A rock is thrown downward into a well that is 7.92 m deep. Part A If the splash is heard 1.17 seconds later, what was the initial speed of the rock? Take the speed of sound in the air to be 343 m/s.

Answer:

[tex]3.6\cdot 10^{-8} N[/tex]

Explanation:

The electrostatic force between the proton and the electron is given by:

[tex]F=k\frac{q_p q_e}{r^2}[/tex]

where

[tex]k=9.00\cdot 10^9 Nm^2 C^{-2}[/tex] is the Coulomb constant

[tex]q_p = 1.6\cdot 10^{-19} C[/tex] is the magnitude of the charge of the proton

[tex]q_e = 1.6\cdot 10^{-19}C[/tex] is the magnitude of the charge of the electron

[tex]r=8\cdot 10^{-11}m[/tex] is the distance between the proton and the electrons

Substituting the values into the formula, we find

[tex]F=(9\cdot 10^9 ) \frac{(1.6\cdot 10^{-19})^2}{(8\cdot 10^{-11})^2}=3.6\cdot 10^{-8} N[/tex]

Answer:

Ratio of the centripetal accelerations = 16

Explanation:

We have ω₁ = 440 rev/min, ω₂ = 110 rev/min

We have centripetal acceleration

         [tex]a=\frac{v^2}{r}=\frac{(r\omega )^2}{r}=r\omega ^2[/tex]

Ratio of centripetal acceleration

         [tex]\frac{a_1}{a_2}=\frac{r\omega_1 ^2}{r\omega_2 ^2}=\left ( \frac{440}{110}\right )^2=4^2=16[/tex]

Ratio of the centripetal accelerations = 16

The centripetal acceleration ratio of a ceiling fan at different angular speeds can be found using the formula a = rω². The ratio a₁/a₂ for angular velocities of 440 rev/min and 110 rev/min is 16.

The ratio of centripetal accelerations a₁/a₂; can be found using the formula a = rω², where a is the centripetal acceleration, r is the distance from the axis of rotation, and ω is the angular velocity.

For ω₁ = 440 rev/min and ω₂ = 110 rev/min, the ratio a₁/a₂ = (rω₁²)/(rω₂²) = (ω₁/ω₂)² = (440/110)² = 16.

Therefore, the ratio of centripetal accelerations a₁/a₂ for the two angular speeds is 16.

Answer:

T would be two times the temperature That it was before the double so if the initial temperature was 20 it would now be 40

Explanation:

If the car initially had twice the speed ΔT would be 4 times the original change in temperature of the breaks.

If we assume no heat loss to the surrounding. Then the loss in kinetic energy (KE) of the car after applying the breaks is completely converted into heat energy (Q) of the breaks.

Loss in KE = Gain in Heat Energy

      ΔKE = ΔQ

Originally,    ΔKE = [tex]\frac{1}{2}mv^{2}[/tex] = ΔQ

Now,  ΔQ is directly proportional to change in temperature

           [tex]\frac{1}{2}mv^{2}[/tex] = ΔQ ∝ ΔT

If the speed is doubled, v' = 2v, then

  ΔKE = [tex]\frac{1}{2}mv^{'} ^{2}[/tex] = [tex]\frac{1}{2}m(4v^{2})[/tex] = 4 ×   [tex]\frac{1}{2}mv^{2}[/tex] = 4ΔQ ∝ 4ΔT

Hence, the temperature increases 4 times is speed is doubled.

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Answer:

[tex]\Delta t = 1.22 hours[/tex]

Explanation:

Initially legal speed to move through the drive way is 55 mi/h

so it is given as

[tex]v_1 = 55 mph = 24.48 m/s[/tex]

final legal speed is given as

[tex]v_2 = 65 mph = 28.93 m/s[/tex]

now we know that time taken to cover the distance of 700 km is given by the formula

[tex]t = \frac{d}{v}[/tex]

so initially time taken to cover that distance is

[tex]t_1 = \frac{700 \times 10^3 m}{24.48 m/s}[/tex]

[tex]t_1 = 28595 s = 7.94 h[/tex]

Now if legal speed is increased then the time taken to move through same distance is given as

[tex]t_2 = \frac{700 \times 10^3}{28.93}[/tex]

[tex]t_2 = 24196 s = 6.72 h[/tex]

so total time saved is given as

[tex]\Delta t = t_1 - t_2 [/tex]

[tex]\Delta t = 1.22 hours[/tex]

Answer:

39.0 m/s

Explanation:

Linear velocity = angular velocity × radius

v = ωr

v = (30.0 rad/s) (1.30 m)

v = 39.0 m/s

The linear velocity of the ball is 39 m/s.

The given parameters;

The velocity of the ball is the tangential velocity of the ball along the circular path of the elbow joint.

v =  ωr

where;

Substitute the given parameters and solve for the value of v;

v = (30 rad/s) x (1.3 m)

v = 39 m/s

Thus, the linear velocity of the ball is 39 m/s.

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The angular acceleration α of the CD is 1.590 rev/min².

The angular acceleration α can be calculated using the formula:

α = [tex]\frac{\Delta \omega}{\Delta t}[/tex]

Where [tex]\(\Delta \omega\)[/tex] is the change in angular velocity and [tex]\(\Delta t\)[/tex] is the change in time. Angular velocity [tex](\(\omega\))[/tex] is related to angular displacement [tex](\(\theta\))[/tex] and time (\(t\)) by the equation:

[tex]\[ \omega = \frac{\theta}{t} \][/tex]

Given that the CD accelerates from rest [tex](\(\omega_0 = 0\))[/tex] to a constant angular velocity of 477 rev/min, and rotates through an angular displacement of 0.250 rev}, we can use the following relationships:

[tex]\[ \omega = \frac{\theta}{t} \][/tex]

[tex]\[ \omega_f = \frac{477}{1} \, \text{rev/min} \][/tex]

[tex]\[ \omega_0 = 0 \, \text{rev/min} \][/tex]

Substituting these values into the first equation, we get:

[tex]\[ \frac{477}{1} = \frac{0.250}{t} \][/tex]

Solving for t, we find t = 0.525 min.

Now, substitute these values into the angular acceleration formula:

α = [tex]\frac{\frac{477}{1} - 0}{0.525}[/tex]

α = 1.590 rev/min²

Therefore, the angular acceleration of the CD is 1.590 rev/min².

Explanation:

The motion of a particle is defined by the relation as:

[tex]x=2t^3-9t^2+12t+10[/tex]........(1)

Differentiating equation (1) wrt t we get:

[tex]v=\dfrac{dx}{dt}[/tex]

[tex]v=\dfrac{d(2t^3-9t^2+12t+10)}{dt}[/tex]

[tex]v=6t^2-18t+12[/tex]............(2)

When v = 2 ft/s

[tex]2=6t^2-18t+12[/tex]

[tex]6t^2-18t+10=0[/tex]

t₁ = 2.26 s

and t₂ = 0.73 s

Put the value of t₁ in equation (1) as :

[tex]x=2(2.26)^3-9(2.26)^2+12(2.26)+10[/tex]

x₁ = 14.23 ft

Put the value of t₂ in equation (1) as :

[tex]x=2(0.73)^3-9(0.73)^2+12(0.73)+10[/tex]

x₁ = 14.74 ft

For acceleration differentiate equation (2) wrt t as :

[tex]a=\dfrac{dv}{dt}[/tex]

[tex]a=\dfrac{d(6t^2-18t+12)}{dt}[/tex]

a = 12 t - 18.........(3)

Put t₁ and t₂ in equation 3 one by one as :

[tex]a_1=12(2.26)-18=9.12\ ft/s^2[/tex]

[tex]a_2=12(0.73)-18=-9.24\ ft/s^2[/tex]

Hence, this is the required solution.

The time taken by the particle is 2.26 s and the magnitude of acceleration and position of the particle at the obtained instant are [tex]9.12 \;\rm ft/s^{2}[/tex] and 14.23 ft respectively.

When an object is observed to move in a straight line, then the motion of the object is known as linear motion. It can be referenced as motion in one dimension.

Given data:

The position of the particle is, [tex]x = 2t^{3}-9t^{2}+12t+10[/tex].

The velocity of the particle is, [tex]v = 2.00 \;\rm ft/s[/tex].

We know that velocity can be obtained by differentiating the position. Then,

[tex]v =\dfrac{dx}{dt}\\\\v = \dfrac{d(2t^{3}-9t^{2}+12t+10)}{dt}\\\\v=6t^{2}-18t+12[/tex]

Now when v = 2.00 ft/s. The time is,

[tex]2 = 6t^{2}-18t+12\\\\6t^{2}-18t+10=0\\\\t = 2.26 \;\rm s[/tex]

Now, the position of the particle is,

[tex]x = 2t^{3}-9t^{2}+12t+10\\\\x = (2 \times 2.26^{3})-(9 \times 2.26^{2})+(12 \times 2.26)+10\\\\x =14.23\;\rm ft[/tex]

And the magnitude of the acceleration is calculated by differentiating the velocity as,

[tex]a = \dfrac{dv}{dt}\\\\a = \dfrac{d(6t^{2}-18t+12)}{dt}\\\\a=12t-18[/tex]

Substituting t = 2.26 s as,

[tex]a = 12(2.26)-18\\\\a=9.12 \;\rm ft/s^{2}[/tex]

Thus, we can conclude that the time taken by the particle is 2.26 s and the magnitude of acceleration and position of the particle at the obtained instant are [tex]9.12 \;\rm ft/s^{2}[/tex] and 14.23 ft respectively.

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Answer:

Work done, W = 401 kJ

Explanation:

It is given that,

Mass of the car, m = 1100 kg

Initial, velocity of the car, u = 27 m/s

Finally it stops, v = 0

According to work energy theorem, the change in kinetic energy is equal to the work done i.e.

[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]

[tex]W=\dfrac{1}{2}\times 1100\ kg\times (0-(27\ m/s)^2)[/tex]

W = -400950 J

or

[tex]W=-401\ kJ[/tex]

-ve sign shows the direction of force and displacement are in opposite direction.

So, 401 kJ work is required to stop a car. Hence, this is the required solution.

To stop an 1100 kg car traveling at 27 m/s, the work required is calculated using the kinetic energy formula KE = ½ mv² and is 401 kJ.

The question asks about the amount of work required to stop a car with a mass of 1100 kg that is traveling at a velocity of 27 m/s. To solve this, we can use the equation for kinetic energy (KE), which is the work needed to accelerate a body of a given mass from rest to its stated velocity. Using the equation KE = ½ mv², we find that KE = ½ (1100 kg) * (27 m/s)² = 399,150 J, which is approximately 401 kJ. Thus, the correct amount of work required to stop the car is 401 kJ.

Answer:

[tex]\beta = 41.68°[/tex]

Explanation:

according to snell's law

[tex]\frac{n_w}{n_g} = \frac{sin\alpha}{sin30 }[/tex]

refractive index of water n_w is 1.33

refractive index of glass  n_g  is 1.5

[tex]sin\alpha = \frac{n_w}{n_g}* sin30[/tex]

[tex]sin\alpha = 0.443[/tex]

now applying snell's law between air and glass, so we have

[tex]\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}[/tex]

[tex]sin\beta = \frac{n_g}{n_a} sin\alpha[/tex]

[tex]\beta = sin^{-1} [\frac{n_g}{n_a}*sin\alpha][/tex]

we know that [tex]sin\alpha = 0.443[/tex]

[tex]\beta = 41.68°[/tex]

Answer:

[tex]v_f = 1.6 m/s[/tex]

Explanation:

As per momentum conservation we know that initial momentum of one box car must be equal to the momentum of all box cars together

here we know that there is no external force on the system of all box cars so there momentum conservation is applicable

[tex]m_1v_{i} = (m_1 + m_2 + m_3 + m_4 + m_5)v_{f}[/tex]

so above equation is momentum of one box car is equal to the momentum of all box cars together

So here we will have

[tex]v_f = \frac{m_1 v_i}{m_1 + m_2 + m_3 + m_4 + m_5}[/tex]

[tex]v_f = \frac{m(8)}{5m}[/tex]

[tex]v_f = \frac{8}{5} = 1.6 m/s[/tex]

By applying the conservation of momentum, the final speed of a boxcar linking with four stationary boxcars would be 1.6 m/s. The mass of each boxcar is 18,537 kg and the initial speed of the moving boxcar is 8 m/s before the collision.

The student's question involves finding the final speed of a set of linked boxcars immediately after a collision. This problem deals with the conservation of momentum, a fundamental concept in physics. According to the law of conservation of momentum, the total momentum of a closed system of objects (which does not interact with external forces) remains constant. In this case, because the boxcars are linking together and moving as one unit post-collision, they must conserve momentum.

To compute the final speed after the collision, we will use the formula:

m1 * v1 + m2 * v2 = (m1 + m2 + m3 + m4 + m5) * vf

where

Given that the initial velocity of the stationary boxcars is 0 (since they are at rest) and knowing all the masses are equal:

m = 18,537 kg

v1 = 8 m/s

v2 = 0 m/s (stationary)

We can calculate the final velocity (vf) by the following:

(18,537 kg * 8 m/s) + (4 * 18,537 kg * 0 m/s) = (5 * 18,537 kg) * vf

Solving this for vf gives us:

vf = (18,537 kg * 8 m/s) / (5 * 18,537 kg)

vf = 1.6 m/s

Therefore, the final speed of the boxcars after the collision would be 1.6 m/s.

Explanation:

The electric field is given by :

[tex]E=k\dfrac{Q}{r^2}[/tex]

Where

k is electrostatic constant, [tex]k=\dfrac{1}{4\pi \epsilon_o}=9\times 10^9\ Nm^2/C^2[/tex]

Q is the magnitude of charge

r is the distance from the charge

The gravitational field equation is given by :

[tex]g=\dfrac{GM}{r^2}[/tex]

G is universal gravitational constant

M is the mass

r is the distance

Final answer:

The equivalent gravitational field equation to E = Q/r² is g = G×M/r², where g is the gravitational field strength, G is the gravitational constant, M is the mass, and r is the distance from the mass.

Explanation:

If an electric field is defined by the equation E = Q/r², where E represents the electric field, Q is the charge, and r is the distance from the point charge, the analogous equation for the gravitational field would be g = G×M/r². In this equation, g is the gravitational field strength, G is the gravitational constant, M is the mass creating the gravitational field, and r is the distance from the center of mass.

The concept of the gravitational field is similar to that of the electric field. Both fields decrease with the square of the distance from the source (charge or mass). The field strength represents the force applied per unit charge for the electric field, measured in newtons per coulomb (N/C), and force applied per unit mass for the gravitational field, measured in newtons per kilogram (N/kg).

Answer:

(4273.7 [tex]\hat{j}[/tex] - 3281.6 [tex]\hat{i}[/tex])

Explanation:

[tex]\underset{v_{i}}{\rightarrow}[/tex] = initial velocity of the baseball before collision = 43 [tex]\hat{i}[/tex] m/s

[tex]\underset{v_{f}}{\rightarrow}[/tex] = final velocity of the baseball after collision = 56 [tex]\hat{j}[/tex] m/s

m = mass of the ball = 145 g = 0.145 kg

t = time of contact of the ball with the bat = 1.90 ms = 0.0019 s

[tex]\underset{F_{avg}}{\rightarrow}[/tex] = Average force vector

Using Impulse-change in momentum equation

[tex]\underset{F_{avg}}{\rightarrow}[/tex] t = m ([tex]\underset{v_{f}}{\rightarrow}[/tex] - [tex]\underset{v_{i}}{\rightarrow}[/tex] )

[tex]\underset{F_{avg}}{\rightarrow}[/tex] (0.0019) = (0.145) (56 [tex]\hat{j}[/tex] - 43 [tex]\hat{i}[/tex])

[tex]\underset{F_{avg}}{\rightarrow}[/tex] = (4273.7 [tex]\hat{j}[/tex] - 3281.6 [tex]\hat{i}[/tex])

Answer:

1119.1 K

Explanation:

From Clausius-Clapeyron equation:

[tex]\frac{dP}{dT}=[/tex]Δ[tex]\frac{h_{v} }{R} (\frac{1}{T^{2} } )dT[/tex]

The equation may be integrated considering the enthalpy of vaporization constant, and its result is:

[tex]ln(\frac{P_{2} }{P_{1} } )=-[/tex]Δ[tex]\frac{h_{v} }{R}*(\frac{1}{T_{2} }-\frac{1}{T_{1} })[/tex]

Isolating the temperature [tex]T_{2}[/tex]

[tex]T_{2}=\frac{1}{\frac{-R}{dhv}*ln(\frac{P_{2} }{P_{1}}) +\frac{1}{T_{1}} }[/tex]

[tex]T_{2}=\frac{1}{\frac{-8.314}{1.00*10^5}*ln(\frac{500}{40}) +\frac{1}{906.15}}[/tex]

[tex]T_{2}=1119.1K[/tex]

Note: Remember to change the units of the enthalpy vaporization to J/mol; and the temperatures must be in Kelvin units.

There is a format mistake with the enthalpy of vaporization, each 'Δ' correspond to that.

Answer:

m = 77.75 g

Explanation:

Here we know that at equilibrium the temperature of the system will be 10 degree C

so heat given by hot latte = heat absorbed by the ice

now we have

heat given by latte = [tex]m s\Delta T[/tex]

[tex]Q_1 = (200)(4.186)(45 - 10)[/tex]

[tex]Q_1 = 29302 J[/tex]

now heat absorbed by ice is given as

[tex]Q_2 = mL + ms\Delta T[/tex]

[tex]Q_2 = m(335 + 4.186(10 - 0))[/tex]

[tex]Q_2 = m(376.86)[/tex]

now by heat balance we have

[tex]Q_1 = Q_2[/tex]

[tex]29302 = m(376.86)[/tex]

[tex]m = 77.75 g[/tex]

To lower the temperature of the hot latte to 10°C, you need to add approximately 124.87 g of ice.

To find out how much ice you need to add to a hot latte to lower its temperature to 10°C, you can use the formula Q = m × Lf, where Q is the heat transferred, m is the mass of the ice, and Lf is the latent heat of fusion of ice. First, calculate the heat that needs to be transferred:

Q = (200 mL) × (0.997 g/mL) × (45°C - 10°C) × (4.184 J/g°C)

Q = 41,817 J

Next, divide the heat transferred by the latent heat of fusion of ice to find the mass of ice needed:

m = 41,817 J ÷ 335 J/g

m ≈ 124.87 g

Therefore, you need to add approximately 124.87 g of ice to the hot latte to lower its temperature to 10°C.

Answer:

[tex]F = 2.2 \times 10^{-3} N[/tex]

Explanation:

Force of gravitation between Astronaut and Moon is given by the Universal law of gravitation

[tex]F = \frac{Gm_1m_2}{r^2}[/tex]

now we will have

[tex]m_1 = 7.355 \times 10^{22} kg[/tex]

[tex]m_2 = 65 kg[/tex]

[tex]r = 3.81 \times 10^8 m[/tex]

now from above formula we have

[tex]F = \frac{(6.67 \times 10^{-11})(65)(7.355 \times 10^{22})}{(3.81 \times 10^8)^2}[/tex]

[tex]F = 2.2 \times 10^{-3} N[/tex]

Answer:

i = 0.5 A

Explanation:

As we know that magnetic flux is given as

[tex]\phi = NBA[/tex]

here we know that

N = number of turns

B = magnetic field

A = area of the loop

now we know that rate of change in magnetic flux will induce EMF in the coil

so we have

[tex]EMF = NA\frac{dB}{dt}[/tex]

now plug in all values to find induced EMF

[tex]EMF = (20)(50 \times 10^{-4})(\frac{6 - 2}{2})[/tex]

[tex]EMF = 0.2 volts[/tex]

now by ohm's law we have

[tex]current = \frac{EMF}{Resistance}[/tex]

[tex]i = \frac{0.2}{0.40} = 0.5 A[/tex]

Using Faraday's Law of electromagnetic induction and Ohm's Law, you can calculate that the magnitude of the induced current in the coil is 0.025 amperes.

The magnitude of the induced current can be calculated using Faraday's Law of electromagnetic induction, which states that the induced electromotive force (EMF) in a closed circuit is equal to the time rate of change of magnetic flux through the circuit. Mathematically, it is denoted as:

ε =-N (ΔΦ/Δt)

where

In this problem, N=20, the initial magnetic field (B1) is 2.0 T, the final magnetic field (B2) is 6.0 T, and the time interval (Δt) is 2.0 s.

Magnetic flux (Φ) is given by the product of the magnetic field and the area each loop encloses:

Φ = BA

The change in magnetic flux (ΔΦ) is then:

ΔΦ = N * A * ΔB

where A=50 cm²=5 x 10⁻⁴ m² (converted to m² from cm²), and ΔB = B2 - B1 = 4.0 T.

So,

ΔΦ = 20 * 5 x 10⁻⁴ * 4 = 0.02 Wb (webers)

Now we can calculate the induced EMF using Faraday's Law:

ε = -(0.02 Wb) / 2.0 s = -0.01 V (volts)

The negative sign merely indicates that the induced current flows in a direction to oppose the change in the magnetic field.

Finally, we calculate the magnitude of the induced current (I) by dividing the induced EMF by the total resistance (R) of the coil using Ohm's law:

I = ε / R = 0.01V/0.40 Ω = 0.025 A (amperes)

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Answer: (a) 1.065 N  (b) 2.13 N

Explanation:

According Newton's 2nd Law of Motion the force [tex]F[/tex] is defined as the variation of linear momentum [tex]p[/tex] in time:

[tex]F=\frac{dp}{dt}[/tex]  (1)

Where the linear momentum is:

[tex]p=mV[/tex]  (2) Being [tex]m[/tex] the mass and [tex]V[/tex] the velocity.

In the case of the rain drops, which initial velocity is [tex]V_{i}=15m/s[/tex] and final velocity is  [tex]V_{f}=0[/tex] (we are told the drops come to rest after striking the roof). The momentum of the drops [tex]p_{drops}[/tex] is:

[tex]p_{drops}=mV_{i}+mV_{f}[/tex]  (3)

If [tex]V_{f}=0[/tex], then:

[tex]p_{drops}=mV_{i}[/tex]  (4)

Now the force [tex]F_{drops}[/tex] exerted by the drops is:

[tex]F_{drops}=\frac{dp_{drops}}{dt}=\frac{d}{dt}mV_{i}[/tex]  (5)

[tex]F_{drops}=\frac{dm}{dt}V_{i}+m\frac{dV_{i}}{dt}[/tex]  (6)

At this point we know the mass of rain per second (mass rate) [tex]\frac{dm}{dt}=0.071 kg/s[/tex] and we also know the initial velocity does not change with time, because that is the velocity at that exact moment (instantaneous velocity). Therefore is a constant, and the derivation of a constant is zero.

This means (6) must be rewritten as:

[tex]F_{drops}=\frac{dm}{dt}V_{i}[/tex]  (7)

[tex]F_{drops}=(0.071 kg/s)(15m/s)[/tex]  (8)

[tex]F_{drops}=1.065kg.m/s^{2}=1.065N[/tex]  (9) This is the force exerted by the rain drops on the roof of the car.

Now let's assume that instead of rain drops, hailstones fall on the roof of the car, and let's also assume these hailstones bounce back up off after striking the roof (this means they do not come to rest as the rain drops).

In addition, we know the hailstones fall with the same velocity as the rain drops and have the same mass rate.

So, in this case the linear momentum [tex]p_{hailstones}[/tex] is:

[tex]p_{hailstones}=mV_{i}+mV_{f}[/tex]   (9)  Being [tex]V_{i}=V_{f}[/tex]

[tex]p_{hailstones}=mV+mV=2mV[/tex]   (10)  

Deriving with respect to time to find the force [tex]F_{hailstones}[/tex] exerted by the hailstones:

[tex]F_{hailstones}=\frac{d}{dt}p_{hailstones}=\frac{d}{dt}(2mV)[/tex]   (10)  

[tex]F_{hailstones}=2\frac{d}{dt}(mV)=2(\frac{dm}{dt}V+m\frac{dV}{dt})[/tex]   (11)  

Assuming [tex]\frac{dV}{dt}=0[/tex]:

[tex]F_{hailstones}=2(\frac{dm}{dt}V)[/tex]   (12)  

[tex]F_{hailstones}=2(0.071 kg/s)(15m/s)[/tex]   (13)  

Finally:

[tex]F_{hailstones}=2.13kg.m/s^{2}=2.13N[/tex] (14)   This is the force exerted by the hailstones  

Comparing (9) and (14) we can conclude the force exerted by the hailstones is two times greater than the force exerted by the raindrops.

The average force exerted by the rain on the roof of a parked car can be calculated using the equation Force = mass × acceleration. If hailstones with the same mass and speed fall on the roof, the average force on the roof would be the same.

(a) To find the average force exerted by the rain on the roof, we can use the equation:

Force = mass × acceleration

The mass of rain per second striking the roof is given as 0.071 kg/s. Since the drops come to rest after striking the roof, the acceleration is equal to the initial velocity of the drops, which is 15 m/s. Therefore, the force is:

Force = 0.071 kg/s × 15 m/s = 1.065 N

(b) If hailstones with the same mass as the raindrops fall on the roof at the same rate and with the same speed, the average force on the roof would be the same as found in part (a). The mass and speed of the hailstones are the same as the raindrops, so the force exerted by the hailstones on the roof would also be 1.065 N.

The total energy is less than the potential energy because of energy conservation. Because you can solve with different linear solutions, I would say it depends on the nature of the system.

Answer:

624.5 N

Explanation:

m = mass of the truck = 2150 kg

v₀ = initial speed of the truck = 55 km/h = 15.3 m/s

v = final speed of the truck = 33 km/h = 9.2 m/s

t = time interval = 21 s

F = magnitude of the average net force

Using Impulse-Change in momentum equation

F t = m (v - v₀ )

F (21) = (2150) (9.2 - 15.3)

F = - 624.5 N

hence the magnitude of net force is 624.5 N

The magnitude of the average net force acting on the truck during the 21.0 s interval is about 626 N

[tex]\texttt{ }[/tex]

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

mass of truck = m = 2150 kg

initial speed of truck = u = 55.0 km/h = 15⁵/₁₈ m/s

final speed of truck = v = 33.0 km/h = 9¹/₆ m/s

time taken = t = 21.0 s

Asked:

average net force = ∑F = ?

Solution:

[tex]\Sigma F = m a[/tex]

[tex]\Sigma F = m ( v - u ) \div t[/tex]

[tex]\Sigma F = 2150 ( 9\frac{1}{6} - 15 \frac{5}{18} ) \div 21.0[/tex]

[tex]\Sigma F = 2150 ( -6\frac{1}{9} ) \div 21.0[/tex]

[tex]\Sigma F = -626 \texttt{ N}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Kinematics

Answer:

Given values of Planck Constant are equivalent in English system and metric system.

Explanation:

Value of Planck's constant is given in English system as 4.14 x 10⁻¹⁵eV s.

Converting this in to metric system .

We have 1 eV = 1.6 x 10⁻¹⁹ J

Converting

     4.14 x 10⁻¹⁵eV s = 4.14 x 10⁻¹⁵x 1.6 x 10⁻¹⁹ = 6.63 x 10⁻³⁴ Joule s

So Given values of Planck Constant are equivalent in English system and metric system.

Answer:

(a) 21.11 J

(b) 3.3 m/s

Explanation:

(a)

m = mass attached to the spring = 3.95 kg

A = amplitude of the oscillation = 19 cm = 0.19 m

T = Time period of oscillation = 0.365 s

U = Total mechanical energy

Total mechanical energy is given as

[tex]U = (0.5)mA^{2}\left ( \frac{2\pi }{T} \right )^{2}[/tex]

[tex]U = (0.5)(3.95)(0.19)^{2}\left ( \frac{2(3.14) }{0.365} \right )^{2}[/tex]

U = 21.11 J

(b)

v = maximum speed of the mass

maximum speed of the mass is given as

[tex]v = A\left ( \frac{2\pi }{T} \right )[/tex]

[tex]v = (0.19)\left ( \frac{2(3.14) }{0.365} \right )[/tex]

[tex]v [/tex] = 3.3 m/s

Answer:

[tex]\lambda = 2.57 \times 10^{-11} m[/tex]

Explanation:

Average velocity of oxygen molecule at given temperature is

[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]

now we have

M = 32 g/mol = 0.032 kg/mol

T = 27 degree C = 300 K

now we have

[tex]v_{rms} = \sqrt{\frac{3(8.31)(300)}{0.032}[/tex]

[tex]v_{rms} = 483.4 m/s[/tex]

now for de Broglie wavelength we know that

[tex]\lambda = \frac{h}{mv}[/tex]

[tex]\lambda = \frac{6.6 \times 10^{-34}}{(5.31\times 10^{-26})(483.4)}[/tex]

[tex]\lambda = 2.57 \times 10^{-11} m[/tex]

Explanation:

It is given that,

Initial speed, v₁ = 14 m/s

Initial force, F₁ = 130 N

We need to find the total horizontal force (F₂) on the driver if the speed on the same curve is 18.0 m/s instead, v₂ = 18 m/s

The centripetal force is given by :

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

[tex]\dfrac{F_1}{F_2}=\dfrac{mv_1^2/r}{mv_2^2/r}[/tex]

[tex]\dfrac{F_1}{F_2}=\dfrac{v_1^2}{v_2^2}[/tex]

[tex]F_2=\dfrac{v_2^2\times F_1}{v_1^2}[/tex]

[tex]F_2=\dfrac{(18\ m/s)^2\times 130\ N}{(14\ m/s)^2}[/tex]

[tex]F_2=214.8\ N[/tex]

So, if the speed is 18 m/s, then the horizontal force acting on the car is 214.8 N. Hence, this is the required solution.

Final answer:

The total horizontal force on the driver when the car's speed increases to 18.0 m/s can be found using the proportionality between the centripetal force and the square of the velocity. By comparing the initial and final conditions, we can calculate the new total horizontal force required for the increased speed.

Explanation:

The question involves finding the total horizontal force on a driver when the speed of the car increases on the same curve. The horizontal force experiences by the driver in a curved path, which is part of a horizontal circle, can be determined by using the formula for centripetal force (Fc = mv²/r). We know from the initial condition that the horizontal force is 130 N when the car is traveling at 14.0 m/s. Since the force depends on the square of the velocity, we can use proportions to find the new force when the speed increases to 18.0 m/s.

Thus, (F2/F1) = (v₂²/v₁²). Substituting the given values, we get (F2/130 N) = (18.0 m/s)² / (14.0 m/s)². Calculating this provides us with the new force F2 which represents the total horizontal force on the driver at the higher speed of 18.0 m/s.

Answer:

The mass's acceleration is 5 m/s^2 in the minus X direction and 9,8 m/s^2 in the minus Y direction.

Explanation:

By applying the second Newton's law in the X and Y direction we found that in the minus X direction an external force of 10 N is exerted, while in the minus Y direction the gravity acceleration is acting:

X-direction balance force: [tex]-10 [N] = m.ax [/tex]

Y-direction balance force: [tex]-m*9,8 \frac{m}{s^2} = m.ay[/tex]

Where ax and ay are the components of the respective acceleration and m is the mass. By solving for each acceleration:

[tex]ax=(-10 [N]) / m [/tex]

[tex]ay=-m*9,8\frac{m}{s^2} / m[/tex]

Note that for the second equation above the mass is cancelled and, the Y direction acceleration is minus the gravity acceleration:

[tex]ay=-9,8\frac{m}{s^2} [/tex]

For the x component aceleration we must replace the Newton unit:

[tex]N =\frac{kg.m}{s^2} [/tex]

[tex]ax= -10 \frac{kg.m}{s^2} / (2 kg) [/tex]

[tex]ax= - 5 \frac{m}{s^2}  [/tex]

The acceleration of a 2 kg mass with a 10 N force exerted in the negative X direction, while free falling, is 5 m/s² in the X direction. The gravitational acceleration in the Y direction remains 9.8 m/s².

To calculate the acceleration of the mass in the situation where a 2 kg mass free falls in the negative Y direction while a 10 N force is exerted in the negative X direction, we must use Newton's Second Law.

The law states that acceleration is equal to the net force divided by the mass: Acceleration (a) = Force (F) / Mass (m). Since the force is applied in the horizontal (X) direction and there's no opposing force mentioned, the acceleration in the X direction can be calculated. However, for the Y direction, the mass is simultaneously experiencing gravitational acceleration.

In this problem, the force exerted on the mass in the negative X direction is 10 N and the mass is 2 kg. So, we first calculate acceleration in the X direction:

ax = F / m = 10 N / 2 kg = 5 m/s²

Since the Y direction involves free fall, the acceleration there is due to gravity, which on Earth is approximately 9.8 m/s² downward or in the negative Y direction. We don't combine these accelerations directly because they are perpendicular to each other. The magnitude of the total acceleration can be found using the Pythagorean theorem, but since the Y direction acceleration is due to gravity and constant, we typically consider only the X direction when discussing the change caused by the applied force.

Answer:

b = 1084.5 kg/s

Explanation:

As we know that the amplitude of damped oscillation is given as

[tex]A = A_o e^{-\frac{bt}{2m}}[/tex]

here we know that

[tex]A = \frac{A_o}{2}[/tex]

after time t = 4.9 minutes

also we know that

[tex]m = 2.30 \times 10^5 kg[/tex]

now we will have

[tex]\frac{A_o}{2} = A_o e^{-\frac{bt}{2m}}[/tex]

[tex]\frac{bt}{2m} = ln2[/tex]

[tex]b = \frac{ln2 (2m)}{t}[/tex]

[tex]b = \frac{2(ln2)(2.30 \times 10^5)}{4.9 \times 60}[/tex]

[tex]b = 1084.5 kg/s[/tex]

Answer:

option (c)

Explanation:

The frequency ranges for infrasonic is below 20 Hz.

The frequency ranges for audible is 20 Hz to 20000 Hz.

The frequency ranges for ultrasonic is above 20000 Hz.

So, infrasonic have frequency below the audible range. We cannot hear this frequency.

Explanation:

Electric force is the force acting between two charged particles. Electric force between electron and proton is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Distance between them, [tex]r=5.3\times 10^{-11}\ m[/tex]

[tex]F=9\times 10^9\times \dfrac{-(1.6\times 10^{-19})^2}{(5.3\times 10^{-11}\ m)^2}[/tex]

[tex]F=-8.2\times 10^{-8}\ N[/tex]

Gravitational force is the force acting between two masses. The gravitational force between the electron and proton is given by :

[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]

Distance between them, [tex]r=5.3\times 10^{-11}\ m[/tex]

[tex]F=6.67\times 10^{-11}\times \dfrac{9.11\times 10^{-31}\ kg\times 1.67\times 10^{-27}\ kg}{(5.3\times 10^{-11}\ m)^2}[/tex]

[tex]F=3.61\times 10^{-47}\ N[/tex]

Hence, this is the required solution.

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time [tex]t_1[/tex], the acceleration vector has direction [tex]\theta_1[/tex] such that

[tex]\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ[/tex]

which indicates the particle is situated at a point on the lower left half of the circle, while at time [tex]t_2[/tex] the acceleration has direction [tex]\theta_2[/tex] such that

[tex]\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ[/tex]

which indicates the particle lies on the upper left half of the circle.

Notice that [tex]\theta_1-\theta_2=90^\circ[/tex]. That is, the measure of the major arc between the particle's positions at [tex]t_1[/tex] and [tex]t_2[/tex] is 270 degrees, which means that [tex]t_2-t_1[/tex] is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

[tex]\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}[/tex]

where [tex]R[/tex] is the radius of the circle and [tex]T[/tex] is the period. We have

[tex]t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s[/tex]

and the magnitude of the particle's acceleration toward the center of the circle is

[tex]\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}[/tex]

So we find that the path has a radius [tex]R[/tex] of

[tex]7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}[/tex]

The acceleration of a particle moving in a circular path with constant speed is given by the centripetal acceleration. Since the velocity is not given, the radius cannot be determined with this information alone.

In circular motion, the acceleration of a particle is given by the centripetal acceleration, which is equal to the square of the speed divided by the radius of the circular path (v^2/r). As the particle moves at a constant speed, we know that the magnitude of acceleration (which gives us the centripetal acceleration) remains the same at both times t1 and t2. Therefore, we can calculate the magnitude of acceleration at both points and set them equal to each other. This will allow us to solve for the radius of the path.

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Answer:

47.17 degree C

Explanation:

mg = 3.5 kg, T1 = 94 degree C, sg = 129 J/kg C

mw = 0.2 kg, T2 = 22 degree C, sw = 4200 J/kg C

Let T be the temperature at equilibrium.

Heat given by the gold = Heat taken by water

mg x sg x (T1 - T) = mw x sw x (T - T2)

3.5 x 129 x (94 - T) = 0.2 x 4200 x (T - 22)

42441 - 451.5 T = 840 T - 18480

60921 = 1291.5 T

T = 47.17 degree C

Answer:

a) 40.2 liters = 40200 cubic centimeters

b) 40.2 liters = 0.0402 cubic meters

Explanation:

Volume of bottle = 40.2 liters.

a)

      1 liter = 1000 cubic centimeter

      40.2 liters = 40.2 x 1000 = 40200 cubic centimeters

b)

      1 liter = 0.001 cubic meter

      40.2 liters = 40.2 x 0.001 = 0.0402 cubic meters