Calculate the power output of a 1.5 g fly as it walks straight up a windowpane at 2.4 cm/s .

Express your answer using two significant figures. (Watts)

(a) 252 N, opposite to the applied force

There are two forces acting on the refrigerator in the horizontal direction:

- the pushing force of 252 N, F, forward

- the frictional force, Ff, pulling backward

In this case, the refrigerator is not moving: this means that its acceleration is zero. According to Newton's second law, this also means that the net force acting on the refrigerator is also zero:

[tex]\sum F = ma = 0[/tex]

So we have

[tex]F-F_f = 0[/tex]

which means that the frictional force is equal in magnitude to the pushing force:

[tex]F_f = F = 252 N[/tex]

and the direction is opposite to the pushing force.

(b) 334.8 N

The force that must be applied to the refrigerator to make it moving is equal to the maximum force of friction, which is given by:

[tex]F_{max} = \mu mg[/tex]

where

[tex]\mu = 0.61[/tex] is the coefficient of static friction

m = 56 kg is the mass of the refrigerator

g = 9.8 m/s^2 is the acceleration of gravity

Substituting:

F_max = (0.61)(56 kg)(9.8 m/s^2)=334.8 N

The force that will move the object is 83 N in the negative x direction.

The frictional force is the force that opposes motion. The force of friction acts in the opposite direction to the force that is moving the object. Now we can obtain the frictional force from;

μs = F/mg

F = μsmg = 0.61 * 56 kg * 9.8 m/s^2 = 335 N (positive  direction)

Secondly;

ma = F - Ff

Where;

ma = resultant force

F = moving force

Ff = frictional force

Now we need to find the net force ma

ma = 252 N - 335 N

ma = -83 N

The force that will move the object is 83 N in the negative x direction.

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Answer:

The length of the string is 0.051 meters

Explanation:

It is given that,

Tension in the string, T = 240 N

Mass of the string, m = 0.086 kg

Speed of the wave, v = 12 m/s

The speed of the wave on the string is given by :

[tex]v=\sqrt{\dfrac{T}{M}}[/tex]

M is the mass per unit length of the string i.e. M = m/l.......(1)

So, [tex]M=\dfrac{T}{v^2}[/tex]

[tex]M=\dfrac{240\ N}{(12\ m/s)^2}[/tex]

M = 1.67 kg/m

The length of the string can be calculated using equation (1) :

[tex]l=\dfrac{m}{M}[/tex]

[tex]l=\dfrac{0.086\ kg}{1.67\ kg/m}[/tex]

l = 0.051 m

So, the length of the string is 0.051 meters. Hence, this is the required solution.

Explanation:

We'll need two equations.

v² = v₀² + 2a(x - x₀)

where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.

x = x₀ + ½ (v + v₀)t

where t is time.

Given:

v = 47.5 m/s

v₀ = 34.3 m/s

x - x₀ = 40100 m

Find: a and t

(47.5)² = (34.3)² + 2a(40100)

a = 0.0135 m/s²

40100 = ½ (47.5 + 34.3)t

t = 980 s

Explanation and answer:

This problem is best answered by drawing a figure as a first step.

ABC is the scaffold.

A downward force of 500N is applied downwards at 1m from end A.

The weight of 800N is exerted by the scaffold uniformly distributed between A & C.

At A and C, ropes are attached to support the load.

Let Fc=tension in rope passing through C.

Take moments about A:

Fc = (500N * 1m +800N*(3+1)/2m / 4m

    = (500 Nm + 1600Nm) / 4m

    = 2100 Nm / 4m

   = 525 N

Answer:

i believe that it is d

Explanation:

In a super heater, the temperature of the steam rises while the pressure remains constant. This process helps to remove the last traces of moisture from the saturated steam.

In a super heater, the conclusion is that option (C) pressure remains constant and temperature rises is the correct choice. A super heater is a device used in a steam power plant to increase the temperature of the steam, above its saturation temperature. The function of the super heater is to remove the last traces of moisture (1 to 2%) from the saturated steam and to increase its temperature above the saturation temperature. The pressure, however, remains constant during this process because the super heater operates at the same pressure as the boiler.

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Answer: v= 160ft/s

a=32ft/s^2 constant

Explanation:

s(t)=400-16t^2 derivative of position is velocity v(t) and derivative of velocity is acceleration a(t) so let s(t)=0 to find the time of flight to reach the ground and take the two derivatives and use the time found and solve. Also acceleration is a constant as it’s gravity.

0=400-16t^2

400=16t^2

25=t^2

t=5s

ds/dt=v(t)=0-32t

dv/dt=a(t)=-32 constant(gravity)

v(t)=-32(5s)= -160ft/s negative sign is only showing direction

The velocity of the object the moment it reaches the ground is 160 ft/s.

The problem can be solved by following steps.

s(t) = 400-16t^2 (given)

Let s(t)=0

Therefore

0= 400-16t^2

400=16t^2

25=t^2

Therefore

t = 5sec

Now as we know that the derivative of the position is Velocity

so v(t) = ds/dt = -32t

where t = 5sec

substitute the value of t in v(t)

Therefore, v(t) = -32(5) = -160

The direction is negative

Hence the velocity is 160ft/s

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The minimum speed of an object in simple harmonic motion occurs when it is at the equilibrium point. Hence the correct answer is option B

In simple harmonic motion, the minimum speed of an object occurs when it is at the equilibrium point. The equilibrium position is where the object would naturally rest in the absence of force. When the object is at the equilibrium point, it has zero acceleration and the speed is at its minimum.

Hence the correct answer is option B

Answer:

The terminal voltage will be 5.99 volt.

(d) is correct option.

Explanation:

Given that,

Voltage = 6.00

Internal r= 0.8322 ohm

Total resistance R =7380 ohm

We need to calculate the current

Using current formula

[tex]I=\dfrac{V}{R+r}[/tex]

Put the value into the formula

[tex]I = \dfrac{6}{7380+0.8322}[/tex]

[tex]I=0.000812\ A[/tex]

We need to calculate the voltage drop due to internal resistance

[tex]V' = Ir[/tex]

[tex]V'=0.000812\times0.8322[/tex]

[tex]V'=0.00067\ volt[/tex]

Now, The terminal voltage will be

[tex]V''=6-V'[/tex]

[tex]V''=6-0.00067[/tex]

[tex]V''=5.99\ volt[/tex]

Hence, The terminal voltage will be 5.99 volt,

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

Data obtained from the question

Mass of copper (M꜀) = 248 g

Volume of water  = 390 mL

Density of water = 1 g/mL

Initial temperature of water (Tᵥᵥ) = 22.6 °C

Equilibrium temperature (Tₑ) = 39.9 °C

Determination of the mass of water

Volume of water = 390 mL

Density of water = 1 g/mL

[tex]Density = \frac{mass}{volume}\\\\1 = \frac{mass}{390}[/tex]

Cross multiply

[tex]Mass = 1 * 390[/tex]

Determination the initial temperature of the copper.

Mass of copper (M꜀) = 248 g

Mass of water (Mᵥᵥ) = 390 g

Initial temperature of water (Tᵥᵥ) = 22.6 °C

Equilibrium temperature (Tₑ) = 39.9 °C

1. Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

2. Specific heat capacity of copper (C꜀) = 0.385 J/gºC

Heat lost by copper = heat gained by water

[tex]Q_{c} = Q_{w} \\ \ M_{c} C_{c}(T_{c}-T_{e}) = M_{w} C_{w}(T_{e}-T_{w})\\248* 0.385(T_{c}-39.9) = 390*4.184(39.9-22.6)\\95.48(T_{c}-39.9) = 1631.76*17.3\\95.48(T_{c}-39.9) = 28229.448[/tex]

Divide both side by 95.48

[tex]T_{c} - 39.9 = \frac{28229.448}{95.48}\\T_{c} - 39.9 = 295.658[/tex]

Collect like terms

[tex]T_{c} = 295.658 + 39.9[/tex]

Therefore, the initial temperature of the piece of copper is 335.6 °C.

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Answer:

The period and frequency of this pendulum are 2.457 s and 0.407 Hz.

Explanation:

Given that,

Length = 1.5 m

Angle = 20°

We need to calculate the period

Using formula of period

[tex]T = 2\pi\sqrt{\dfrac{L}{g}}[/tex]

Where, T = time period

g = acceleration due to gravity

l = length

Put the value into the formula

[tex]T=2\times3.14\sqrt{\dfrac{1.5}{9.8}}[/tex]

[tex]T=2.457\ sec[/tex]

We need to calculate the frequency

[tex]T = \dfrac{1}{f}[/tex]

[tex]f=\dfrac{1}{T}[/tex]

Put the value of T

[tex]f=\dfrac{1}{2.457}[/tex]

[tex]f =0.407\ Hz[/tex]

Hence, The period and frequency of this pendulum are 2.457 s and 0.407 Hz.

Answer:

The power of the motor is 153000 watts.

Explanation:

It is given that,

Mass of the car, m = 1700 kg

Initially, it is at rest, u = 0

Final velocity of the car, v = 30 m/s

Time taken, t = 5 s

We need to find the power of a motor. Work done per unit time is called power of the motor. We know that the change in kinetic energy is equal to the work done i.e.

[tex]P=\dfrac{W}{t}=\dfrac{\Delta E}{t}[/tex]

[tex]P=\dfrac{\dfrac{1}{2}mv^2}{t}[/tex]

[tex]P=\dfrac{\dfrac{1}{2}\times 1700\ kg\times (30\ m/s)^2}{5\ s}[/tex]

P = 153000 watts

So, the power of the motor is 153000 watts. Hence, this is the required solution.

Probably not. The elevator has security system just like any device. It will automatically trigger breaks in case it detects some sort of free fall.

Person inside free falling elevator could experience 0G or no gravity effect something similar to what astronauts experience in earth's orbit.

The only damage that could happen to you is when the breaks are released you won't feel 0G anymore and fall about 4ft to the floor of the elevator.

Hope this helps.

r3t40

Explanation:

It is given that,

Spring constant, k = 81 N/m

We need to find the force required to :

(a) Compress the spring by 6 cm i.e. x₁ = 6 cm = -0.06 m

It can be calculated using Hooke's law as :

F = - k(-x₁)

[tex]F=81\ N/m\times 0.06\ m[/tex]

F = 4.86 N

(b) Expand the spring by 17 cm i.e. x₂ = 17 cm = +0.17 m

So, F = -kx₂

[tex]F=-81\ N/m\times 0.17\ m[/tex]

F = -13.77 N

Hence, this is the required solution.

Final answer:

The force required to compress a spring with a spring constant of 81 N/m by 6 cm is 4.86 N, and the force required to expand the same spring by 17 cm is 13.77 N.

Explanation:

The force required to compress or expand a spring can be determined by Hooke's Law, which states that the force (F) exerted by a spring is directly proportional to the displacement (x) from its equilibrium position, and is given by the equation F = kx, where k is the spring constant.

To calculate the force required to:

Compress the spring by 6 cm:

First convert 6 cm to meters (6 cm = 0.06 m). Then apply Hooke's Law: F = kx = 81 N/m times 0.06 m = 4.86 N.

Expand the spring by 17 cm:

First convert 17 cm to meters (17 cm = 0.17 m). Then apply Hooke's Law: F = kx = 81 N/m times 0.17 m = 13.77 N.

In both cases, the magnitude of force is reported, as the question specifies, ignoring the sign which indicates the direction of the force.

Answer:

The current in the heater is 12.5 A

Explanation:

It is given that,

Power of electric heater, P₁ = 1400 W

Power of toaster, P₂ = 1150 W

Power of electric grill, P₃ = 1560 W

All three appliances are connected in parallel across a 112 V emf source. We need to find the current in the heater. We know that in parallel combination of resistors the current flowing in every branch of resistor divides while the voltage is same.

Electric power, [tex]P_1=V\times I_1[/tex]

[tex]I_1=\dfrac{P_1}{V}[/tex]

[tex]I_1=\dfrac{1400\ W}{112\ V}[/tex]

[tex]I_1=12.5\ A[/tex]

So, the current in the heater is 12.5 A. Hence, this is the required solution.  

Answer:

The charge passing through a battery is [tex]1.22\times10^{5}\ C[/tex].

Explanation:

Given that,

Current = 7.2 A

Time = 4.7 hours

We need to calculate the charge

The charge is the product of current and time.

Using formula of charge

[tex]Q=it[/tex]

Where, Q = charge

i = current

t = time

Put the value into the formula

[tex]Q=7.2\times4.7\times3600[/tex]

[tex]Q=1.22\times10^{5}\ C[/tex]

Hence, The charge passing through a battery is [tex]1.22\times10^{5}\ C[/tex].

Answer:

Option (a)

Explanation:

If a body is thrown upwards, it's velocity goes on decreasing with constant rate. It is because an acceleration is acting on the body which is equal to acceleration due to gravity and acting downwards. The value of acceleration due to gravity is constant and always acting downwards.

The reaction quotient (Q) for the given reaction, calculated using initial concentrations of the reactants and the product, is approximately 0.092. This value suggests the reaction will move forward, producing more NH3 to reach equilibrium.

The reaction quotient, commonly referred to as 'Q', is a value used to determine the direction in which a reaction will proceed. It is calculated similarly to the equilibrium constant but uses the initial concentrations instead. For this reaction, the equation for Q would be [NH3]^2 / ([N2] * [H2]^3) based on the balanced chemical equation.

The initial concentrations given in the question are 0.50 M for N2, 3.00 M for H2, and 1.98 M for NH3. To find Qc, substitute these concentrations into our Q equation to get (1.98)^2 / (0.50 * 3.00^3), which simplifies to approximately 0.092.

If Qc < Kc, the reaction will proceed in the forward direction to reach equilibrium, so in this case, since our Qc (0.092) is less than Kc (0.291), the reaction will produce more NH3 to reach equilibrium.

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Explanation:

If the distance between the bottom of the ladder and the wall is x, then:

cos θ = x / 10

Taking derivative with respect to time:

-sin θ dθ/dt = 1/10 dx/dt

Substituting for θ:

-sin (acos(x / 10)) dθ/dt = 1/10 dx/dt

Given that x = 6 and dx/dt = 1.1:

-sin (acos(6/10)) dθ/dt = 1/10 (1.1)

-0.8 dθ/dt = 0.11

dθ/dt = -0.1375

The angle is decreasing at 0.1375 rad/s.

Answer:

The time is 5.21 s.

Explanation:

Given that,

Force F = 646 N

Mass m = 82 kg

Velocity v = 41 m/s

We need to calculate the acceleration

Using formula of force

F= ma

[tex]a = \dfrac{F}{m}[/tex]

[tex]a = \dfrac{646}{82}[/tex]

[tex]a =7.87\ m/s^2[/tex]

We need to calculate the time

Now, using equation of motion

[tex]v = u+at[/tex]

[tex]41=0+7.87 t[/tex]

[tex]t = \dfrac{41}{7.87}[/tex]

[tex]t = 5.21\ s[/tex]

Hence, The time is 5.21 s.

Answer:

Number of electrons, [tex]n=2.7\times 10^{19}[/tex]

Explanation:

It is given that,

Charge, q = 4.33 C

We need to find the number of electrons that make 4.33 C of charge. According to quantization of charge as :

[tex]q=ne[/tex]

n = number of electrons

e = electron's charge

[tex]n=\dfrac{q}{e}[/tex]

[tex]n=\dfrac{4.33\ C}{1.6\times 10^{-19}\ C}[/tex]

[tex]n=2.7\times 10^{19}[/tex]

So, the number of electrons are [tex]2.7\times 10^{19}[/tex] Hence, this is the required solution.  

Answer:

The correct answer is :A and C

Explanation:

According to the rule

1- Zeroes at the beginning of a number are never significant.

2- Zeroes at the end of a number are significant if there is a decimal point.

(a) 0.00012 s and 12000 s

0.00012 has 2 significant figure and 12000 has 2 significant figures

(b) 11.0 m and 11.00 m

11.0 has 3 significant figures and 11.00 has 4 significant figure.

(c)  0.0250 m and 0.205

0.0250 has 3 significant figure and 0.250 has 3 significant figures

(d) 250.0 L and 2.5×10−2L

250.0 has 4 significant figures and 2.5x10−2 has 2 sig figures.

So (a) and (c) pairs have same number of sig figures.

Final answer:

The pair out of the provided options that contains the same number of significant figures is 0.0250 m and 0.205 m, both with three significant figures.

Explanation:

When comparing significant figures in different numbers, we must apply the rules for determining the number of significant figures to each number.

0.00012 s has two significant figures (the leading zeros are not significant).

12000 s likely has two significant figures (unless further decimal places are indicated or if it's written in scientific notation, such as 1.2×10⁴ which would have three significant figures).

11.0 m and 11.00 m have three and four significant figures respectively (trailing zeros after a decimal are significant).

0.0250 m has three significant figures (leading zeros are not significant, but trailing zeros after a decimal are).

0.205 m has three significant figures (leading zeros are not significant).

250.0 L has four significant figures (trailing zeros after a decimal are significant).

2.5×10⁻² L has two significant figures (in scientific notation, all digits are significant).

Therefore, the pairs that contain the same number of significant figures are 0.0250 m and 0.205 m, both with three significant figures.

Answer:

Upward

Explanation:

For charged particles immersed in an electric field:

- if the particle is positively charged, the direction of the force is the same as the direction of the electric field

- if the particle is negatively charged, the direction of the force is opposite to the direction of the electric field

In this problem, we have an electron - so a negatively charged particle - so the direction of the force is opposite to that of the electric field.

Since the electric field is directed downward, therefore, the electric force on the electron will be upward.

Answer:

The ball to hit the ground in 0.53 s.

Explanation:

Given that,

Height = 1.4 m

Time t = 0.53 s

Initial speed = 35 m/s

We need to calculate the time when the ball to hit the ground

Using equation of motion

[tex]s_{y}=u_{y}t-\dfrac{1}{2}gt^2+h_{0}[/tex]

Where, s= vertical height

u= vertical velocity

t = time

h = height

Put the value in equation

[tex]0=0-\dfrac{1}{2}\times9.8\times t^2+1.4[/tex]

[tex]t^2=\dfrac{1.4}{4.9}[/tex]

[tex]t=\sqrt{\dfrac{1.4}{4.9}}[/tex]

[tex]t=0.53\ s[/tex]

Hence, The ball to hit the ground in 0.53 s.

Final answer:

A baseball thrown horizontally with an initial speed will take the same amount of time to fall as a dropped object from the same height, which is 0.53 seconds, because the horizontal speed does not affect the vertical fall time.

Explanation:

The time it takes for an object to fall from a height solely depends on the force of gravity and the initial vertical speed. Since the baseball is thrown horizontally with an initial speed of 35 m/s, this speed does not affect the vertical fall time. The ball will hit the ground in the same duration as any object dropped from the same height without any initial vertical velocity, provided that air resistance is negligible. The previously stated object took 0.53 s to fall from a height of 1.4 m, therefore the baseball will also take 0.53 seconds to hit the ground.

Answer:

2.76 N

Explanation:

m = mass of the object = 1.5 kg

v₀ = initial velocity at t = 0, = 0 i + 5 j

v = final velocity of the object at t = 5, = 6 i + 12 j

t = time interval = 5 sec

a = acceleration of the object = ?

Acceleration of the object is given as

[tex]a = \frac{v - v_{o}}{t}[/tex]

inserting the values

a = ((6 i + 12 j) - (0 i + 5 j))/5

a = (6 i + 7 j)/5

a = 1.2 i + 1.4 j

magnitude of the acceleration is given as

|a| = √((1.2)² + (1.4)²)

|a| = 1.84 m/s²

magnitude of the resultant force is given as

|F| = m |a|

|F| = (1.5) (1.84)

|F| = 2.76 N

To find the magnitude of the resultant force, we use Newton's second law of motion. Evaluating the acceleration at 2.0s gives a magnitude of 24.8 m/s^2. Using the formula F = ma, the magnitude of the resultant force is 37.2 N.

To find the magnitude of the resultant force acting on the object during the time interval, we need to use Newton's second law of motion, which states that the force is equal to the mass of the object multiplied by its acceleration.

First, we need to find the acceleration of the object. We can use the formula:

a(t) = 5.0i + 2.0tj - 6.0t^2 km/s^2

By evaluating a(2.0 s), we get a magnitude of 24.8 m/s^2.

Now, we can use Newton's second law:

F = ma

Substituting the values, we get:

F = 1.5 kg * 24.8 m/s^2

F = 37.2 N

Therefore, the magnitude of the resultant force acting on the object during this time interval is 37.2 N.

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The time it takes a beam of light to get from the Sun to the Earth is [tex]4.99\times 10^2secs[/tex]

The formula for calculating the average speed is expressed according to the formula:

[tex]Speed= \frac{distance}{time}[/tex]

Given the speed of light as [tex]3.0 \times 10^8m/s[/tex]

Distance from earth to the sum is [tex]149.6\times 10^9m[/tex]

Substitute the given parameters into the formula to get the time "t"

[tex]t=\frac{d}{t} \\t=\frac{149.6\times10^9m}{3.0\times10^8}\\t=49.87\times 10\\t =498.7secs[/tex]

Hence the time it takes a beam of light to get from the Sun to the Earth is [tex]4.99\times 10^2secs[/tex]

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(a) -0.211 m

At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is

[tex]\theta=65.4^{\circ}[/tex]

The projection of the length of the pendulum along the vertical direction is

[tex]L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm[/tex]

the full length of the pendulum when the mass is at the lowest position is

L = 36.1 cm

So the y-displacement of the mass is

[tex]\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m[/tex]

(b) 0.347 J

The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

[tex]\Delta U = mg \Delta y[/tex]

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

g = 9.8 m/s^2 is the acceleration due to gravity

[tex]\Delta y = 0.211 m[/tex] is the vertical displacement of the pendulum

So, the work done by gravity is

[tex]W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J[/tex]

And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.

(c) Zero

The work done by a force is:

[tex]W=Fd cos \theta[/tex]

where

F is the magnitude of the force

d is the displacement

[tex]\theta[/tex] is the angle between the direction of the force and the displacement

In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula

[tex]\theta=90^{\circ}, cos \theta = 0[/tex]

and so the work done is zero.

The pendulum falls 0.212 m, the work done by gravity is 0.349 J, and the work done by string tension is 0 J.

The first part of the question asks for the vertical displacement (y-displacement) of the pendulum. The length of the pendulum is the hypotenuse of a right triangle, and the vertical displacement is the adjacent side, so we can use the cosine function to solve: y = L*cos(θ). Plugging in the given values: y = 0.361 m * cos(65.4 degrees) = 0.149 m. So the fall is the length of the pendulum minus this displacement: 0.361 m - 0.149 m = 0.212 m.

The second part of the question asks for the work done by gravity. The work done by gravity is equal to the weight of the pendulum times the vertical distance it falls (Work = m*g*y), or 0.168 kg * 9.8 m/s² * 0.212 m = 0.349 J.

The final part of the question asks for the work done by string tension. The tension force always acts perpendicular to the direction of displacement, meaning it does no work on the pendulum, as work is defined as force times the displacement in the direction of the force. Therefore, the work done by the tension in the string is 0 J.

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Answer:

[tex]1.22\cdot 10^3 L[/tex]

Explanation:

We can solve the problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas is directly proportional to its temperature:

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where here we have

[tex]V_1 = 1.09\cdot 10^3 L[/tex] is the initial volume

[tex]T_1 = 23^{\circ}+ 273 = 296 K[/tex] is the initial temperature

[tex]V_2[/tex] is the final volume

[tex]T_2 = 59^{\circ}+ 273 =332 K[/tex] is the final temperature

Solving for V2, we find

[tex]V_2 = \frac{V_1 T_2}{T_1}=\frac{(1.09 \cdot 10^3 L)(332 K)}{296 K}=1.22\cdot 10^3 L[/tex]

Answer:

The magnetic field produced by a long straight current-carrying wire is directly proportional to the current in the wire and inversely proportional to the distance from the wire.

Explanation:

The magnetic field produced by a long straight current-carrying wire is given by :

[tex]B=\dfrac{\mu_0I}{2\pi d}[/tex]............(1)

Where

[tex]\mu_o[/tex] = permeability of free space, [tex]\mu_o=4\pi\times 10^{-7}\ T-m/A[/tex]

I = current flowing in the wire

d = distance from wire

From equation (1), it is clear that the magnetic field produced by a long straight current-carrying wire is directly proportional to the current flowing and inversely proportional to the distance from the wire. So, the correct option is (d).

The magnetic field produced by a long straight current-carrying wire is proportional to the current in the wire and inversely proportional to the distance from the wire, demonstrated by Ampere's Law and the right-hand rule.

The magnetic field produced by a long straight current-carrying wire is D) proportional to the current in the wire and inversely proportional to the distance from the wire. This relationship is described by Ampere's Law and the formula B = μI/2πr, where B is the magnetic field strength, μ is the permeability of free space, I is the current in the wire, and r is the distance from the wire. This suggests that as the current increases, the magnetic field strength increases, and as the distance from the wire increases, the magnetic field strength decreases.

Furthermore, the direction of the magnetic field is given by the right-hand rule. If you point the thumb of your right hand in the direction of the current, your fingers will curl in the direction of the magnetic field loops, which form concentric circles around the wire.

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Answer:

The current will be 3.23 A.

Explanation:

Given that,

Current I = 4.68 A

Voltage V = 220 volt

We need to calculate the resistance

Using ohm's law

[tex]V = I R[/tex]

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

Where,

V = voltage

I = current

R = resistance

Put the value into the formula

[tex]R = \dfrac{220}{4.68}[/tex]

[tex]R = 47\ \Omega[/tex]

We need to calculate the current

If the voltage drops by 31%

Voltage will be

[tex]V'=V-V\times31%[/tex]

[tex]V'=220-220\times\dfrac{31}{100}[/tex]

[tex]V'=151.8\ volt[/tex]

Now, the current will be

[tex]I = \dfrac{151.8}{47}[/tex]

[tex]I=3.23\ A[/tex]

Hence, The current will be 3.23 A.