A record of travel along a straight path is as follows:

1. Start from rest with constant acceleration of 2.04 m/s2 for 11.0 s.
2. Maintain a constant velocity for the next 2.85 min.
3. Apply a constant negative acceleration of −9.73 m/s2 for 2.31 s.

(a) What was the total displacement for the trip?

(b) What were the average speeds for legs 1, 2, and 3 of the trip, as well as for the complete trip?

(C)COMPLETE TRIP:

Answer:

[tex]p_2 = 664081 N/m^{2}[/tex]

Explanation:

from the ideal gas law we have

PV = mRT

[tex]P = \rho RT[/tex]

[tex]\rho = \frac{P}{RT}[/tex]

HERE  R is gas constant for dry air  =  287  J K^{-1} kg^{-1}

[tex]\rho = \frac{7.00 10^{5}}{287(18+273)}[/tex]

[tex]\rho = 8.38 kg/m^{3}[/tex]

We know by ideal gas law

[tex]\rho = \frac{m_1}{V_1}[/tex]

[tex]m_1 = \rho V_1 = 8.38 *2*10^{-3}[/tex]

[tex]m_1 = 0.0167 kg[/tex]

for m_2

[tex]m_2 = \rho V_i - V_removed[/tex]

[tex]m_2 = 8.38*(.002 - 10^{-4})[/tex]

[tex]m_2 = 0.0159 kg[/tex]

WE KNOW

PV = mRT

V, R and T are constant therefore we have

P is directly proportional to mass

[tex]\frac{p_2}{p_1}=\frac{m_2}{m_1}[/tex]

[tex]p_2 = p_1 * \frac{m_2}{m_1}[/tex]

[tex]p_2 =7*10^{5} * \frac {.0159}{0.0167}[/tex]

[tex]p_2 = 664081 N/m^{2}[/tex]

This problem can be solved using Boyle's Law, which relates the pressure and volume of a gas at a constant temperature. The question asks for the new pressure of a bicycle tire after letting out a certain volume of air. The answer is approximately 7.37 x 10⁵ Pa.

The subject of this question is gas laws, specifically Boyle's Law which states that the pressure and volume of a gas have an inverse relationship when the temperature is kept constant. Assuming the temperature and volume of the tire remain constant before and after you let out the air, when a volume of 100 cm³ (which we will convert to 0.1 L for consistency) of air is let out, the new total volume of the gas is 1.9 L.

According to Boyle's Law, P1*V1 = P2*V2, where P1 and V1 represent the initial pressure and volume, and P2 and V2 represent the final pressure and volume. Plugging the values into this equation, we get:

(7.00 x 10⁵ Pa)(2.00 L) = P2 * (1.9 L)

Which gives us:

P2 = (7.00 x 10⁵ Pa * 2.00 L) / 1.9 L

Therefore, the pressure in the bike tire after letting out 100 cm³ of gas is approximately 7.37 x 10⁵ Pa.

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The International Space Station's operational weight in orbit is effectively zero due to its state of continuous free-fall around Earth, even though the gravitational force at its altitude is not significantly less than on Earth's surface.

The question relates to the weight of the International Space Station (ISS) when in orbit. Weight in physics is defined as the force exerted on an object due to gravity, calculated as the product of mass and gravitational acceleration (g). On Earth's surface, g is approximately 9.81 m/s2, but this value decreases with altitude due to the equation g = GME / r2, where G is the gravitational constant, ME is Earth's mass, and r is the distance from Earth's center. At the ISS's altitude (> 350 km), g is about 8.75 m/s2. However, it's crucial to understand that while the ISS has a significant mass, leading to a large weight calculation on Earth, its apparent weight in orbit is effectively zero due to it being in a continuous free-fall state around Earth, experiencing microgravity. This explains why astronauts appear weightless, even though the actual gravitational force at that altitude is not much less than on Earth's surface. Therefore, while the ISS has a calculable weight based on its mass and Earth's gravitational pull at its altitude, its operational weight in orbit, in terms of the experience within it, is zero.

Answer:

The distance between the person and the building is 68.48 meters.

Explanation:

It is given that,

Angle of elevation, θ = 30.6 degrees

Height of building, MP = 42 m

Height of person, AB = 1.5 m

We need to find the distance between person and building. It is given by BP.

Since, MN + NP = 42

So, MN = 40.5 m

Using trigonometric equation as :

[tex]tan\theta=\dfrac{MN}{AN}[/tex]

[tex]tan(30.6)=\dfrac{40.5}{AN}[/tex]

AN = 68.48 meters.

So, the distance between the person and the building is 68.48 meters. Hence, this is the required solution.

To determine the distance from a building, we use trigonometry and the formula adjacent = opposite / tangent(angle), taking into account the height of the building minus your height. The distance is calculated to be approximately 68.88 meters.

To find out how far you are from the building, we need to calculate the horizontal distance from the building's base to the point where you are standing. To do this, we can use trigonometry, specifically the tangent function which relates the angle of elevation to the opposite side and the adjacent side of a right-angle triangle. We need to consider the height of the building minus your height to find the correct opposite side.

Since the building is 42.0 meters tall, and you are 1.50 meters tall, the effective height we are looking at is 42.0 m - 1.50 m = 40.5 m. The angle of elevation you are looking at is 30.6 degrees. By using the formula tangent (angle) = opposite / adjacent, we can rearrange this to find the adjacent side (the distance from you to the building): adjacent = opposite / tangent (angle).

Therefore, the distance from you to the building is approximately adjacent = 40.5 m / tan(30.6°). Plugging in the values, we get:

Distance = 40.5 m / tan(30.6°) ≈ 40.5 m / 0.588 ≈ 68.88 m.

So, you are approximately 68.88 meters away from the building.

Answer:

Final velocity, v = 1.74 m/s

Explanation:

Given that,

Mass of car 1, m₁ = 22509 kg

Velocity of car 1, v₁ = 4.21 m/s

Mass of car 2, m₂ = 31647 kg

It is stationary, v₂ = 0

Let v be the velocity of the two cars after they collide and attach. It can be calculated using law of conservation of momentum as :

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]

[tex]v=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}[/tex]

[tex]v=\dfrac{22509\ kg\times 4.21\ m/s+0}{22509\ kg+31647\ kg}[/tex]

v = 1.74 m/s

So, the velocity of two cars after the collision is 1.74 m/s. Hence, this is the required solution.

Explanation:

It is given that,

Mass of the shot, m = 7.27 kg

Time taken to accelerate, t = 1.3 s

It is shot from rest to 16 m/s and it raises to a height of 0.9 m. We need to find the power output of the shot-putter. It is given by :

[tex]P=\dfrac{energy}{time}[/tex]

Energy = kinetic energy + potential energy

[tex]E=\dfrac{1}{2}\times 7.27\ kg\times (16\ m/s)^2+7.27\ kg\times 9.8\ m/s^2\times 0.9\ m[/tex]

E = 994.68 J

Power, [tex]P=\dfrac{994.68\ J}{1.3\ s}[/tex]

P = 765.13 Watts

We know that, 1 horse power = 745.7 watts

Or P = 1.02 horse power

Hence, this is the required solution.

Answer : The mass of cereal in two units are, 34.815 oz and 978.075 grams respectively.

Explanation :

As there are two systems for measuring the units which are English and Metric system.

The Metric System measured the things in meters, grams, liters, etc and adds prefixes like kilo, milli and centi to the count orders of the magnitude.

The English system measured the things in feet, inches, pounds, mile, etc.

As we are given the mass of cereal in 978 grams that means 987 grams is in metric unit. Now we have to convert metric unit into English unit.

Conversion factor used :  (1 oz = 28.350 g)

As, 28.350 grams = 1 oz

So, 987 grams = [tex]\frac{987\text{ grams}}{28.350\text{ grams}}\times 1oz[/tex]

                        =  34.815 oz

As we are given the mass of cereal in 34.5 oz that means 34.5 oz is in metric unit. Now we have to convert metric unit into English unit.

Conversion factor used :  (1 oz = 28.350 g)

As, 1 oz = 28.350 grams

So, 34.5 oz = [tex]\frac{34.5\text{ oz}}{1\text{ oz}}\times 28.350\text{ grams}[/tex]

                  =  978.075 grams

Therefore, the mass of cereal in two units are, 34.815 oz and 978.075 grams respectively.

The conversion factor from ounces to grams, divide 978 grams by 34.5 ounces, resulting in approximately 28.3 g/oz when rounded to three significant figures.

To find a conversion factor between English and metric units using the information that a box of cereal has a mass of 978 grams and 34.5 ounces, you would divide the number of grams by the number of ounces.

Conversion factor = 978 g / 34.5 oz ≈ 28.34783 g/oz

However, we need to consider significant figures in our answer. The number 34.5 has three significant figures, and the number 978 has three significant figures as well. Therefore, we can justify three significant figures in our conversion factor, giving us 28.3 g/oz as the conversion with significant figures properly accounted for.

Answer:

49.85 V

Explanation:

u = 0, s = 5.62 cm, t = 1.15 x 10^-6 s

Let the electric field is E and voltage is V.

Use second equation of motion

s = ut + 1/2 a t^2

5.62 x 10^-2 = 0 + 0.5 a x (1.15 x 10^-6)^2

a = 8.5 x 10^10 m/s^2

m x a = q x E

E = m x a / q

E = (1.67 x 10^-27 x 8.5 x 10^10) / (1.6 x 10^-19)

E = 887.19 V/m

V = E x s

V = 887.19 x 5.62 x 10^-2 = 49.85 V

Answer:

Spongebob: Bye Mr. Krabs! Bye Squidward! BYE SQUIDWARD!

Patrick: (clearly triggered) Why'd you say "bye squidward" twice?

Spongebob: I LiKe SqUiDwArD

Answer:

The amplifier gain is 30 dB.

(c) is correct option.

Explanation:

Given that,

Output power = 100 mW

Input power = 0.1 mW

We need to calculate the amplifier gain in dB

Using formula of power gain

[tex]a_{p}=10\ log_{10}(A_{p})[/tex]....(I)

We calculate the [tex]A_{p}[/tex]

[tex]A_{p}=\dfrac{P_{o}}{P_{i}}[/tex]

[tex]A_{p}=\dfrac{100}{0.1}[/tex]

[tex]A_{p}=1000[/tex]

Now, put the value of  [tex]A_{p}[/tex] in equation (I)

[tex]a_{p}=10\ log_{10}(1000)[/tex]

[tex]a_{p}=10\times log_{10}10^{3}[/tex]

[tex]a_{p}=10\times 3log_{10}10[/tex]

[tex]a_{p}=30\ dB[/tex]

Hence, The amplifier gain is 30 dB.

The gain of an amplifier, given a power output of 100 mW and an input power of 0.1 mW, can be calculated using the gain formula in decibels, which results in a gain of 30 dB.

In this context, the gain of the amplifier can be calculated using the formula for Gain in decibels (dB), which is 10 times the log base 10 of the output power divided by the input power. Therefore, we first divide 100 mW by 0.1 mW, which gives us 1000. Taking the log base 10 of 1000 returns 3, and multiplying 3 by 10 gives us a gain of 30 dB.

So the correct answer to your question: 'An amplifier has a power output of 100 mW when the input power is 0.1 mW, what is the amplifier gain?' is option c which states that the gain is 30 dB.

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Answer:

[tex]T_C=256.2^{\circ}C[/tex]

Explanation:

Given that,

Efficiency of heat engine, [tex]\eta=40\%=0.4[/tex]

Temperature of hot source, [tex]T_H=427^{\circ}C[/tex]

We need to find the temperature of cold sink i.e. [tex]T_C[/tex]. The efficiency of heat engine is given by :

[tex]\eta=1-\dfrac{T_C}{T_H}[/tex]

[tex]T_C=(1-\eta)T_H[/tex]

[tex]T_C=(1-0.4)\times 427[/tex]

[tex]T_C=256.2^{\circ}C[/tex]

So, the temperature of the cold sink is 256.2°C. Hence, this is the required solution.

Answer:

Explanation:

When the pendulum falls freely the net acceleration due to gravity is zero.

As we know that the time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity, thus the time period becomes infinity.

In freefall, the pendulum's effective acceleration due to gravity becomes zero, causing the pendulum to not swing, and its period becomes theoretically infinite and immeasurable.

Effect of Freefall on a Pendulum's Period

When considering simple pendulum motion in an elevator under normal conditions, we can determine its periodic time (T) using the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. This equation illustrates that the period of the pendulum (T) is affected by two variables: the length of the pendulum (L) and the acceleration due to gravity (g).

When the elevator is in free fall, the effective acceleration g becomes zero because the elevator and the pendulum are both in a state of free fall with the same acceleration due to gravity. Therefore, in this scenario, the pendulum would experience weightlessness and would not oscillate, resulting in an infinite theoretical oscillation period, making the concept of a period inapplicable.

The period is normally independent of mass or amplitude for small angles, but since freefall changes the acceleration experienced by the pendulum to zero, it significantly affects the pendulum's oscillation, negating the normal conditions for calculating a pendulum's period.

The speed of an electron with a kinetic energy of 1.25 eV is approximately 1.57 x 10⁶ m/s, and the speed of a proton with the same kinetic energy is approximately 5.29 x 10⁵ m/s.

To calculate the speed of an electron and a proton with a kinetic energy of 1.25 electron volts (eV), we can use the kinetic energy formula and relate it to the speed of the particles. The kinetic energy (KE) of a particle is given by:

KE = (1/2) * m * v²

Where:

KE = kinetic energy

m = mass of the particle

v = speed of the particle

We are given the kinetic energy in electron volts (eV), but we need to convert it to joules (J) since the mass is given in kilograms (kg). The conversion factor is 1 eV = 1.60219 x 10⁻¹⁹ J.

So, the kinetic energy in joules is:

KE = 1.25 eV * 1.60219 x 10⁻¹⁹ J/eV = 2.0027375 x 10⁻¹⁹ J

Now, we can rearrange the kinetic energy formula to solve for the speed (v):

v = √((2 * KE) / m)

For an electron:

Mass of electron (mₑ) = 9.11 x 10⁻³¹ kg

v(electron) = √((2 * 2.0027375 x 10⁻¹⁹ J) / (9.11 x 10⁻³¹ kg))

Calculating this gives us the speed of the electron.

For a proton:

Mass of proton (m_p) = 1.67 x 10⁻²⁷ kg

v(proton) = √((2 * 2.0027375 x 10⁻¹⁹ J) / (1.67 x 10⁻²⁷ kg))

Calculating this gives us the speed of the proton.

Now, let's calculate these speeds.

After performing the calculations, the speed of the electron is approximately 1.57 x 10⁶ m/s, and the speed of the proton is approximately 5.29 x 10⁵ m/s.

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Answer:

Well..

Explanation:

That's impossible. I know because I once weighed 11.1 kN, and I was temporarily immobile. It's probably the same for a car, and therefore it can not be "traveling" anywhere at all.. unless you put the car on an airplane or a boat or something.

Answer:

115 m/s, 414 km/hr

Explanation:

There are two forces acting on a skydiver: gravity and air resistance (drag).  At terminal velocity, the two forces are equal and opposite.

∑F = ma

D − mg = 0

D = mg

Drag force is defined as:

D = ½ ρ v² C A

where ρ is the fluid density,

v is the velocity,

C is the drag coefficient,

and A is the cross sectional surface area.

Substituting and solving for v:

½ ρ v² C A = mg

v² = 2mg / (ρCA)

v = √(2mg / (ρCA))

We're given values for m and A, and we know the value of g.  We need to look up ρ and C.

Density of air depends on pressure and temperature (which vary with elevation), but we can estimate ρ ≈ 1.21 kg/m³.

For a skydiver falling headfirst, C ≈ 0.7.

Substituting all values:

v = √(2 × 80.0 kg × 9.8 m/s² / (1.21 kg/m³ × 0.7 × 0.140 m²))

v = 115 m/s

v = 115 m/s × (1 km / 1000 m) × (3600 s / hr)

v = 414 km/hr

The terminal velocity of the skydiver in m/s and km/h is;  115m/s  and  414 km/h

Using Given data :

mass of skydiver ( M ) = 80 kg

Cross sectional surface area ( A ) = 0.14 m^2

p ( fluid density ) ≈ 1.21 kg/m³.

C ( drag coefficient ) = 0.7

Determine the terminal velocity of the skydiver

At terminal velocity drag force and gravity is equal and opposite therefore canceling out each other

∑ F = ma

Drag force - Mg = 0

therefore;  D = Mg ----- ( 1 )

where D ( drag force ) = 1/2 pv² C A ---- ( 2 )

p = fluid density , C = drag coefficient , A = cross sectional area

v = velocity

Back to equations 1 and 2  ( equating them )

1/2 pv² CA = Mg ---- ( 3 )

v² = 2mg / ( p C A )

∴ V = √ ( 2mg / (p C A ))

V = √ ( 2 * 80 * 9.8 ) / ( 1.21 * 0.7 * 0.140 ))

    = 115 m/s  

also  V = 414 km/h

Hence we can conclude that the terminal velocity of the skydiver is in m/s and km/h are 115m/s and  414 km/h

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Answer:

Weight in planet X = 0.413 N

Explanation:

Weight = Mass x Acceleration due to gravity.

W = mg

Mass, m = 100 g = 0.1 kg

We have equation of motion s = ut + 0.5 at²

Displacement, s = 10 m

Initial velocity, u = 0 m/s

Time, t = 2.2 s

Substituting

        s = ut + 0.5 at²

        10 = 0 x 2.2 + 0.5 x a x 2.2²        

        a = 4.13 m/s²

Acceleration due to gravity, a = 4.13 m/s²

W = mg = 0.1 x 4.13 = 0.413 N

Weight in planet X = 0.413 N

In Planet X, a 100-g ball is released from rest from a height of 10.0 m and it takes 2.2 s for it to reach the ground. The weight of the ball on the surface of Planet X is 0.41 N.

In physics, gravitational acceleration is the acceleration of an object in free fall within a vacuum (and thus without experiencing drag).

A 100-g (m) ball is released from rest from a height of 10.0 m (s) and it takes 2.2 s (t) for it to reach the ground. We can calculate the gravitational acceleration using the following kinematic equation.

s = 1/2 × g × t²

g = 2 s / t² = 2 (10.0 m) / (2.2 s)² = 4.1 m/s²

We will use Newton´s second law of motion.

w = m × g = 0.100 kg × 4.1 m/s² = 0.41 N

In Planet X, a 100-g ball is released from rest from a height of 10.0 m and it takes 2.2 s for it to reach the ground. The weight of the ball on the surface of Planet X is 0.41 N.

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Answer:

Answer to the question is: 1837.65 millimeters of mercury are equal to 245 kPa.

Explanation:

1 kPa are equal to 7.50062 millimeters of mercury.

Final answer:

To convert the pressure from 245 kPa to mmHg, first convert kPa to atm, then multiply by the conversion factor from atm to mmHg. The pressure is 1837.68 mmHg.

Explanation:

To convert the pressure in a container from kilopascals (kPa) to millimeters of mercury (mmHg), we use the conversion factor that 1 atmosphere (atm) is equivalent to 760 mmHg. First, we convert the given pressure in kilopascals to atmospheres:

1 atm = 101.325 kPa

So, to convert 245 kPa to atm, we divide 245 kPa by 101.325 kPa/atm:

245 kPa / 101.325 kPa/atm = 2.418 atm

Next, we convert atmospheres to millimeters of mercury (mmHg) using the conversion factor:

2.418 atm x760 mmHg/atm = 1837.68 mmHg

Therefore, the pressure in the container is 1837.68 mmHg.

Answer:

[tex]C = 2.48 \times 10^{-4} Farad[/tex]

Explanation:

As per the equation of voltage on capacitor we know that

[tex]V = V_{max}(1 - e^{-\frac{t}{\tau}})[/tex]

now we know that voltage reached to its 80% of maximum value in 4 second time

so we will have

[tex]0.80 V_{max} = V_{max}(1 - e^{-\frac{4}{\tau}})[/tex]

[tex]0.20 = e^{-\frac{4}{\tau}}[/tex]

[tex]-\frac{4}{\tau} = ln(0.20)[/tex]

[tex]-\frac{4}{\tau} = -1.61[/tex]

[tex]\tau = 2.48[/tex]

as we know that

[tex]\tau = RC[/tex]

[tex](10 k ohm)(C) = 2.48[/tex]

[tex]C = 2.48 \times 10^{-4} Farad[/tex]

Answer:

rate of change in temperature of copper is more than the rate of change in temperature of aluminium.

so here copper will reach to our body temperature first

Explanation:

As we know that rate of energy absorb by the two sphere is same

so here we will have

[tex]\frac{dQ}{dt} = ms\frac{\Delta T}{\Delta t}[/tex]

now for copper sphere we will have

[tex]\frac{dQ}{dt} = 50(0.4)\frac{\Delta T}{\Delta t}[/tex]

[tex]\frac{\Delta T}{\Delta t} = \frac{1}{20}\frac{dQ}{dt}[/tex]

now for Aluminium sphere we will have

[tex]\frac{dQ}{dt} = 25(0.9)\frac{\Delta T}{\Delta t}[/tex]

[tex]\frac{\Delta T}{\Delta t} = \frac{1}{22.5}\frac{dQ}{dt}[/tex]

So rate of change in temperature of copper is more than the rate of change in temperature of aluminium.

so here copper will reach to our body temperature first

Answer:

work =p×v =3.27×10^5×1.5×10^-3 =490.5 joule

efficiency =w/q in

:. qin= w/efficiency =490.5/0.225=2180 joule

qout =q in - work =1689.5 joule

q out is work given as heat

The engine must give up 1689.5 J of heat to the low-temperature reservoir after calculating the total work done by the gas and accounting for the engine's efficiency.

To find the amount of work the engine gives up as heat, we first calculate the total work done by the gas using the formula W = PΔV, where W is work, P is pressure, and ΔV is the change in volume. Given a constant pressure of 3.27 x 105 Pa and a change in volume of 1.50 x 10-3 m3, the work done is:

W = PΔV = 3.27 x 105 Pa x 1.50 x 10-3 m3 = 490.5 J.

The efficiency of the engine is the ratio of the useful work output to the total work input, given by  ext_eta = useful work / total work. The equation that relates efficiency, work done (W), and heat given up (Q) is  ext_eta = W / (Q + W). We rearrange the equation to solve for Q:

Q = W /  ext_eta - W

Substituting the known values:

Q = 490.5 J / 0.225 - 490.5 J = 2180 J - 490.5 J = 1689.5 J.

Therefore, the engine must give up 1689.5 J of heat to the low-temperature reservoir.

Answer:

q = 1.815 \times 10^{-8} C

Charge on one plate is positive in nature and on the other plate it is negative in nature.

Explanation:

E = 8.20 x 10^5 V/m, A = 25 cm^2, d = 22.45 mm

According to the Gauss's theorem in electrostatics

The electric field between the two plates

[tex]E = \frac{\sigma }{\varepsilon _{0}}[/tex]

[tex]{\sigma }= E \times {\varepsilon _{0}}[/tex]

[tex]{\sigma }= 8.20 \times 10^{5} \times {8.854 \times 10^{-12}[/tex]

[tex]{\sigma }= 7.26 \times 10^{-6} C/m^{2}[/tex]

Charge, q = surface charge density x area

[tex]q = 7.26 \times 10^{-6} \times 25 \times 10^{-4}[/tex]

q = 1.815 \times 10^{-8} C

Answer:

2.07 m/s

Explanation:

m = 1.07 x 10^5 kg, u1 = 3.41 m/s, u2 = 1.4 m/s

Let the speed of three coupled car after collision is v

Use conservation of momentum

m x u1 + 2 m x u2 = 3 m x v

u1 + 2 u2 = 3 v

3.41 + 2 x 1.4 = 3 v

v = 2.07 m/s

Answer:

option (B)

Explanation:

q = 3 c, m = 4 x 10^-3 kg, g  9.8 m/s^2,

In the equilibrium condition, the weight of the charge particle is balanced by the electrostatic force.

q E = mg

Electrostatic force = m g = 4 x 10^-3 x 9.8 = 0.0392 N

Answer:

0.94 m/s^2 downwards

Explanation:

m = 70 kg, m g = 70 x 9.8 = 686 N

R = 620 N

Let the acceleration be a, as the apparent weight decreases so the elevator is moving downwards with an acceleration a.

mg - R = ma

686 - 620 = 70 x a

a = 0.94 m/s^2

Answer:

(a) 0.33 second

(b) 6 cm/s

Explanation:

Frequency, f = 3 waves per second

wavelength, λ = 2 cm = 0.02 m

(a) The period of wave is defined as the time taken by the wave to complete one oscillation. It is the reciprocal of frequency.

T = 1 / f = 1 / 3 = 0.33 second

(b) the relation between wave velocity, frequency and wavelength is given by

v = f x λ

v = 3 x 0.02 = 0.06 m /s

v = 6 cm /s

Final answer:

The period of the waves generated by a grasshopper in water is 0.333 seconds, and the wave velocity is 0.06 m/s.

Explanation:

Calculating the Period and Wave Velocity

When dealing with waves generated by a grasshopper floating in water, two key properties to determine are the period of the waves and the wave velocity.

(a) The Period of the Waves

The period (T) of a wave is the amount of time it takes for one complete wave cycle to pass. It can be calculated as the inverse of the frequency (f), which is the rate at which waves are generated. The formula to find the period is:

T = 1/f

In this case, the grasshopper generates waves at a frequency of 3 waves per second (3 Hz). Therefore, the period is:

T = 1/3 Hz = 0.333 seconds

(b) The Wave Velocity

The velocity (v) of a wave is determined by multiplying the frequency (f) of the wave by its wavelength (λ). The formula for wave velocity is:

v = f × λ

Here, the wavelength given is 2 cm, which we need to convert to meters (since the SI unit for velocity is m/s). Thus:

λ = 2 cm = 0.02 m

The velocity of the waves generated by the grasshopper is then:

v = 3 Hz × 0.02 m = 0.06 m/s

Answer:

4.4 cm

Explanation:

B = 1.98 mT = 1.98 x 10^-3 T, v = 1.53 x 10^7 m/s, m = 9.1 x 10^-31 kg

q = 1.6 x 10^-19 C

(a) The force due to the magnetic field is balanced by the centrpetal force

mv^2 / r = q v B

r = m v / q B

r = (9,1 x 10^-31 x 1.53 x 10^7) / (1.98 x 10^-3 x 1.6 x 10^-19)

r = 0.044 m = 4.4 cm

Answer:

Work done by the gravitational force is 37 mJ.

Explanation:

It is given that,

Mass of the raindrop, [tex]m=3.26\times 10^{-5}\ kg[/tex]

It falls from a height of, h = 115 m

It falls vertically at constant speed under the influence of gravity and air resistance. We need to find the  work done on the raindrop by the gravitational force. It is given by :

[tex]W=mgh[/tex]

[tex]W=3.26\times 10^{-5}\ kg\times 9.8\ m/s^2\times 115\ m[/tex]

W = 0.0367 J

or

W = 0.037 J = 37 mJ

So, the work done on the raindrop by the gravitational force is 37 mJ. Hence, this is the required solution.

Final answer:

The work done on a raindrop of mass 3.26 times [tex]10^{-5}[/tex] kg by the gravitational force.

Explanation:

To calculate the work done on a raindrop by the gravitational force as it falls, we use the formula for work: W = mgh, where W is the work done, m is the mass of the object, g is the acceleration due to gravity (9.8 m/s2), and h is the height the object falls through.

In this case, the mass m of the raindrop is 3.26 times 10-5 kg, and the height h is 115 m. So:

W = (3.26 times 10-5 kg)(9.8 [tex]m/s^2[/tex])(115 m) = 0.0368134 J

Therefore, the work done on the raindrop by the gravitational force as it falls 115 m is approximately 0.0368134 joules.

Answer:

The wavelength of the light is 633 nm.

Explanation:

Given that,

Distance between the two slits d= 0.025 cm

Distance between the screen and slits D = 120 cm

Distance between the slits y= 1.52 cm

We need to calculate the angle

Using formula of double slit

[tex]\tan\theta=\dfrac{y}{D}[/tex]

Where, y = Distance between the slits

D = Distance between the screen and slits

Put the value into the formula

[tex]\tan\theta=\dfrac{1.52}{120}[/tex]

[tex]\theta=\tan^{-1}\dfrac{1.52}{120}[/tex]

[tex]\theta=0.725[/tex]

We need to calculate the wavelength

Using formula of wavelength

[tex]d\sin\theta=n\lambda[/tex]

Put the value into the formula

[tex]0.025\times\sin0.725=5\times\lambda[/tex]

[tex]\lambda=\dfrac{0.025\times10^{-2}\times\sin0.725}{5}[/tex]

[tex]\lambda=6.326\times10^{-7}\ m[/tex]

[tex]\lambda=633\ nm[/tex]

Hence, The wavelength of the light is 633 nm.

The angular velocity of the CD is 754.4 rad/s. The time it takes for the CD to rotate through 90 degrees is 0.0021 seconds. The average angular acceleration of the CD is 188.6 rad/s².

To calculate the angular velocity of the CD, we can convert the given 7200 rev/min to radians per second. Since one revolution is equal to 2π radians, we can use the conversion factor to find the angular velocity. Thus, the angular velocity of the CD is 754.4 rad/s.

To calculate the time it takes for the CD to rotate through 90 degrees, we need to find the fraction of the total rotation that corresponds to 90 degrees. Since a full rotation is 360 degrees or 2π radians, 90 degrees is equal to π/2 radians. We can then use the formula Δθ = ωΔt to find the time it takes, where Δθ is the angle in radians, ω is the angular velocity, and Δt is the time. Rearranging the formula, we have Δt = Δθ/ω. Substituting the values, we get Δt = π/2 / 754.4 = 0.0021 seconds.

The average angular acceleration can be found using the formula α = (ωf - ωi) / Δt, where α is the angular acceleration, ωf is the final angular velocity, ωi is the initial angular velocity, and Δt is the time interval. The CD starts from rest and reaches full speed in 4 seconds, so the initial angular velocity is 0. Using the given full speed of 7200 rev/min, we can convert it to radians per second and use it as the final angular velocity. Thus, the average angular acceleration is α = (754.4 rad/s - 0 rad/s) / 4 s = 188.6 rad/s².

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Answer:

d

Explanation:

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Answer:

10042.6 ohm

Explanation:

f = 10 kHz = 10000 Hz, L = 36 mH = 0.036 H, R = 10 kilo Ohm = 10000 ohm

C = 5 nF = 5 x 10^-9 F

XL = 2 x π x f x L

XL = 2 x 3.14 x 10000 x 0.036 = 2260.8 ohm

Xc = 1 / ( 2 x π x f x C) = 1 / ( 2 x 3.14 x 10000 x 5 x 10^-9)

Xc = 3184.7 ohm

Total impedance is Z.

Z^2 = R^2  + (XL - Xc)^2

Z^2 = 10000^2 + ( 2260.8 - 3184.7 )^2

Z = 10042.6 ohm