A 61 kg skier starts from rest at the top of a 1200 m long trail which drops a total of 227 m from top to bottom. At the bottom, the skier is moving 11 m/s. How much energy was dissipated by friction?

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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Taking the drag force of air into account, it takes longer for the ball to travel to the top of its motion compared to the time it takes to fall back down.

When a ball is thrown straight up, the drag force of air acts against its motion. As the ball ascends, the drag force opposes its upward velocity, causing a reduction in the net force acting on the ball. This results in a slower upward acceleration and a longer time to reach the top of its motion.

On the descent, the gravitational force aids the motion, overcoming the drag force. The downward velocity increases, and the time taken to fall back down is less than the time taken to ascend.

In the absence of air resistance, the ascent and descent times would be equal. However, with air resistance considered, the ascent time is prolonged, impacting the overall motion of the ball.

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

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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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:

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:

Option B is the correct answer.

Explanation:

Let us consider 40 meter above ground as origin.

Initial velocity = 17 m/s

Final velocity = 24 m/s

Acceleration = 9.81 m/s

We have equation of motion v² = u² + 2as

Substituting

         24² = 17² + 2 x 9.81 x s

           s = 14.63 m

Distance traveled by rock = 14.63 m down.

Height of rock from ground = 40 - 14.63 = 25.37 m = 25.4 m

Option B is the correct answer.

To convert the thermal conductivity value of 0.3 Btu/(h ft°F) to W/(m °C), multiply by 694.7. To convert the surface heat transfer coefficient value of 105 Btu/(h ft2 oF) to W/(m2 °C), multiply by 11,545.

To convert the thermal conductivity value of 0.3 Btu/(h ft°F) to W/(m °C), we can use the following conversion factors:

Using these conversion factors, we can convert the thermal conductivity value as follows:

0.3 Btu/(h ft°F) = (0.3 * 1055.06 J)/(h * 0.3048 m * 0.5556 °C) = 694.7 W/(m °C)

Similarly, to convert the surface heat transfer coefficient value of 105 Btu/(h ft2 oF) to W/(m2 °C), we can use the following conversion factors:

Using these conversion factors, we can convert the surface heat transfer coefficient value as follows:

105 Btu/(h ft2 oF) = (105 * 1055.06 J)/(h * 0.0929 m2 * 0.5556 °C) = 11,545 W/(m2 °C)

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The conversion of the thermal conductivity value yields approximately 0.52 W/(m°C), and the surface heat transfer coefficient conversion yields approximately 596.15 W/(m²°C). Conversion factors for BTU to J, ft to m, and °F to °C are critical in these calculations.

To convert thermal conductivity from BTU/(h ft°F) to W/(m°C):

To convert surface heat transfer coefficient from BTU/(h ft²°F) to W/(m²°C):

Answer:

P₁ = 192 W and P₂ = 384 W

Explanation:

It is given that,

Resistor 1, [tex]R_1=12\ \Omega[/tex]

Resistor 2, [tex]R_1=6\ \Omega[/tex]

Voltage, V = 48 V

Power dissipated by resistor 1 is given by :

[tex]P_1=\dfrac{V^2}{R_1}[/tex]

[tex]P_1=\dfrac{(48)^2}{12}[/tex]

P₁ = 192 watts

Power dissipated by resistor 2 is given by :

[tex]P_2=\dfrac{V^2}{R_1}[/tex]

[tex]P_2=\dfrac{(48)^2}{6}[/tex]

P₂ = 384 watts

So, the power dissipated by both the resistors is 192 watts and 384 watts respectively. Hence, this is the required solution.

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:

Volumetric flow rate = 0.024 m³/s  

Explanation:

Volumetric flow rate = Discharge = Q

We have expression for discharge, Q = Area ( A) x Velocity (v)

Velocity , v = 3 m/s

[tex]\texttt{Area, A}=\frac{\pi d^2}{4}=\frac{\pi \times (10\times 10^{-2})^2}{4}=7.85\times 10^{-3}m^2[/tex]

Substituting

  Discharge, Q =Av = 7.85 x 10⁻³ x 3 = 0.024 m³/s  

Volumetric flow rate = 0.024 m³/s  

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

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]

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:

(a) No. of protons = atomic number = 7

No. of electrons = no. of protons = 7

No. of neutrons = mass no. - atomic no. = 6

(b) No. of protons = atomic number = 16

No. of electrons = no. of protons = 16

No. of neutrons = mass no. - atomic no. = 19

(c) No. of protons = atomic number = 29

No. of electrons = no. of protons = 29

No. of neutrons = mass no. - atomic no. = 34

(d) No. of protons = atomic number = 38

No. of electrons = no. of protons = 38

No. of neutrons = mass no. - atomic no. = 51  

Explanation:

No. of protons = No. of electrons = atomic number

No. of neutrons = Mass no. - atomic no.

(a) 13 N 7

Here atomic number is 7, mass number is 13

No. of protons = atomic number = 7

No. of electrons = no. of protons = 7

No. of neutrons = mass no. - atomic no. = 13 - 7 = 6

(b) 35 S 16

Here atomic number is 16, mass number is 35

No. of protons = atomic number = 16

No. of electrons = no. of protons = 16

No. of neutrons = mass no. - atomic no. = 35 - 16 = 19

(c) 63 Cu 29

Here atomic number is 29, mass number is 63

No. of protons = atomic number = 29

No. of electrons = no. of protons = 29

No. of neutrons = mass no. - atomic no. = 63 - 29 = 34

(d) 89 Sr 38

Here atomic number is 38, mass number is 89

No. of protons = atomic number = 38

No. of electrons = no. of protons = 38

No. of neutrons = mass no. - atomic no. = 89 - 38 = 51

Final answer:

The isotopes ¹N⁷, ³⁵S¹⁶, ⁶³Cu₂₉, and ₈₉Sr₃₈ have 7, 16, 29, and 38 protons, respectively, with neutrons calculated by subtracting the number of protons from the mass number. Lastly, the number of electrons is equal to the number of protons in a neutral atom.

Explanation:

The number of protons, neutrons, and electrons for each isotope can be determined using the atomic number, mass number, and charge. The atomic number represents the number of protons (and also the number of electrons in a neutral atom), the mass number represents the sum of protons and neutrons, and the charge indicates how many electrons have been lost or gained.

Each isotope comprises an equal number of protons and electrons in their neutral states. Neutrons are the difference between the mass number and the number of protons.

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]

Answer:

He made it, he across safely the river without falling ing.

Explanation:

Vt= 7.8 m/s

r= 11m

m= 80.5 kg

Vt= ω * r

ω= 0.71 rad/s

ac= ω² * r

ac= 5.54 m/s²

F= m * ac

F= 445.97 N  < 1,000 N

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

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: (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.

Answer:

The pendulum clock does it loses 83.0304 seconds a day at the new location.

Explanation:

T1= 1 s

T2= ?

g1= 9.8135 m/s

g2= 9.7943 m/s

L=?

L= (T1/2π)²  * g1=

L= 0.24857m

T2= 2π * √(L/g2)

T2= 1.000961 s

ΔT= T2 - T1

ΔT = 0.000961 s

seconds lost a day = 24 * 3600 * ΔT

seconds lost a day= 24 * 3600 * 0.000961 s

seconds lost a day= 83.0304 s

A pendulum clock loses accuracy when it is moved from one location to another due to changes in the acceleration due to gravity. The period of a pendulum is dependent on its length and the acceleration due to gravity. To calculate how many seconds a day the clock loses at the new location, we can subtract the new period from the original period and convert it to seconds.

A pendulum clock loses accuracy when it is moved from one location to another due to changes in the acceleration due to gravity, not changes in the length of the pendulum. In this case, the clock was moved from a location where g = 9.8135 m/s2 to a new location where g = 9.7943 m/s2. The clock loses accuracy because the period of a pendulum is dependent on its length and the acceleration due to gravity.

The period of a pendulum can be calculated using the equation: T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

So, to calculate how many seconds a day the clock loses at the new location, we can subtract the new period from the original period and convert it to seconds. Since the original period is 2.00000 s, we have:

ΔT = Toriginal - Tnew

ΔT = 2.00000 s - Tnew

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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:

The electric potential difference is 32500 volt.

Explanation:

Given that,

Charge[tex]q=2\times10^{-4}C[/tex]

Energy = 6.5 J

We need to calculate the electric potential difference

Potential difference :

Potential difference is equal to the energy divide by charge.

Using formula of potential difference

[tex]V=\dfrac{E}{Q}[/tex]

[tex]V=\dfrac{6.5}{2\times10^{-4}}[/tex]

[tex]V=32500\ volt[/tex]

Hence, The electric potential difference is 32500 volt.

The electric potential difference is 32,500 volts.

To determine the electric potential difference, we can use the formula:

V = ΔE / q

where V is the electric potential difference, ΔE is the energy and q is the test charge. In this case, the energy given is 6.5J and the test charge is 2 x 10-4C. Plugging these values into the formula, we get:

V = 6.5J / (2 x 10-4C)

V = 32,500V

Therefore, the electric potential difference is 32,500 volts.

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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.

Answer:

[tex]\lambda = 260.9nm[/tex]

Explanation:

As we know that work function is defined as the minimum energy of photons required to produce the photo electric effect

so here we have

[tex]\phi = \frac{hc}[\lambda}[/tex]

now we know that

[tex]h = 6.63\times 10^{-34}[/tex]

[tex]c = 3\times 10^8 m/s[/tex]

[tex]\lambda[/tex] = wavelength of photons

now from above formula

[tex]7.59 \times 10^{-19} = \frac{(6.36 \times 10^{-34})(3\times 10^8)}{\lambda}[/tex]

[tex]\lambda = \frac{(6.36 \times 10^{-34})(3\times 10^8)}{7.59 \times 10^{-19}}[/tex]

[tex]\lambda = 260.9nm[/tex]

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:

0.4 times the radius of moon

Explanation:

gravity on moon is equal to the one sixth of gravity on earth.

g' = g / 6

where, g' is the gravity on moon and g be the gravity on earth.

As the earth shrinks, the mass of earth remains same.

The acceleration due to gravity is inversely proportional to the square of radius of planet.

g' ∝ 1/R'²   .....(1)

Where, R' is the radius of moon.

g ∝ 1/R²      ..... (2)

Where, R be the radius of earth.

Divide equation (1) by (2)

g / g' = R'² / R²

Put g' = g / 6

6 = R'² / R²

2.5 = R' / R

R = R' / 2.5 = 0.4 R'

Thus, the radius of earth should be 0.4 times the radius of moon.

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:

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:

106.83

Explanation:

N = 332, l = 14 cm = 0.14 m, i = 0.88 A, B = 0.28 T

Let ur be the relative permeability

B = u0 x ur x n x i

0.28 = 4 x 3.14 x 10^-7 x ur x 332 x 0.88 / 0.14       ( n = N / l)

ur = 106.83