The sound from a trumpet radiates uniformly in all directions in air with a temperature of 20∘C. At a distance of 5.18 m from the trumpet the sound intensity level is 54.0 dB . The frequency is 561 Hz .

What is the pressure amplitude at this distance?

Final answer:

The magnitude of an electric field in which the electric force equals a proton's weight is calculated to be approximately 102 N/C, considering the proton's charge and mass along with Earth's gravitational acceleration.

Explanation:

To determine the magnitude of an electric field in which the electric force equals a proton's weight, we must equate the electric force to the gravitational force (weight) acting on the proton. The electric force (FE) experienced by a charge in an electric field (E) is given by FE = qE, where q is the charge of the proton. The weight of the proton (W) can be found using W = mg, where m is the mass of the proton and g is the acceleration due to gravity (approximately 9.81 m/s2 on Earth). The charge of a proton is approximately 1.6 x 10-19 C, and its mass is approximately 1.67 x 10-27 kg.

Setting FE equal to W, we have qE = mg. Solving for E, the electric field magnitude, gives us E = mg/q. Thus, substituting the known values, we find:

E = (1.67 x 10-27 kg * 9.81 m/s2) / 1.6 x 10-19 C,

this gives us E approximately equal to 102 N/C. This is the magnitude of the electric field in which the electric force on a proton is equal to the proton's weight.

Final answer:

The magnitude of the electric field can be found by setting the electric force equal to the proton's weight and solving for the electric field. The magnitude of the electric field in this case is approximately 1.03 x 10¹² N/C.

Explanation:

The magnitude of the electric field in which the electric force it exerts on a proton is equal in magnitude to the proton's weight can be found by setting the two forces equal to each other:

FE = mg

Where FE is the electric force, m is the mass of the proton, and g is the acceleration due to gravity. Since the proton's mass is known to be approximately 1.67 x 10⁻²⁷ kg, and the acceleration due to gravity is approximately 9.8 m/s², we can calculate the magnitude of the electric field as follows:

E = mg/q

Where q is the charge of the proton, which is approximately 1.6 x 10⁻¹⁹ C. Plugging in the values:

E = (1.67 x 10⁻²⁷ kg) x (9.8 m/s²) / (1.6 x 10⁻¹⁹ C)

E ≈ 1.03 x 10¹² N/C

Answer:

    The increase in the internal energy = 350 J

Explanation:

Given that

Q= 275  J

W= - 125 J

W' = 50 J

W(net)= -125  + 50 = -75 J

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take change in the  internal energy =ΔU

We know that

Q= ΔU + W(net)

275 = ΔU -75

ΔU= 275 + 75 J

ΔU=350 J

The increase in the internal energy = 350 J

The internal energy of the gas is the energy contained by the gas. The increase in the internal energy of the gas in the given container is 350 J.

The internal energy of the gas can be calculated by the formula

Q= ΔU + W(net)

Where,

Q - energy absorbed by the system = 275  J

ΔU - internal energy of the gas= ?

W(net)=  W - W' = -125  + 50 = -75 J

Put the values in the formula,

275 = ΔU -75

ΔU= 275 + 75 J

ΔU=350 J

Therefore, the increase in the internal energy of the gas in the given container is 350 J.

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Answer:

[tex]\lambda_w=0.6509\ m[/tex]

Explanation:

Given:

Now the angular frequency of the spring oscillation:

[tex]\omega=\sqrt{\frac{k}{m} }[/tex]

[tex]\omega=\sqrt{\frac{9.3}{13.9} }[/tex]

[tex]\omega=0.81796\ rad.s^{-1}[/tex]

Now according to the question the wave is created after each cycle of the spring oscillation.

So the time period of oscillation:

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

[tex]T=\frac{0.81796}{2\pi}[/tex]

[tex]T=0.130182\ s[/tex]

Now the wave length of the water wave:

[tex]\lambda_w=v_w.T[/tex]

[tex]\lambda_w=5\times 0.130182[/tex]

[tex]\lambda_w=0.6509\ m[/tex]

Answer:

Wavelength will be 38.388 m

Explanation:

We have given mass m = 13.9 kg

Spring constant K= 9.3 N/m

Velocity v = 5 m /sec

Angular frequency is given by [tex]\omega =\sqrt{\frac{k}{m}}[/tex]

So [tex]\omega =\sqrt{\frac{9.3}{13.9}}=0.817[/tex]

Now we have to find frequency for further calculation

So frequency will be equal to [tex]f=\frac{\omega }{2\pi }=\frac{0.817}{2\times 3.14}=0.130Hz[/tex]

Now we have to find wavelength, it is ratio of velocity and frequency

There is a relation between frequency velocity and wavelength

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

[tex]\lambda =\frac{v}{f}=\frac{5}{0.130}=38.388m[/tex]

Answer:

The second ball

Explanation:

Both balls are under the effect of gravity, accelerating with exactly the same value. The first ball is dropped, therefore its initial velocity is zero. Since the second ball has horizontal and vertical velocity components, its initial velocity is given by:

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]

The vertical component is zero, however, it has a horizontal velocity, so its initial speed is not zero, therefore the secong ball has the greater speed at ground level.

The second ball will reach a greater total speed at ground level due to its initial horizontal speed added to its vertical speed. But, both balls will hit the ground at the same time if the horizontal speed of the second ball is not too high.

In the described situation, gravity is the only vertical force acting on both balls, hence they will reach the ground at the same speed in the vertical direction. But, the second ball has additional horizontal speed. Therefore, with this horizontal component added to the vertical detachment speed due to gravity, the second ball will have a greater total speed by the Pythagorean theorem: (total speed)^2 = (vertical speed)^2 + (horizontal speed)^2. However, the question of which hits the ground first depends on the initial horizontal speed of the second ball. If the horizontal speed is not too high, both balls should hit the ground at the same time.

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Answer:

Explanation:

Uniform polarization of the heart membranes is associated with shocks, when passing a large current through the heart fail to cure. Therefore, the correct option is option A.

The heart is an organ that acts as a blood pump. In spiders as well as annelid worms, it is a straight tube. In mollusks, it is a little more complex structure with one or even more collecting chambers (atria) and a primary pumping chamber (ventricle).

The heart of fish is indeed a folded tube with three to four enlarged regions that resemble the chambers inside the heart of a mammal. Uniform polarization of the heart membranes is associated with shocks, when passing a large current through the heart fail to cure.

Therefore, the correct option is option A.

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Answer:

540.8m/s

Explanation:

From the information giving, the total energy is conserved and the momentum is conserved.

To determine the speed of the ball after the collision, we use the energy conservation rule, I.e

Kinetic Energy of ball after collision = energy to rise to attain height

1/2mv²=mgh

where m,mass of ballistic pendulum=1.5kg,

v=velocity of ballistic pendulum after collision,

g=gravitational acceleration

h=height attain=11cm=0.11m

if we substitute we arrive at

v=√(2gh)

v=√(2*9.8*0.11)

v=1.47m/s.

since we have determine the velocity of the ballistic pendulum after collision, we now use conservation of momentum to determine the initial speed of the bullet.

since

initial momentum=final momentum

mₓ₁vₓ₁+mₐ₁vₐ₁=mₓ₂vₓ₂+mₐ₂vₐ₂

were mₓ₁vₓ₁,mₓ₂vₓ₂ =mass and velocity of ballistic pendulum before and after collision

mₐ₁vₐ₁,mₐ₂vₐ₂=mass and velocity of bullet before and after collision

if we substitute values,we arrive at

(1.5kg*0m/s)+(0.0057kg*vₐ₁)=(1.5kg*1.47m/s)+(0.0057kg*154m/s)

vₐ₁= 3.0828/0.0057

vₐ₁=540.8m/s

Answer:

0.47 m/s²

Explanation:

Assuming the string is inelastic, m₃ will accelerate downward at a rate of -a, and m₄ will accelerate upward at a rate of +a.

Draw a four free body diagrams, one for each hanging mass and one for each wheel.

For m₃, there are two forces: weight force m₃g pulling down, and tension force T₃ pulling up.  Sum of forces in the +y direction:

∑F = ma

T₃ − m₃g = m₃(-a)

For m₄, there are two forces: weight force m₄g pulling down, and tension force T₄ pulling up.  Sum of forces in the +y direction:

∑F = ma

T₄ − m₄g = m₄a

For m₁, there are two forces: tension force T₃ pulling down, and tension force T pulling right.  Sum of the torques in the counterclockwise direction:

∑τ = Iα

T₃r₃ − Tr₃ = (m₁r₃²) (a/r₃)

T₃ − T = m₁a

For m₂, there are two forces: tension force T₄ pulling down, and tension force T pulling left.  Sum of the torques in the counterclockwise direction:

∑τ = Iα

Tr₄ − T₄r₄ = (m₂r₄²) (a/r₄)

T − T₄ = m₂a

We now have 4 equations and 4 unknowns.  Let's add the third and fourth equations to eliminate T:

(T₃ − T) + (T − T₄) = m₁a + m₂a

T₃ − T₄ = (m₁ + m₂) a

Now let's subtract the second equation from the first:

(T₃ − m₃g) − (T₄ − m₄g) = m₃(-a) − m₄a

T₃ − m₃g − T₄ + m₄g = -(m₃ + m₄) a

T₃ − T₄ = (m₃ − m₄) g − (m₃ + m₄) a

Setting these two expressions equal:

(m₁ + m₂) a = (m₃ − m₄) g − (m₃ + m₄) a

(m₁ + m₂ + m₃ + m₄) a = (m₃ − m₄) g

a = (m₃ − m₄) g / (m₁ + m₂ + m₃ + m₄)

Plugging in values:

a = (1.64 kg − 1.27 kg) (9.8 m/s²) / (2.5 kg + 2.3 kg + 1.64 kg + 1.27 kg)

a = 0.47 m/s²

The difference in torque and mass applied determines the acceleration of

the system.

The acceleration is approximately 0.4703 m/s²

Reasons:

The mass of the left wheel = 2.5 kg

Radius of the left wheel = 24.03 cm

Mass of the right wheel = 2.3 kg

Radius of the right wheel = 31.38 cm

Mass on the left = 1.64 kg

Mass on the right = 1.27 kg

Acceleration due to gravity, g ≈ 9.8 m/s²

Required:

The acceleration of the system

Solution:

The given acceleration of the

T₃ - m₃·g = m₃·(-a)...(1)

T₄ - m₄·g = m₄·a...(2)

[tex]T_3 \cdot r_1 - T \cdot r_1 = I \cdot \alpha = m_1 \cdot r_1^2 \times \dfrac{a}{r_1} = \mathbf{m_1 \cdot r_1 \cdot a}[/tex]

T₃ - T = m₁·a...(3)

[tex]T \cdot r_2 - T_4 \cdot r_2 = \mathbf{ I \cdot \alpha} = m_2 \cdot r_2^2 \times \dfrac{a}{r_2} = m_2 \cdot r_2 \cdot a[/tex]

T - T₄ = m₂·a...(4)

Add equation (3) to equation (4) gives;

T₃ - T + (T - T₄) = T₃ - T₄ = m₁·a + m₂·a

Subtracting equation (2) from equation (1) gives;

(T₃ - m₃·g) - (T₄ - m₄·g) = m₃·a + m₄·a

T₃ - T₄ = -m₃·a - m₄·a - (m₄·g - m₃·g)

Which gives;

m₁·a + m₂·a = -m₃·a - m₄·a - (m₄·g - m₃·g)

a·(m₁ + m₂) = -a·(m₃ + m₄) - (m₄·g - m₃·g)

-(m₄·g - m₃·g) = (m₃·g - m₄·g) = a·(m₃ + m₄) + a·(m₁ + m₂) = a·(m₃ + m₄ + m₁ + m₂)

[tex]a = \mathbf{\dfrac{m_3 \cdot g - m_4 \cdot g}{(m_3 + m_4 + m_1 - m_2)}}[/tex]

[tex]a = \dfrac{1.64 \times 9.8 - 1.27\times 9.8}{(1.27 +1.64 + 2.5 + 2.3)} \approx 0.4703[/tex]

The acceleration is a ≈ 0.4703 m/s²

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Answer:

The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.

Explanation:

Hi there!

The position of the patrol car at a time "t" can be calculated using this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the patrol car at a time "t"

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:

x = x0 + v · t

Where "v" is the constant velocity.

First, let´s convert the velocity units into m/s:

140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s

105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s

We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.

When the patrol car catches the speeder, the position of both cars is the same:

position of the patrol car = position of the speeder

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:

x = v · t

x = 38.9 m/s · 1 s = 38.9 m

When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t

Let´s agrupate terms and equalize to zero:

-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0

-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0

Solving the quadratic equation for t using the quadratic formula:

t = 8.24 s  (the other solution is discarded because it is negative)

The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.

Answer:

The detailed explanations is attached below

Explanation:

What is applied is the De brogile equation and the equation showing a relationship between Energy, speed of light and wavelength.

The explanation is as attached below.

The weight of the measuring stick is balanced by the 1-kg rock suspended from it at the 0.75m mark according to the principle of torques. This results in the weight of the measuring stick being approximately 13.07 Newtons.

The problem you've asked falls right into the area of physics related to torques and forces. The 1-kg rock generates a force of approximately 9.8 N downwards (using the approximate gravity strength of 9.8 m/s^2). According to the principle of torques, for the stick to be balanced, the force at the 0.75m mark (1 - 0.25 = 0.75m), which is the weight of the stick, should balance the total torque generated by the weight of the rock. That means the weight of the stick (in Newtons) times 0.75 should equal the weight of the rock times 1. This gives us the equation: 0.75*W = 1*9.8, where W is the weight of the stick. Solving this gives us the weight of the stick as approximately 13.07 N.

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Answer:

(a) t = 2.97s

(b) h = 43.3 m

Explanation:

Let t be the time it takes to fall a distance h, then t - 1 (s) is the time it takes to fall a distance of h - 0.56h = 0.44 h

For the ball to fall from rest a distance of h after time t

[tex]h = gt^2/2[/tex]

Also for the ball to fall from rest a distance of 0.44h after time (t-1)

[tex]0.44h = g(t-1)^2/2[/tex]

We can substitute the 1st equation into the 2nd one

[tex]0.44gt^2/2 = g(t-1)^2/2[/tex]

and divide both sides by g/2

[tex]0.44t^2 = (t-1)^2[/tex]

[tex]0.44t^2 = t^2 - 2t + 1[/tex]

[tex]0.56t^2 - 2t + 1 = 0[/tex]

[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{2\pm \sqrt{(-2)^2 - 4*(0.56)*(1)}}{2*(0.56)}[/tex]

[tex]t= \frac{2\pm1.33}{1.12}[/tex]

t = 2.97 or t = 0.6

Since t can only be > 1 s we will pick t = 2.97 s

(b) [tex]h = gt^2/2 = 43.3 m[/tex]

Answer:

A. [tex]KE_h=\frac{1}{16}\times KE_l[/tex]

B. the lighter block travels at a speed 4 times faster than the heavier block.

C. b The heavy block must be pushed 4 times farther than the light block.

Explanation:

Given that the blocks are  acted upon by equal force.

A.

Then the kinetic energy of the blocks depends on their individual velocity.

And velocity is related to momentum through Newton's second law of motion:

[tex]\frac{d}{dt} .p=F[/tex]

[tex]\frac{d}{dt} (m.v_l)=F[/tex]

considering that the time for which the force acts on each mass is equal.

[tex]dv_l=\frac{F.dt}{m}[/tex]

For the heavier block:

[tex]dv_{_h} =\frac{F.dt}{4m}[/tex]

Therefore:

Kinetic energy of lighter block:

[tex]KE_l=\frac{1}{2}\times m.(\frac{F.dt}{m} )^2[/tex]

[tex]KE_l=\frac{1}{2m} \times (F.dt)^2[/tex]

Kinetic energy of heavier block:

[tex]KE_h=\frac{1}{2} \times m.(\frac{F.dt}{4m} )^2[/tex]

[tex]KE_h=\frac{1}{16}\times (\frac{1}{2m} \times (F.dt)^2)[/tex]

[tex]KE_h=\frac{1}{16}\times KE_l[/tex]

B.

From the above calculations and assumptions we observe that the lighter block travels at a speed 4 times faster than the heavier block.

C.

Since the lighter block is having the speed 4 times more than the heavier block so it must be pushed 4 times farther because the speed is directly proportional to the  distance.

A) The statement that is true from the attached link about the kinetic energy of the heavier block after the push is;

B: the kinetic energy of the heavier block is equal to the kinetic energy of the lighter block

B) Compared to the speed of the heavierblock, the speed of the light block travel is; twice as fast.

C) If we assume that both blocks have the same speed after being pushed with the same force F_vec, what we can say about the distances the two blocks are pushed is;

The heavyblock must be pushed 4 times farther than the lightblock.

A) The formula for the work done on each block is gotten from the formula;

W = F × d

We are told that the two blocks were pushed the same distance and with the same force F_vec being exerted on them. Thus, the work done on each of the blocks will be the same.

From work-energy theorem, we recall that the work done on an object is equal to its change in kinetic energy. Thus;

W = ΔKE

Where;

ΔKE = Final KE - Initial KE

Thus;

W = Final KE - Initial KE

Since the blocks were initially at rest, then it means that;

Initial KE = 0

Thus;

W = Final KE - 0

W = Final KE

The work done on each block is the same and as a result their final kinetic energies will be the same.

B) We are told that one of the blocks is ¼ times as heavy as the other block.

Thus;

½(m)v_h² = ½(¼mv_l²)

v_h² = ¼v_l²

Where;

v_h is speed of heavy block

v_l is speed of light block

Thus, taking square root of both sides gives;

v_h = ½v_l

Thus, the speed of the light block is twice as fast.

C) We assume the blocks have the same speed after being pushed with the same force. Thus;

It means that heavier block must be pushed four times farther than lighter block.

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Answer:

v = 26.7 mph

Explanation:

During the first 5 hours, at a constant speed of 20 mph, we find the total displacement to be as follows:

Δx₁ = v₁*t₁ = 20 mph*5 h = 100 mi

Assuming we can neglect the displacement during the speeding up from 20 to 60 mph, we can find the the total displacement at 60 mph as follows:

Δx₂ = v₂*t₂ = 60 mph*1 h = 60 mi

So, the total displacement during all the trip wil be:

Δx = Δx₁ + Δx₂ = 100 mi + 60 mi = 160 mi

So we can find the the average velocity during the 6-hour period, applying the definition of average velocity, as follows:

v = Δx / Δt = 160 mi / 6 h = 26.7 mph

Answer:

Frequency, [tex]f=6.57\times 10^{15}\ Hz[/tex]

Explanation:

It is given that, the electron moves in a circular orbit of radius 0.053 nm around a stationary proton. The electric force acting on the electron is balanced by the centripetal force as :

[tex]\dfrac{kq^2}{r^2}=\dfrac{mv^2}{r}[/tex]

v is the speed of electron

[tex]v=\sqrt{\dfrac{ke^2}{mr}}[/tex]

[tex]v=\sqrt{\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{9.1\times 10^{-31}\times 0.053\times 10^{-9}}}[/tex]

[tex]v=2.18\times 10^6\ m/s[/tex]

The speed of electron is given by :

[tex]v=\dfrac{2\pi r}{t}[/tex]

[tex]t=\dfrac{2\pi r}{v}[/tex]

[tex]t=\dfrac{2\pi \times 0.053\times 10^{-9}}{2.18\times 10^6}[/tex]

[tex]t=1.52\times 10^{-16}\ s[/tex]

We know that the number of revolutions per second is called frequency of electron. It is given by :

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

[tex]f=\dfrac{1}{1.52\times 10^{-16}}[/tex]

[tex]f=6.57\times 10^{15}\ Hz[/tex]

So, the total number of revolutions per second make by the electron is [tex]f=6.57\times 10^{15}\ Hz[/tex]. Hence, this is required solution.

Answer:

1.80 Meter

Explanation:

The bottom of a water (liquid) container appears to be raised up in comparison to its actual depth because of the refraction. How shallow will the bottom appear depends on the refractive index of the liquid.

The refractive index(n) and apparent and real depth are related as given below:

[tex]n = \frac{Real Depth}{Apparent Depth}[/tex]

[tex]Apparent Depth = \frac{Real Depth}{n}[/tex]

Given,

n = 1.333

Real depth = 2.40 m

[tex]Apparent Depth = \frac{2.40}{1.33}[/tex]

Thus, Apparent depth = 1.80 Meter

Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

Answer:

225 N

Explanation:

From Pascal's principle,

F/A = f/a ...................... Equation 1

Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.

Making f the subject of the equation,

f = F(a)/A ..................... Equation 2

Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².

Substituting into equation 2

f = 15000(3/200)

f = 225 N.

Hence the downward force that must be applied to small piston = 225 N

The downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN is 225N.

Given that,

Based on the above information, the calculation is as follows:

Here we use the Pascal's principle,

[tex]F\div A = f\div a[/tex] ........... (1)

Here

 F denoted Force exerted on the larger piston,

f denoted force that applied to the smaller piston,

A denoted cross-sectional area of the larger piston,

And, a denoted cross-sectional area of the smaller piston.

Now

[tex]f = 15000 \times (3\div 200)[/tex]

=  225 N

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Answer:

(a). The weight of the system is 336.32 N.

(b).  The molar volume is 16.6 m³/k mol.

The mass based volume is 0.145 m³/kg.

Explanation:

Given that,

Weight of octane = 0.3 kmol

Volume = 5 m³

(a). Molecular mass of octane

[tex]M=114.28\ g/mol[/tex]

We need to calculate the mass of octane

Mass of 0.3 k mol of octane is

[tex]M=114.28\times0.3\times1000[/tex]

[tex]M=34.284\ kg[/tex]

We need to calculate the weight of the system

Using formula of weight

[tex]W=mg[/tex]

Put the value into the formula

[tex]W=34.284\times9.81[/tex]

[tex]W=336.32\ N[/tex]

(b). We need to calculate the molar volume

Using formula of molar volume

[tex]\text{molar volume}=\dfrac{volume}{volume of moles}[/tex]

Put the value into the formula

[tex]\text{molar volume}=\dfrac{5}{0.3}[/tex]

[tex]\text{molar volume}=16.6\ m^3/k mol[/tex]

We need to calculate the mass based volume

Using formula of mass based volume

[tex]\text{mass based volume}=\dfrac{volume}{mass}[/tex]

Put the value into the formula

[tex]\text{mass based volume}=\dfrac{5}{34.284}[/tex]

[tex]\text{mass based volume}=0.145\ m^3/kg[/tex]

Hence, (a). The weight of the system is 336.32 N.

(b).  The molar volume is 16.6 m³/k mol.

The mass based volume is 0.145 m³/kg.

Answer:

Layer of glass rod rubbed with silk.

Explanation:

Some of the atoms in the surface layer of a glass rod positively charged by rubbing it with a silk cloth have lost electrons, leaving a net positive charge because of the unneutralized protons of their nuclei. A negatively charged object has an excess of electrons on its surface.

Final answer:

Conducting materials like metals have freely moving electrons that enable the flow of charge, supported by experimental evidence like studies on conductivity and direct experiments confirming the presence of electrons as charge carriers.

Explanation:

Conducting materials such as metals have freely moving electrically charged particles within them. These free electrons can move through the material allowing the flow of charge. Experimental evidence supporting this idea includes studies on conductivity, Ohm's law for electromagnetics, and direct experiments like the one by Tolman and Stewart confirming that charge carriers in metals are indeed electrons.

Final answer:

When the charge on each of two small spheres is halved, the force of attraction between them, according to Coulomb's Law, will be quartered (Option D). This is because the force is proportional to the product of the charges, which would be reduced to a quarter.

Explanation:

The question is related to Coulomb's Law, which describes the electrostatic force between two charges. According to Coulomb's Law, the force F between two charges is proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance (r^2) between them: F = k * (q1*q2) / r^2, where k is Coulomb's constant.

If the charge on each of the two small spheres is halved, we have new charges q1/2 and q2/2. Substituting these into the formula, we get the new force F' as F' = k * ((q1/2)*(q2/2)) / r^2. Simplifying this, F' = (1/4) * k * (q1*q2) / r^2, which is one quarter of the original force F. Therefore, the force of attraction between the spheres will be quartered when the charge on each sphere is halved.

Answer:

Volume will increase by factor 2

So option (A) will be correct answer  

Explanation:

Let initially the volume is V pressure is P and temperature is T

According to ideal gas equation [tex]PV=nRT[/tex], here n is number of moles and R is gas constant

So [tex]V=\frac{nRT}{P}[/tex]....................eqn 1

Now pressure is doubled and temperature is quadrupled

So new volume [tex]V_{new}=\frac{nR4T}{2P}=\frac{2nRT}{P}[/tex] ........eqn 2

Now comparing eqn 1 nad eqn 2

[tex]V_{new}=2V[/tex]

So volume will increase by factor 2

So option (A) will be correct answer

Answer:

Resultant velocity will be equal to 6.10 m/sec

Explanation:

We have given a motorbike is traveling with 5 m/sec in east

And a current is flowing at a rate of 3.5 m /sec in north

We know that east and north is perpendicular to each other

So resultant velocity will be vector sum of both velocity

So resultant velocity [tex]v=\sqrt{5^2+3.5^2}=6.10m/sec[/tex]

So resultant velocity will be equal to 6.10 m/sec

Answer:

option B

Explanation:

given,

Length of field = 360 ft

                        = 360/3.281 = 109.72 m    ∵ 1 m = 3.281 ft

width of field = 160 ft

                      = 160/ 3.281 = 48.76 m

width of mower = 0.5 m

number of rounds required

   = [tex]\dfrac{48.76}{0.5}[/tex]

   = 97.5

we know,  1 km/h = 0.278 m/s

time taken for each round is equal to

[tex]t = \dfrac{length}{speed}[/tex]

[tex]t = \dfrac{109.72}{0.278}[/tex]

     t = 394.67 s

total time,

T = 394.67 x 94.5 = 37296.91 s

[tex]T = \dfrac{37296.91}{3600}\ hours[/tex]

[tex]T =11\ hours[/tex]

Hence. the correct answer is option B

Answer:

[tex]0.16098\times 10^{-3}\ g/cm^3[/tex]

Explanation:

P =Pressure = 9846 Pa

V = Volume

n = Amount of substance = 1

T = Temperature = 21.1°C

[tex]\rho[/tex] = Density

R = Gas constant = 8.314 J/mol K

M = Molar mass of argon = 40 g/mol

From ideal gas law we have the relation

[tex]PV=nRT[/tex]

Multiply density on both sides

[tex]PV\rho=nR\rho T\\\Rightarrow PM=nR\rho T\\\Rightarrow \rho=\dfrac{PM}{nRT}\\\Rightarrow \rho=\dfrac{9846\times 40\times 10^{-3}}{8.314\times (21.1+273.15)}\\\Rightarrow \rho=0.16098\ kg/m^3\\\Rightarrow \rho=0.16098\times 10^{-3}\ g/cm^3[/tex]

The density of argon gas is [tex]0.16098\times 10^{-3}\ g/cm^3[/tex]

To find the density of argon gas inside a Geiger tube, we use the Ideal Gas Law and convert the given units, plugging in these values yields 1.65 g/L.

The density, ρ, of a gas can be calculated following the Ideal Gas Law formula, which is PV = nRT, where P is pressure, V is volume, n is the number of moles of the gas, R is the universal gas constant, and T is the temperature in Kelvin. To solve for density, we can use the equation ρ = m/V, where m is the mass of the gas and V is the volume. We can also express this in terms of the Ideal Gas Law, leading to ρ = (n×M)/(RT/P), where M is molar mass.

Given the molar mass of Argon is approximately 39.948 g/mol and the universal gas constant R is 8.314462618 J/(mol*K), first convert the temperature from Celsius to Kelvin (T = 21.1 °C + 273.15 = 294.25K), and the pressure from Pascal to atm (1 Pa = 0.00000986923 atm, thus 9846 Pa = 9846× 0.00000986923 = 0.0972 atm).

Plugging the values into the density formula, we get ρ = (1 mol× 39.948 g/mol) / ((8.314462618 J/(mol×K)×294.25K)/ 0.0972 atm) = 1.65 g/L.

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Answer:

[tex]v_{avg}=8.5\ m/s[/tex]

Explanation:

given,

initial speed of the car, v₁ = 17 m/s

final speed of the car, v₂ = 0 m/s

car stops in time = 3 s

we need to calculate average speed

[tex]v_{avg}=\dfrac{v_1 + v_2}{2}[/tex]

[tex]v_{avg}=\dfrac{17 + 0}{2}[/tex]

[tex]v_{avg}=\dfrac{17}{2}[/tex]

[tex]v_{avg}=8.5\ m/s[/tex]

average speed of the car during interval of 3 s is 8.5 m/s

Answer:

α = 9.18 rad/s²

Explanation:

given,

mass of the solid disk = 24.3 Kg

radius of the disk = 0.314 m

Force, F₁ = 90 N

           F₂ = 125 N

net force acting on the disk

F = 125 - 90

F = 35 N

Torque

τ = F . r

τ = 35 x 0.314

τ = 11 N.m

we know that

τ = I α

moment of inertia of the solid disk

[tex]I = \dfrac{1}{2}MR^2[/tex]

[tex]I = \dfrac{1}{2}\times 24.3\times 0.314^2[/tex]

   I = 1.198 kg.m²

now,

11 = 1.198 x α

α = 9.18 rad/s²

the angular acceleration of the disk is equal to 9.18 rad/s²

To find the angular acceleration of the disk, use the formula: angular acceleration = (net torque) / (moment of inertia). Calculating the torques exerted by the forces and the moment of inertia will allow us to find the answer.

The angular acceleration of the disk can be found using the formula:

angular acceleration = (net torque) / (moment of inertia)

where the moment of inertia of a disk is given by:

moment of inertia = (1/2) * mass * radius^2

In this case, the net torque on the disk is the difference between the torques exerted by the two forces:

net torque = torque(top force) - torque(bottom force)

Each torque can be calculated using the formula:

torque = force * radius

Substituting the given values into these formulas, we can find the angular acceleration of the disk.

So, the angular acceleration of the disk is approximately 18.26 rad/s^{2}

.

Answer:

K E = 300 J

Explanation:

given,

Force of pushing crate, F = 100 N

distance of push = 10 m

frictional force = 70 N

kinetic energy is gained by the crate = ?

using work energy theorem

K E = work done

K E  = F . s

K E  = (100 - 70) x 10

K E = 30 x 10

K E = 300 J

kinetic energy is gained by the crate is equal to 300 J.

Final answer:

The kinetic energy gained by the crate is 300 J, calculated as the net work done on the crate, which is the work done by the applied force (1000 J) minus the work done by friction (700 J).

Explanation:

To determine how much kinetic energy is gained by the crate, we can use the work-energy theorem which states that the work done on an object is equal to the change in kinetic energy of the object. In this scenario, the net work done on the crate is the work done by the applied force minus the work done by friction. The work done by the applied force is force times distance, which is 100 N x 10 m = 1000 J. The work done by friction is 70 N x 10 m = 700 J. Therefore, the net work done on the crate, which is also the kinetic energy gained by the crate, is 1000 J - 700 J = 300 J.

Answer: On the basis of speed they are all equivalent.

Yellow light = Fm radio wave = Green light = X-ray = AM radio wave = Infrared wave

Explanation:

Yellow light, Fm radio wave, Green light ,X-ray, AM radio wave and Infrared wave are all electromagnetic waves, and all electromagnetic waves move at the same vacuum speed which is the speed of light and is approximately 3.0x10^8 m/s.

They only differ in wavelength and frequency

c = λf

c (speed of light) = λ (wavelength) x f (frequency)

Therefore; on the basis of speed they are all equivalent.

Yellow light = Fm radio wave = Green light = X-ray = AM radio wave = Infrared wave

All electromagnetic waves, including yellow light, FM radio waves, green light, X-rays, AM radio waves, and infrared waves travel at the same speed in a vacuum, known as the speed of light, which is 3.00 × 10^8 m/s. Therefore, they are equivalent in speed.

The student has asked to rank various types of electromagnetic waves based on their speed in a vacuum. In a vacuum, all electromagnetic waves travel at the same speed, which is the speed of light, and is one of the fundamental constants of nature. The speed of light (c) in a vacuum is 3.00 × 108 meters per second (m/s).

Therefore, the ranking of the electromagnetic waves from fastest to slowest for yellow light, FM radio wave, green light, X-ray, AM radio wave, and infrared wave would be:

Since all of these waves travel at the same speed in a vacuum, they can be considered equivalent in terms of speed. However, they differ in wavelength and frequency.

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Answer:

F = 0.0034 N

Explanation:

Given:

[tex]B = 5.2*10^(-5) T\\Q = 57 degrees\\I_{wire} = 12 A\\L_{wire} = 10 m[/tex]

The angle between B and wire = 90 - 57 = 33 degrees

Using formula:

[tex]F = B*I*L*sin (90-Q)\\F = (5.2*10^(-5)*(12)*(10)*sin (33)\\F = 0.0034 N[/tex]

(A) The magnetic force exerted on the wire is 3.4×10⁻³N

(B) The direction of the force is to the west.

Given that the magnetic field B = [tex]5.2\times10^{-5}T[/tex] which points in the direction  57° below a horizontal line pointing north.

Length of the wire L = 11m

current in the wire I = 11A

The angle between the wire and the magnetic field is θ = (90-57) = 33°

(A) The magnetic force on a finite wire of length L carrying a current I is given by:

[tex]F=BILsin\theta\\\\F=5.2\times10^{-5}\times11\times11\sin33\\\\F=3.4\times10^{-3}N[/tex]

(B) The direction of the force is given by dl×B, now B is at 57° with the north direction and the wire is verticle, so the direction of the field will be to the west.

Learn more about magnetic force:

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Answer:

a)   [tex] = 9.30\times 10^{-3} N[/tex]

b) [tex]= 2.49\times 10^{-3} N[/tex]

Explanation:

Given data;

magnetic field [tex]= 0.60 \times 10^{-4}  T[/tex]

current I = 11 A

length of wire L = 14.6 m

Angle[tex] \theta = 75 A[/tex]

a) Magnetic force due to current

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

 [tex] = 0.60 \times 10^{-4} \times 11 \times 14.6 sin 75^o[/tex]

 [tex] = 9.30\times 10^{-3} N[/tex]

B) magnitude of force due upward current direction

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

  [tex]= 0.60 \times 10^{-4} \times 11 \times 14.6 sin (75^o + 90^o)[/tex]

[tex]= 2.49\times 10^{-3} N[/tex]

The magnetic force on a wire depends on the direction of the current relative to the Earth's magnetic field. In scenario (a), the force is calculated using an angle of 165° between the wire and magnetic field. For (b), the angle is 75°, and the force's magnitude is computed with the sin of that angle.

Magnetic Force on a Current-Carrying Wire

The magnetic force on a current-carrying wire is given by F = I x L x B x sin(θ), where I is the current, L is the length of the wire, B is the magnetic field strength, and θ is the angle between the wire and the direction of the magnetic field.

a) When the current is directed horizontally toward the east and the Earth's magnetic field is pointing 75° below the horizontal in a north-south plane, we can calculate the force using the cross product of the current direction and magnetic field. The angle between the wire and the magnetic field is 90° + 75° = 165°. Therefore, the magnitude of the force is F = 11 A * 14.6 m * 0.60 * 10^-4 T * sin(165°) and its direction is perpendicular to both the current and the magnetic field, following the right-hand rule.

b) If the current is directed vertically upward, the force is perpendicular to the wire and the magnetic field. The angle between the current direction and the magnetic field is now 75°, so the magnitude of the force is F = 11 A * 14.6 m * 0.60 * 10^-4 T * sin(75°), and its direction is again determined by the right-hand rule.