the junction. What is the magnitude of the current in wire 3?

To calculate the current in the wire 3:

First, let's name the currents in the circuit. So:

The current in the wire 1 is I₁ = 0.40 A, the current in the wire 2 is I₂ = 0.57 A, and the current of wire 3 is I₃ = ?.

Second,  we make a scheme of the circuit, with the current I₁ of wire 1 into the junction, the current I₂ of wire 2 out of the junction, and let's suppose that the current I₃ of the wire 3 into the junction. (See the image attached)

Using Kirchhoff's Current Law, the sum of the currents I₁ and I₃ into the junction is equal to the current I₂ out of junction.

I₁ + I₃ = I₂

0.40 A + I₃ = 0.57 A

I₃ = 0.57 A - 0.40A

I₃ = 0.17 A

Checking the Kirchhoff's Current Law, the sum of all currents is equal to 0 :

I₁ + I₃ = I₂

I₁ - I₂ + I₃ = 0

0.40 A - 0.57 A + 0.17 A = 0

Let's suppose that the current I₃ of the wire 3 out the junction.

I₁ = I₂ + I₃

0.40 A = 0.57 A + I₃

I₃ = 0.40 A - 0.57 A

I₃ = -0.17 A which means that the current flow  in the opposite direction that we selected.

The magnitude of the current in wire 3 entering the junction is 0.17A. This result is based on the principle of conservation of electric current which states that the total current entering the junction must equal the total current exiting the junction.

This question involves the principle of the conservation of electric current, also known as Kirchhoff's Current Law. According to this concept, at any junction point the total current entering the junction is equal to the total current leaving the junction. In this case, if wire 1 has a current of 0.4A entering the junction and wire 2 has a current of 0.57A leaving the junction, the current in wire 3 would be the difference between these two values.

Since current can't be negative, if the current in wire 1 (entering the junction) is less than the current in wire 2 (exiting), the current in wire 3 must be entering the junction to balance the current flow.

Therefore, the magnitude of the current in wire 3 would be 0.57A - 0.4A = 0.17A. This means wire 3 has a current of 0.17A entering the junction.

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Answer:

Speed of car, v₁ = 55 m/s

Explanation:

It is given that,

Mass of Semi, m₁ = 9565 kg

Initial velocity of semi, u₁ = 55 m/s

Mass of car, m₂ = 992 kg

Initial velocity of car, u₂ = 0 (at rest)

Since, the collision between two objects is elastic and all the momentum is transferred to the car i.e final speed of semi, v₂ = 0

Let the speed of the car is v₁. Using conservation of linear momentum as :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

[tex]9565\ kg\times 55\ m/s+992\ kg\times 0=9565\ kg\times v_1+0[/tex]

v₁ = 55 m/s

Hence, the car move forward with a speed of 55 m/s.

Answer:

1.67 m

Explanation:

Ux = 3 m/s, ax = -2.7 m/s^2, vx =0

Let the distance travelled before stopping along x axis is x.

Use third equation of motion along x axis

Vx^2 = ux^2 + 2 a x

0 = 9 - 2 × 2.7 × x

X = 1.67 m

Answer:

The sound level is 34 dB.

(A) is correct option.

Explanation:

Given that,

Distance from a source of sound [tex]r_{1}=20 m[/tex]

Sound level [tex]L_{1}= 40\ dB[/tex]

Distance from a source of sound[tex]r_{2}=40\ m[/tex]

We need to calculate the sound level

Using equation

[tex]L_{2}=L_{1}-|20 log(\dfrac{r_{1}}{r_{2}})|[/tex]

[tex]L_{2}=40-|20 log(\dfrac{20}{40})|[/tex]

[tex]L_{2}=40-|20 log(0.5)|[/tex]

[tex]L_{2}=33.9\ dB[/tex]

[tex]L_{2}=34\ dB[/tex]

Hence, The sound level is 34 dB.

The sound level decreases by 20 dB when the distance from the source is doubled because the intensity of sound decreases with the square of the distance. Thus, the sound level at 40 m would be 20 dB. The answer is option C.

When the observer moves from a distance of 20 m to 40 m away from the source of sound, the sound intensity decreases. This is due to the inverse square law, which states that the intensity of sound is inversely proportional to the square of the distance from the source. Since the observer doubles the distance from the source, the intensity of the sound will be reduced to one-fourth. Because each 10-fold decrease in intensity corresponds to a reduction of 10 dB in sound level, halving the distance corresponds to an intensity ratio of 1/4, which is two factors of 10 (or 102 times less intense). Thus, the sound level decreases by 20 dB (2 factors of 10).

Therefore, at a distance of 40 m, the sound level would be 20 dB less than the original 40 dB, leading to a new sound level of 20 dB. The answer is option C: 20 dB.

Answer:

Spring constant, k = 47 N/m

Explanation:

It is given that,

Force applied to a spring, F = 34 N

A very light ideal spring moves 0.73 m from equilibrium position i.e. x = 0.73 m

We have to find the force constant or spring constant of the spring. It can be calculated using Hooke's law. According to him, the force acting on the spring when it compresses or stretches is given by :

[tex]F=-kx[/tex]  (-ve sign shows opposite direction)

[tex]k=\dfrac{F}{x}[/tex]

[tex]k=\dfrac{34\ N}{0.73\ m}[/tex]

k = 46.5 N/m

or

k = 47 N/m

Hence, the spring constant of the spring is 47 N/m.

Utilizing Hooke's law, which posits that the force needed to compress or extend a spring is proportional to the distance, the spring constant can be calculated to be approximately 46.6 N/m by dividing the force (34N) by the displacement (0.73m). So, the closest option amongst the provided ones is (A) 47N/m.

In the situation described where a force of 34N stretches a very light ideal spring 0.73 m from equilibrium, you would want to find the spring's force constant, often denoted as k. This involves a principle in physics known as Hooke's law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance. The equation for Hooke's law is F=kx.

To find the spring constant (k), you rearrange the equation for Hooke's Law to give you: k = F/x. Substituting the force (F = 34N) and displacement from equilibrium (x = 0.73m) into the equation gives: k = 34N / 0.73m. Calculating this gives a spring constant of approximately 46.6 N/m. So, none of the given options (A) 47N/m  (B) 38N/m  (C) 53N/m  (D) 25N/m are exactly correct, but Option A is the closest.

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Answer:

4.15 x 10^6 N

Explanation:

Area, A = 1.43 cm^2 = 1.43 x 10^-4 m^2

mass, m = 60.5 kg

Weight, F = m g = 60.5 x 9.8 = 592.9 N

Pressure = Force / Area

P = Weight / Area

P = 592.9 / (1.43 x 10^-4)

P = 4.15 x 10^6 N

Answer:

Capacitance of the second capacitor = 2C

Explanation:

[tex]\texttt{Capacitance, C}=\frac{\varepsilon_0A}{d}[/tex]

Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.

Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.

We have    

              [tex]\texttt{Capacitance, C}_1=\frac{\varepsilon_0A_1}{d_1}=C[/tex]

Similarly for capacitor 2

               [tex]\texttt{Capacitance, C}_2=\frac{\varepsilon_0A_2}{d_2}=\frac{\varepsilon_0A_1}{\frac{d_1}{2}}=2\times \frac{\varepsilon_0A_1}{d_1}=2C[/tex]

Capacitance of the second capacitor = 2C

Answer:

Explanation:

In case of rotational motion, every particle is rotated with a same angular velocity .

The relation between linear velocity and angular velocity is

V = r w

Where v is linear velocity, r is the radius of circular path and w be the angular velocity.

Here w is same for all the particles but every particle has different radius of circular path I which they are rotating, so linear velocity is differnet for all.

Final answer:

The angle theta required for the water main to turn from north to east can be determined by treating the grades as the rise over run and using vector addition with trigonometry to find the resultant direction and its angle east of north.

Explanation:

To determine the angle theta (θ) required for the water main to make the turn from north to east with a 12.5% grade in the north direction and a 20% grade in the east direction, we need to use vector addition and trigonometry. The water main's path can be viewed as containing two components: one in the north direction and one in the east direction. We will represent these as vectors and find their resultant to determine the required angle between them.

The grades can be understood as the rise over run, which means the north vector has a rise of 12.5 units for every 100 units of run, and the east vector has a rise of 20 units for every 100 units of run. To find the resultant vector, we can use the arctangent function:

θ = arctan(opposite/adjacent) = arctan(east grade / north grade) = arctan(0.20 / 0.125) = arctan(1.6)

Once we calculate θ, this will give us the angle between the north direction and the resultant direction of the water main. Keep in mind that this angle should be measured east of north to reflect the correct orientation of the water main's turn.

Answer:

Mass of the solution  = 54.75 g

Explanation:

Mass of solid dissolved = 4.75 g

Mass of the solution = Mass of solid dissolved + Mass of water.

Mass of water = Volume x Density

Volume = 50 mL = 50 cm³

Density = 1 g/cm³

Mass of water = 50 x 1 = 50 g

Mass of the solution = 4.75 + 50 = 54.75 g

Answer:

The frequency of square coil is 47 Hz.

(A) is correct option.

Explanation:

Given that,

Radius =10 cm

Side = 20 cm

Magnetic field = 1.5 T

Frequency = 60 Hz

We need to calculate the the maximum induced voltage

[tex]V_{m}=BAN\omega[/tex]

Where, B = magnetic field

A = area of cross section

N = number of turns

[tex]\omega=2\pi f[/tex]= angular frequency

Put the value into the formula

[tex]V_{m}=1.5\times3.14\times(10\times10^{-2})^2\times1\times2\times3.14\times60[/tex]

[tex]V_{m}=17.75\ V[/tex]

If the square coil have the same induced voltage.

Area of square of coil [tex]A =(20\times10^{-2})^2[/tex]

[tex]A=0.04\ m^2[/tex]

Now, The angular velocity of square coil

[tex]\omega=\dfrac{V_{m}}{AB}[/tex]

[tex]\omega=\dfrac{17.75}{0.04\times1.5}[/tex]

[tex]\omega=295.8\ \dfrac{rad}{s}[/tex]

Now, frequency of rotation

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

Put the value into the formula of frequency

[tex]f=\dfrac{295.8}{2\times3.14}[/tex]

[tex]f=47\ Hz[/tex]

Hence, The frequency of square coil is 47 Hz.

Answer:

The moment of inertia of the wheel is 0.593 kg-m².

Explanation:

Given that,

Force = 82.0 N

Radius r = 0.150 m

Angular speed = 12.8 rev/s

Time = 3.88 s

We need to calculate the torque

Using formula of torque

[tex]\tau=F\times r[/tex]

[tex]\tau=82.0\times0.150[/tex]

[tex]\tau=12.3\ N-m[/tex]

Now, The angular acceleration

[tex]\dfrac{d\omega}{dt}=\dfrac{12.8\times2\pi}{3.88}[/tex]

[tex]\dfrac{d\omega}{dt}=20.73\ rad/s^2[/tex]

We need to calculate the moment of inertia

Using relation between torque and moment of inertia

[tex]\tau=I\times\dfrac{d\omega}{dt}[/tex]

[tex]I=\dfrac{I}{\dfrac{d\omega}{dt}}[/tex]

[tex]I=\dfrac{12.3}{20.73}[/tex]

[tex]I= 0.593\ kg-m^2[/tex]

Hence, The moment of inertia of the wheel is 0.593 kg-m².

Answer:

0.593 kg-m²

Explanation:

edg.

Explanation:

The expression for an Ideal Gas is:  

[tex]P.V=n.R.T[/tex]    (1)

Where:  

[tex]P[/tex] is the pressure of the gas  

[tex]V[/tex] is the volume of the gas  

[tex]n[/tex] the number of moles of gas  

[tex]R[/tex] is the gas constant  

[tex]T[/tex] is the absolute temperature of the gas  

Finding [tex]V[/tex]:

[tex]V=\frac{n.R.T}{P}[/tex]    (2)

If we are told the the amount of gas [tex]n[/tex] and pressure [tex]P[/tex] remain constant, but we increase the temperature [tex]T[/tex] by a factor of three; we will have to rewrite (2) with the new temperature [tex]T_{N}[/tex]:

[tex]T_{N}=3T[/tex]

[tex]V=\frac{n.R.3T}{P}[/tex]    (3)

[tex]V=3\frac{n.R.T}{P}[/tex]    (4)

Now, if we compare (2) with (4), it is clearly noticeable the volume of the gas has increased by a factor of 3.

The volume will triple ( increase by a factor of three )

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The basic formula of pressure that needs to be recalled is:

Pressure = Force / Cross-sectional Area

or symbolized:

[tex]\large {\boxed {P = F \div A} }[/tex]

P = Pressure (Pa)

F = Force (N)

A = Cross-sectional Area (m²)

Let us now tackle the problem !

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In this problem , we will use Ideal Gas Law as follows:

Given:

Initial Temperature of The Gas = T₁

Final Temperature of The Gas = T₂ = 3T₁

Asked:

Final Volume of The Gas = V₂ = ?

Solution:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

[tex]\frac{PV_1}{T_1} = \frac{PV_2}{3T_1}[/tex]

[tex]\frac{V_1}{1} = \frac{V_2}{3}[/tex]

[tex]3(V_1) = 1(V_2)[/tex]

[tex]V_2 = 3V_1[/tex]

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The volume will triple ( increase by a factor of three )

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Pressure

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure, Volume , Ideal , Gas , Law

Explanation:

Average acceleration is the change in velocity over the change in time:

a = (v − v₀) / t

In the x direction:

aₓ = (6.11 cos (-54.2°) − 5.33 cos (37.9°)) / 2.00

aₓ = -0.316 m/s²

In the y direction:

aᵧ = (6.11 sin (-54.2°) − 5.33 sin (37.9°)) / 2.00

aᵧ = -4.11 m/s²

To find the average acceleration, decompose the given velocities into their x and y components. Then calculate the change in velocities in both directions divided by the time interval.

The average acceleration of a particle can be found using the change in velocity and the time interval. The given velocities are in magnitude and direction, so we need to decompose these into their x and y components. The initial and final x-components are Vix = 5.33 m/s cos(37.9°) and Vfx = 6.11 m/s cos(54.2°), respectively. The average x-direction acceleration (ax) can be calculated as ax = (Vfx - Vix)/time. Similarly, the initial and final y-components are Viy = 5.33 m/s sin(37.9°) and Vfy = 6.11 m/s sin(54.2°), respectively. The y-direction acceleration (ay) is ay = (Vfy - Viy)/time.

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Answer:

1047 miles

Explanation:

The radius of the Earth is

[tex]r = 4000[/tex] (miles)

So its circumference, which is the total length of the equator, is given by

[tex]L=2\pi r= 2\pi(4000)=25133 mi[/tex]

Now we know that the Earth rotates once every 24 hours. So the distance through which the equator moves in one hour is equal to its total length divided by the number of hours, 24:

[tex]L' = \frac{25133 mi}{24h}=1047 mi[/tex]

Answer:

29.0 g

Explanation:

The mass of the piece of gold is given by:

m = dV

where

m is the mass

d is the density

V is the volume of the piece of gold

The density of gold is

d = 19.3 g/cm^3

while the volume of the sample is equal to the volume of displaced water, so

V = 64.5 mL - 63.0 mL = 1.5 mL

And since

1 mL = 1 cm^3

the volume is

V = 1.5 cm^3

So the mass of the piece of gold is:

m = (19.3 g/cm^3)(1.5 cm^3)=29.0 g