A bobsled, moving at 32 m/s, decelerates to 22 m/s at a rate of 4.8 m/s2. Determine the distance traveled by the bobsled during this time.

Answer:

Option c

Explanation:

Soil tilth refers to the physical condition of the soil, particularly with regards to the suitability of the soil for the growth of crop.

The determinants of the tilth in the soil incorporates the arrangement and steadiness of aggregated particles of the soil, pace of water penetration, level of air circulation, dampness content and seepage.

Thus option C follows the definition of the soil tilth.

Answer:

Option (C)

Explanation:

The soil tilth is defined as one of the physical property that determines the necessary condition for the better growth of plants and crops.

This property depends upon various factors such as-

These are the main features that characterize the soil tilth, and fulfilling the above required necessary condition, a plant grows in a much better way.

Thus, the correct answer is option (C).

Even though GTAW welding is slower and requires a higher skill level as compared to other manual welding processes, it is still in demand because it can be used to make extremely high-quality welds in applications where weld integrity is critical.​

Answer: Option D

Explanation:

Among the various sorts of welding forms accessible today, gas tungsten bend welding, or GTAW is commonly viewed as the most moving welding technique to ace. In spite of the fact that it is additional challenging than other welding strategies, in any event, when drilled with the consideration of a specialist, the improved quality and nature of welds created with GTAW.

It can offer a pragmatic option in contrast to less complex welding techniques, especially for restricted segments of hardened steel and non-ferrous metals, for example, copper, aluminium, and magnesium combinations.

One of the principal modern applications for GTAW welding started inside the aeronautic trade. Current fields where GTAW abilities are most sought after incorporate the ship fitting exchange, as aluminium welding assumes a significant job in the development of a ship's superstructure, and the assembling and fix of bikes.

Answer:

[tex]y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})[/tex]

Explanation:

As we know that the wave equation is given as

[tex]y = A sin(\omega t - k x + \phi_0)[/tex]

now we have

[tex]A = 0.19 m[/tex]

[tex]\lambda = 2.6 m[/tex]

so we have

[tex]k = \frac{2\pi}{\lambda}[/tex]

[tex]k = \frac{2\pi}{2.6}[/tex]

[tex]k = 2.42  per m[/tex]

also we have

[tex]T = 1.2 s[/tex]

so we have

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

[tex]\omega = \frac{2\pi}{1.2}[/tex]

[tex]\omega = 5.23 rad/s[/tex]

now we know that at t = 0 and x = 0 wave is at y = 0.19 m

so we have

[tex]\phi_0 = \frac{\pi}{2}[/tex]

so we have

[tex]y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})[/tex]

Answer:71 dB

Explanation:

Given

sound Level [tex]\beta _1=124 dB[/tex]

distance [tex]r_1=5.01 m[/tex]

From sound Intensity

[tex]\beta =10dB\log (\frac{I_1}{I_0})[/tex]

[tex]124=10dB\log (\frac{I_1}{I_0})[/tex]

[tex]12.4=\log (\frac{I_1}{I_0})[/tex]

[tex]I_1=(1\times 10^{-12})\times 10^{12.4}[/tex]

[tex]I_1=2.51 W/m^2[/tex]

we know Intensity [tex]I\propto ^\frac{1}{r^2}[/tex]

[tex]I_1r_1^2=I_2r_2^2[/tex]

[tex]I_2=I_1(\frac{r_1}{r_2})^2[/tex]

[tex]I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2[/tex]

[tex]I_2=1.24\times 10^{-5} W/m^2[/tex]

Sound level corresponding to [tex]I_2[/tex]

[tex]\beta _2=10\log (\frac{I_2}{I_0})[/tex]

[tex]\beta _2=10\log (\frac{1.24\times 10^{-5}}{1\times 10^{-12}})[/tex]

[tex]\beta _2=70.93\approx 71 dB[/tex]

Answer:

becomes narrower

Explanation:

Confidence intervals for the population mean μ and population proportion p becomes narrower as the size of the sample increases.

As,the sample size increases,standard error decreases,so margin of error decreases and hence width of CI decreases.

Answer:

(A) Vf = 13.8 m/s

(B)  Vf = 15.1 m/s      

Explanation:

length of rope (L) = 35 m

angle to the vertical = 44 degrees

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) from conservation of energy

final kinetic energy + final potential energy = initial kinetic energy + initial potential energy

0.5m(Vf)^{2} + mg(Hf) =  0.5m(Vi)^{2} + mg(Hi)

where

m = mass

Hi = initial height = 35 cos 44 = 25.17

Hf = final height = length of vine = 35 m

Vi = initial velocity = 0 since he starts from rest

Vf = final velocity

the equation now becomes

0.5m(Vf)^{2} + mg(Hf) = mg(Hi)

0.5m(Vf)^{2} = mg (Hi - Hf)

0.5(Vf)^{2} = g (Hi - Hf)

0.5(Vf)^{2} = 9.8 x (25.17 - 35)

0.5(Vf)^{2} = - 96.3  (the negative sign tells us the direction of motion is downwards)

Vf = 13.8 m/s

(B) when the initial velocity is 6 m/s the equation remains

      0.5m(Vf)^{2} + mg(Hf) =  0.5m(Vi)^{2} + mg(Hi)

       m(0.5(Vf)^{2} + g(Hf)) =  m(0.5(Vi)^{2} + g(Hi))

      0.5(Vf)^{2} + g(Hf) = 0.5(Vi)^{2} + g(Hi)

      0.5(Vf)^{2} = 0.5(Vi)^{2} + g(Hi) - g(Hf)

       0.5(Vf)^{2} = 0.5(6)^{2} + (9.8 x (25.17 - 35))

        0.5(Vf)^{2} =  -114.3  ( just as above, the negative sign tells us the direction of motion is downwards)      

       Vf = 15.1 m/s

Answer:

a) [tex]v_{f} \approx 0.328\,\frac{m}{s}[/tex], b) [tex]v_{f} \approx 6.009\,\frac{m}{s}[/tex]

Explanation:

Let consider that bottom has a height of zero. The motion of Tarzan can be modelled after the Principle of Energy Conservation:

[tex]U_{g,1} + K_{1} = U_{g,2} + K_{2}[/tex]

The final speed is:

[tex]K_{2} = U_{g,1} - U_{g,2} + K_{1}[/tex]

[tex]\frac{1}{2}\cdot m \cdot v_{f}^{2} = m\cdot g \cdot L\cdot (\cos \theta_{2}-\cos \theta_{1}) + \frac{1}{2}\cdot m \cdot v_{o}^{2}[/tex]

[tex]v_{f}^{2} = 2 \cdot g \cdot L \cdot (\cos \theta_{2} - \cos \theta_{1}) + v_{o}^{2}[/tex]

[tex]v_{f} = \sqrt{v_{o}^{2}+2\cdot g \cdot L \cdot (\cos \theta_{2}-\cos \theta_{1})}[/tex]

a) The final speed is:

[tex]v_{f} = \sqrt{(0\,\frac{m}{s} )^{2}+2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (35\,m)\cdot (\cos 0^{\textdegree}-\cos 44^{\textdegree})}[/tex]

[tex]v_{f} \approx 0.328\,\frac{m}{s}[/tex]

b) The final speed is:

[tex]v_{f} = \sqrt{(6\,\frac{m}{s} )^{2}+2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (35\,m)\cdot (\cos 0^{\textdegree}-\cos 44^{\textdegree})}[/tex]

[tex]v_{f} \approx 6.009\,\frac{m}{s}[/tex]

The mass of Actinium 227 left is 0.0078 kg

Explanation:

The amount of mass left of a radioactive isotope after time t is given by the equation:

[tex]m(t) = m_0 (\frac{1}{2})^{-\frac{t}{\tau_{1/2}}}[/tex]

where

[tex]m_0[/tex] is the initial amount of the sample

t is the time

[tex]\tau_{\frac{1}{2}}[/tex] is the half-life of the isotope

For the sample of Actinium 227 in this problem,

[tex]m_0 = 2 kg[/tex]

[tex]\tau_{1/2}=20 years[/tex]

t = 160 years

Substituting into the equation,

[tex]m(160) = (2 kg) (\frac{1}{2})^{-\frac{160}{20}}=0.0078 kg[/tex]

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Answer:

Part a)

[tex]A = \frac{N}{s^4}[/tex]

[tex]B = \frac{N}{s^2}[/tex]

PART B)

[tex]I = 0.393 Ns[/tex]

Explanation:

PART A)

As we know that the force is given as

[tex]F = At^4 + B t^2[/tex]

here we know that each term of the equation must have same dimensions

so we will have

[tex]At^4 = N[/tex]

[tex]A = \frac{N}{s^4}[/tex]

similarly for other term

[tex]Bt^2 = N[/tex]

[tex]B = \frac{N}{s^2}[/tex]

PART B)

Impulse given by the force is given as

[tex]impulse = \int Fdt[/tex]

now we have

[tex]I = \int (At^4 + Bt^2)dt[/tex]

[tex]I = \int (4.50 t^4 + 8.75 t^2) dt[/tex]

[tex]I = \frac{4.50(0.5)^5}{5} + \frac{8.75(0.5)^3}{3}[/tex]

[tex]I = 0.028 + 0.36[/tex]

[tex]I = 0.393 Ns[/tex]

The SI units of the constants A and B are kg·m/s·6 and kg·m/s·4 respectively, essential for ensuring dimensional consistency in the force equation. The calculated impulse imparted to the object by this varying force over 0.500 s is 12.9 N·s.

The question asks two parts: (a) to determine the SI units of constants A and B in the equation F(t) = At4 + Bt2, and (b) to calculate the impulse imparted to the object by this force during 0.500 s. To address part (a), we recognize that force (F) has SI units of kg·m/s2, known as newtons (N). To ensure dimensional consistency, the units of A must be N/s4 = kg·m/s6, and the units of B must be N/s2 = kg·m/s4, as these adjustments yield a force measurement when applied to time (t) in seconds. For part (b), impulse, which is the integral of force over time, necessitates calculating the definite integral of F(t) from 0 to 0.500 s. Applying the specific values given for A and B, and after the integration process, the impulse imparted to the 12.25-kg object is found to be 12.9 N·s, horizontally.

The first confirmed detections of extrasolar planets were made in the 1990s, with the first planet discovered orbiting a main-sequence star similar to our Sun detected in 1995.

The first confirmed detections of extrasolar planets, or planets outside our own solar system, occurred in the 1990s. Before this time, the existence of such planets was believed, but had yet to be confirmed. The breakthrough came in 1992 when two planets were detected orbiting a pulsar, a type of neutron star. However, the first confirmed extrasolar planet orbiting a main-sequence star similar to our Sun, was discovered in 1995. This marked a significant event in the field of astronomy and has led to the discovery of thousands more extrasolar planets since.

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The first confirmed detections of extrasolar planets happened in the 1990s. The major breakthrough came in 1995, when astronomers discovered a planet orbiting the regular star 51 Pegasi, heralding a new era in the search for planets outside of our solar system.

The first confirmed detections of extrasolar planets, or planets outside of our own solar system, occurred in the 1990s. Before this, while many astronomers and theorists speculated about the existence of planets around other stars, none had indeed been confirmed. This changed dramatically when in 1995, Didier Queloz and Michel Mayor of the Geneva Observatory discovered a planet around a regular star, 51 Pegasi. This pioneering discovery proved that our solar system was not alone in the universe, leading to the detection of thousands of extrasolar planets in the following decades. The detection techniques they proposed, specifically the Doppler and transit techniques, have enabled astronomers to observe the effects of planets on the stars they orbit without directly seeing the planets themselves.

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Answer: Thermal comductivity (K) is 3.964x 10 ^-3 W/m.k

Explanation:

Thermal comductivity K = QL/A∆T

Q= Amount of heat transferred through the material in watts = 75W

L= Distance between two isothermal planes = 0.740mm

A= Area of the surface in square metres = 2m^2

∆T= Temperature change = (37-30) °C.

Solving this : K =( 75 x 0.740 x 10^-3)/ 2 x (37-30)

K = 3.964x 10 ^-3 W/m.k

The question pertains to calculating the thermal conductivity of human skin, a Physics concept linked to heat transfer. By using Fourier's Law of heat conduction, and rearranging the formula to solve for thermal conductivity using given data, an approximate thermal conductivity can be obtained.

The query is related to the determination of thermal conductivity of human skin based on the known parameters. This phenonmenon belongs to the domain of Physics, specifically heat transfer. Here, thermal conductivity is the measure of a material's ability to conduct heat. In this scenario, you have to consider the heat conduction through the skin, which relies on Fourier's Law of heat conduction. It can be represented as:

Q = (k*A*(T1 - T2))/d

Where, Q is the heat transfer rate, k is the thermal conductivity, A is the area of heat transfer, T1 and T2 are the initial and final temperatures, and d is the thickness of the material, in this case, the skin.

From the given data, you can plug in the values into this formula. However, our primary objective is to find out 'k'. Rearranging the formula to find k gives us:

k = (Q * d) / (A * (T1 - T2))

Now, if we put all the given values into the formula, we get:

k = (75 W * 0.00074 m) / (2 m^2 * (37°C - 30°C))

Solving this would provide us with the estimate of thermal conductivity of the skin.

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The work done is 1425 J

Explanation:

The work done by a force to move an object is given by

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

where

F is the magnitude of the force

d is the displacement

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

For the suitcase in this problem, we have:

F = 95.0 N is the force applied

d = 15.0 m is the displacement

[tex]\theta=0[/tex], since the force is parallel to the displacement

Substituting, we find

[tex]W=(95.0)(15.0)(cos 0)=1425 J[/tex]

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To find the final angular velocity of the system, we need to apply the principle of conservation of angular momentum. The final angular momentum can be obtained by equating the initial angular momentum and the final angular momentum of the system. Solving for the final angular velocity gives us a value of approximately 38.54 rad/s.

To find the final angular velocity of the system, we need to apply the principle of conservation of angular momentum. The initial angular momentum of the system is given by:

Li = Itωt + Irωr

Where It and Ir are the moments of inertia of the turntable and the runner respectively, and ωt and ωr are their respective angular velocities.

Since the runner comes to rest relative to the turntable, ωr = 0. Therefore, the final angular momentum of the system is:

Lf = Itωt

Using the conservation of angular momentum principle, we can set Li equal to Lf:

Itωt = Itωt

Substituting the given values:

81.0kg × m² × 0.190rad/s = It × ωt

Solving for ωt, we find that the final angular velocity of the system is approximately 38.54 rad/s.

To find the final angular velocity of the runner and turntable system, we apply conservation of angular momentum. The final angular velocity is calculated to be 0.611 rad/s. This involves determining the initial angular momenta of both the runner and the turntable, then using the total moment of inertia to find the final velocity.

To find the final angular velocity of the system when the runner comes to rest relative to the turntable, we need to apply the principle of conservation of angular momentum. The initial angular momentum of the system (runner plus turntable) must equal the final angular momentum since there are no external torques acting.

Step-by-Step Calculation :

Determine the initial angular momentum of the runner:

Determine the initial angular momentum of the turntable:

Calculate the total initial angular momentum:

Find the final angular velocity:

The final angular velocity of the system is 0.611 rad/s.

The average power is 20.3 kW

Explanation:

First of all, we calculate the work done on the car: the work-energy theorem states that the work done on the car is equal to the change in kinetic energy of the car, so we have

[tex]W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]

where

W is the work done

[tex]K_i, K_f[/tex] are the initial and final kinetic energy of the car

[tex]m=1.50\cdot 10^3 kg = 1500 kg[/tex] is the mass of the car

u = 0 is the initial velocity

v = 18.0 m/s is the final velocity

Substituting,

[tex]W=\frac{1}{2}(1500)(18)^2=2.43\cdot 10^5 J[/tex]

Now we can find the average power developed by the car's engine, which is given by

[tex]P=\frac{W}{t}[/tex]

where

[tex]W=2.43\cdot 10^5 J[/tex] is the work done

t = 12.0 s is the time taken

Substituting,

[tex]P=\frac{2.43\cdot 10^5 J}{12.0}=20,250 W = 20.3 kW[/tex]

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Answer:

True

Explanation:

t = Time taken = 6 s

u = Initial velocity = 0

v = Final velocity

a = Acceleration = 34.3 m/s²

[tex]v=u+at\\\Rightarrow v=0+34.3\times 6\\\Rightarrow v=205.8\ m/s[/tex]

After the six seconds the acceleration from the rocket stops. After the fuel is exhausted the velocity of the rocket will be 205.8 m/s which will reduce by the acceleration due to gravity (i.e., 9.81 m/s²).

Free fall is the state of a body where the only the force of gravity is acting on it other forces are not acting on it.

Hence, the statement here is true.

Answer:

Explanation:

To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum

I₁ ω₁ + I₂ ω₂ = ( I₁  + I₂ ) ω

I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .

I₁ = 1/2 mr²

= .5 x 175 x 2.13²

= 396.97 kgm²

I₂ = m r²

= 55.4 x 2.13²

= 251.34 mgm²

ω₁ = .651 rev /s

= .651 x 2π rad /s

ω₂ = tangential velocity of man / radius of disc

= 3.51 / 2.13

= 1.65 rad/s

I₁ ω₁ + I₂ ω₂ = ( I₁  + I₂ ) ω

396.97 x  .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω

ω = 3.14 rad /s

kinetic energy = 1/2 I ω²

= 3196 J

Answer:

a) W = 10995.6 J

b) W = - 9996 J

c) Kf = 999.6 J

d) v = 5.77 m/s

Explanation:

Given

m = 60 Kg

h = 17 m

a = g/10

g = 9.8 m/s²

a) We can apply Newton's 2nd Law as follows

∑Fy = m*a     ⇒     T - m*g = m*a     ⇒    T = (g + a)*m

where T is the force exerted by the cable

⇒    T = (g + (g/10))*m = (11/10)*g*m = (11/10)*(9.8 m/s²)*(60 Kg)

⇒    T = 646.8 N

then we use the equation

W = F*d = T*h = (646.8 N)*(17 m)

W = 10995.6 J

b) We use the formula

W = m*g*h     ⇒    W = (60 Kg)(9.8 m/s²)(-17 m)

⇒    W = - 9996 J

c) We have to obtain Wnet as follows

Wnet = W₁ + W₂ = 10995.6 J - 9996 J

⇒    Wnet = 999.6 J

then we apply the equation

Wnet = ΔK = Kf - Ki = Kf - 0 = Kf    

⇒  Kf = 999.6 J

d) Knowing that

K = 0.5*m*v²    ⇒    v = √(2*Kf / m)

⇒    v = √(2*999.6 J / 60 Kg)

⇒    v = 5.77 m/s

Answer:

Explanation:

mass =60kg d = 17m  a=g/10

(a) work done on the astronaut by the force from the helicopter = fd

but f =m(g+a)

 w= m( g+g/10)d

wt = 11/10 mgd

w =11/10 * 60 *9.8 * 17 = 10995.6J  = 1IKJ

(b) workdone  by her weight = -mgh

   = 60*9.8* 17 = -9996J

(C) Kinetic energy = wt + w

                             = (10995.6 - 9996)J = 999.6J

(d) Kinetic energy =1/2m[tex]v^{2}[/tex]

hence velocity = [tex]\sqrt{2ke/m}[/tex] = 5.777m/s

Answer

Given,

Flow rate of river is equal to 300,000 L/s.

Width of river = 18 m

and depth of river = 22 m

a) Average speed of river

     Q = 300,000 L/s

         = 300 m³/s

     Q = Av

     [tex]v = \dfrac{Q}{A}[/tex]

     [tex]v = \dfrac{300}{18 \times 22}[/tex]

     [tex]v = \dfrac{300}{396}[/tex]

     [tex]v = 0.757\ m/s[/tex]

b) Average speed when river is widen to 63 m and depth is increased to

     [tex]v = \dfrac{Q}{A}[/tex]

     [tex]v = \dfrac{300}{63 \times 42}[/tex]

     [tex]v = \dfrac{300}{2646}[/tex]

     [tex]v = 0.113\ m/s[/tex]

Final answer:

The average speed of the river in the gorge is 0.75 m/s, and downstream of the falls, when the river widens and its depth increases, the average speed decreases to 0.125 m/s.

Explanation:

To calculate the average speed of the river at Huka Falls, we can use the formula for flow rate, which is the volume of fluid passing a point in the river per unit of time:

Flow rate (Q) = Area (A)  imes Velocity (V)

(a) Average speed in the gorge: Given that the flow rate (Q) is 300,000 liters per second (which is equal to 300 cubic meters per second, since 1,000 liters is equal to 1 cubic meter), and the cross-sectional area of the river in the gorge (A) is 20 meters wide  imes 20 meters deep (400 square meters), we can solve for velocity (V) using the formula:

Q = A  imes V

300 m³/s = 400 m²  imes V

V = 300 m³/s \/ 400 m²

V = 0.75 meters per second

(b) Average speed downstream of the falls: Downstream, the river widens to 60 meters and deepens to an average of 40 meters, so the cross-sectional area is 2,400 square meters. Using the same flow rate, we can find the new velocity:

Q = A  imes V

300 m³/s = 2,400 m²  imes V

V = 300 m³/s \/ 2,400 m²

V = 0.125 meters per second

Answer:

 t = 5.4 s

Explanation:

from the question we are given :

height (s) = 20 m

mass of paint (Mp) = 4 kg

mass of nails (Mn) = 3 kg

acceleration due to gravity (g) = 9.8 m/s^{2}

            weight of nails = mass of nails x g = 3 x 9.8 = 29.4 N

            weight of paint = mass nails x g = 4 x 9.8 = 39.2 N

             net force = 39.2 - 29.4 = 9.8 N

             9.8 = (3 +4) x a

              a = 1.4 m/s^{2}

          where U is the initial velocity and is 0 since the can was initially at            

            rest

           20 = (0 x t) + (0.5 x 1.4 x t^{2})

            20 = 0.7 x t^{2}

             t^{2} = 28.6

             t = 5.4 s

Final answer:

The escape velocity of a spherical asteroid with a given radius and given gravitational acceleration can be calculated using a special formula. The distance a particle travels from the surface when it leaves it at a given radial velocity can be determined using the projectile height equation. The speed at which an object hits an asteroid after falling from a given height can be calculated using the falling object terminal velocity equation.

Explanation:

(a) Escape velocity can be defined as the minimum speed required for an object to escape the gravitational pull of a celestial body. To calculate the escape velocity on a spherical asteroid we can use the formula:

escape velocity = sqrt(2 * acceleration due to gravity * radius).

Using the given values, the escape velocity of a spherical asteroid is:

exhaust rate = sqrt(2 * 3.00 m/s2 * 500 000 m) = 6928 m/s.

(b) To calculate the distance from the surface that a particle will travel when it leaves the surface of the asteroid with a radial velocity of 1000 m/s, we can use the formula for the height of the projectile:

Height = (radial velocity)2 / (2 * acceleration due to gravity).

After entering the specified values, the particle travels the distance:

Altezza = (1.000 m/s)2 / (2 * 3,00 m/s2) = 166.666,67 m.

(c) To calculate the speed at which an object hits an asteroid as it falls 1000 km above the surface, we can use the equation for the final velocity of the falling object:

Final velocity = sqrt (initial velocity2 + 2 * acceleration due to gravity * height).

Substituting the given values, the object hits the asteroid with a speed of:

Final velocity = square(0 + 2 * 3.00 m/s2 * 1,000,000 m) = square(6,000,000 m2/s2) = 2,449 m/s.

Answer:

c)The gases have the same average kinetic energy.

Explanation:

As we know that the kinetic energy of gas is given as

[tex]K = \frac{1}{2}mv^2[/tex]

here we know that

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

so we have

[tex]K = \frac{1}{2}m (\frac{3RT}{M})[/tex]

now we have

[tex]K = \frac{3}{2}n RT[/tex]

now mean kinetic energy per molecule is given as

[tex]K_{avg} = \frac{3}{2}KT[/tex]

so this is independent of the mass of the gas

so average kinetic energy will remain same for both the gas molecules

Explanation:

Here velocity of ball and bat are in opposite direction.

Velocity of ball = 40 m/s

Velocity of bat = 18 m/s

Since they are in opposite direction relative velocity is given by,

           Relative velocity = 40 + 18 = 58 m/s

Distance to home plate = 18 m

We have

                Displacement = Velocity x Time

                           18 = 58 x Time

                          Time = 0.3 seconds

Option A is the correct answer.

Answer:

8.91 [tex]\frac{m}{sec^{2} }[/tex]

Explanation:

Given

We know the formula

[tex]g=4\times(pie)^{2}\times \frac{L}{T^{2} }[/tex]

Where

[tex]g=4\times(pie)^{2} \times \frac{L}{T^{2}}[/tex]

[tex]g= 39.47842\times \frac{1}{2.105^{2} }[/tex]

[tex]g=8.91 \frac{m}{sec^{2} }sec[/tex]

The acceleration due to gravity on Venus is 0.6 m/s².

To calculate the acceleration due to gravity on Venus, we can use the formula for the period of a simple pendulum:

T = 2π√(L/g)

Where T is the period of the pendulum, L is the length of the string, and g is the acceleration due to gravity.

Given that the period is 42.1 seconds and the length is 1.00 meter,To calculate the acceleration due to gravity on Venus, we can use the formula for the period of a simple pendulum: we can rearrange the formula to solve for g:

g = (4π²L)/T²

Plugging in the values, we get:

g = (4π²*1.00)/(42.1)² = 0.6 m/s²

Therefore, the acceleration due to gravity on Venus is 0.6 m/s².

Answer:

a. The temperature of the ice/water system remains constant.

c. Heat energy enters the ice/water system.

Explanation:

As we know that when ice coverts into the water then it is known as phase transfer. In the phase transfer process temperature and the pressure remains constant but the heat enters into the system. This heat is responsible for melting the ice. The heat is taken by ice is known as latent heat.

Therefore the option "a" and "c" are correct.

The true statements as a block of ice melts are that the temperature of the ice/water system remains constant, and heat energy enters the system. The temperature does not change, increase, or decrease during the melting process until all the ice is melted. So the correct options are a and c.

As a block of ice melts, several key processes occur within the ice/water system:

Therefore, from the options provided:

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Final answer:

A wooden block that floats in an unknown liquid with only one-third submerged indicates that the unknown liquid is more dense than water.

Explanation:

When the wooden block is placed in an unknown liquid and only one-third of it is submerged, it suggests that the block is less dense than the unknown liquid. This is due to the principle of buoyancy, which states that an object will displace a volume of fluid that is equal to its own weight. Since the wooden block floats higher in the unknown liquid compared to how it floats in water, where it must submerge more to displace enough water to equal its weight, we can conclude that the unknown liquid is more dense than water.

Answer:

a) V = 0.085 m^3

b) m = 17 kg

Explanation:

1) Data given

mb = 68 kg (mass for the block)

20% of the block volume is floating

100-20= 80% of the block volume is submerged

2) Notation

mb= mass of the block

Vw= volume submerged

mw = mass water displaced

V= total volume for the block

3) Forces involved (part a)

For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)

Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :

B = (mass water displaced) g = (mw) g   (1)

The definition of density is :

[tex] \rho_w = \frac{m_w}{V_w} [/tex]

If we solve for mw we got [tex]m_w = \rho_w V_w [/tex]  (2)

Replacing equation (2) into equation (1) we got:

[tex] B = \rho_w V_w g [/tex] (3)

On this case Vw represent the volume of water displaced = 0.8 V

If we replace the values into equation (3) we have

0.8 ρ_w V g = mg  (4)

And solving for V we have

 V =  (mg)/(0.8 ρ_w g )

We cancel the g in the numerator and the denominator we got

V = (m)/(0.8 ρ_w)

V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3

4) Forces involved (part b)

For this case we have bricks above the block, and we want the maximum mass for the bricks without causing  it to sink below the water surface.

We can begin finding the weight of the water displaced when the block is just about to sink (W1)

W1 = ρ_w V g

W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N

After this we can calculate the weight of water displaced before putting the bricks above (W2)

W2 = 0.8 x 833 N = 666.4 N

So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)

W3 = W1 -W2 = 833-666.4 N = 166.6 N

And finding the mass fro the definition of weight we have

m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg

Final answer:

The volume of the block of wood and the maximum mass of bricks it can carry without sinking are determined using principles of buoyancy and uniform density.

Explanation:

(a) To determine the volume of the block of wood, we use the fact that 20.0% of its volume is above the water. Given a uniform density, this means the submerged volume is 80% of the total volume. Hence, the volume of the block is 0.8 x (68.0 kg / 920 kg/m³) = 0.472 m³.

(b) The maximum mass of bricks that can be loaded without sinking the block can be calculated using the concept of buoyancy. The buoyant force equals the weight of the water displaced, so the mass of the bricks should not exceed the mass of the water displaced by the volume of the submerged part of the block. Therefore, the maximum mass of bricks is 0.472 m³ x 1100 kg/m³ x 9.8 m/s² = 5148.64 kg.

Answer:

Explanation:

Kinetic energy of the block

= 1/2 m v²

= .5 x 3 x 4 x 4

= 24 J

Negative work of - 24 J is required to be done on this object to bring it to rest.

magnitude of acceleration due to frictional force

= force / mass

2 / 3

= 0 .67 m /s²

Let the body slide by distance d before coming to rest so work done by force = Kinetic energy

=  2 x d = 24

d = 12 m

The magnitude of the work done to bring the block to rest is 2 times the distance the block slides. The magnitude of the acceleration of the block is 2/3 m/s^2. The block slides a distance of 8/3 m before it comes to rest.

The work done on an object is given by the equation:

Work = Force x Distance

In this case, the work done on the block to bring it to rest is equal to the force applied multiplied by the distance the block slides.

Given that the force exerted by the surface is 2 Newtons, we can calculate the magnitude of the work done:

Work = 2 N x Distance

To determine the distance the block slides, we need to calculate its deceleration using Newton's second law:

Force = mass x acceleration

Since the friction force is constant and in the opposite direction of motion, we can write:

2 N = 3 kg x acceleration

Solving for acceleration, we find:

acceleration = 2 N / 3 kg

With the acceleration calculated, we can use the kinematic equation:

vf^2 = vi^2 + 2ad

Since the final velocity is 0 (block comes to rest), the equation simplifies to:

0 = (4 m/s)^2 + 2(-acceleration)d

Solving for distance, we find:

d = (4 m/s)^2 / (2 x -acceleration)

Now, we can substitute the calculated acceleration into the equation to find the distance the block slides.

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Answer:

the initial speed of the object is 6.26 m/s

Explanation:

given information:

distance, s = 10 m

the coefficient of kinetic friction, μ = 0.2

we use the equation where the kinetic energy is equal to the friction force.

kinetic energy, KE = [tex]\frac{1}{2} mv^{2}[/tex]

friction work,  W = F(friction) s

KE = W

[tex]\frac{1}{2} mv^{2}[/tex] = F(friction) s

where, F(friction) = μ N, N is normal force (N = m g)

                            = μ m g

so,

[tex]\frac{1}{2} mv^{2}[/tex] = μ m g s

[tex]\frac{1}{2} v^{2}[/tex] = μ g s

[tex]v^{2}[/tex] = 2 μ g s

  = 2 (0.2) (9.8) (10)

  = 39.2

hence,

v = [tex]\sqrt{3.92}[/tex]

  = 6.26 m/s

The initial speed [tex]\( v_i \)[/tex] of the sled was approximately[tex]\( 6.26 \, \text{m/s} \).[/tex]

The initial speed of the object can be found using the work-energy principle, which states that the work done by friction will be equal to the change in kinetic energy of the sled.

First, let's calculate the work done by friction. The force of friction [tex]\( F_{\text{friction}} \)[/tex] is given by the product of the normal force \( N \) and the coefficient of kinetic friction [tex]\( \mu_k \):[/tex]

[tex]\[ F_{\text{friction}} = \mu_k N \][/tex]

[tex]\[ W = F_{\text{friction}} d = \mu_k m g d \][/tex]

[tex]\[ \Delta KE = 0 - \frac{1}{2} m v_i^2 = -\frac{1}{2} m v_i^2 \][/tex]

 According to the work-energy principle, the work done by friction is equal to the change in kinetic energy:

[tex]\[ \mu_k m g d = -\frac{1}{2} m v_i^2 \][/tex]

Solving for \( v_i \), we get:

[tex]\[ v_i^2 = -2 \mu_k g d \][/tex]

[tex]\[ v_i = \sqrt{-2 \mu_k g d} \][/tex]

 Given that [tex]\( \mu_k = 0.20 \), \( g = 9.81 \, \text{m/s}^2 \), and \( d = 10 \, \text{m} \),[/tex] we can plug in these values:

[tex]\[ v_i = \sqrt{-2 \times 0.20 \times 9.81 \times 10} \][/tex]

[tex]\[ v_i = \sqrt{39.24} \][/tex]

[tex]\[ v_i \approx 6.26 \, \text{m/s} \][/tex]

Therefore, the initial speed \( v_i \) of the sled was approximately[tex]\( 6.26 \, \text{m/s} \).[/tex]

The answer is: [tex]6.26 \, \text{m/s}.[/tex]

Answer:

a) 40,000 N/m

b) f = 6.37 Hz

c) v = 4,8 m/s

Explanation:

part a)

First in order to estimate the spring constant k, we need to know the expression or formula to use in this case:

k = ΔF / Δx

Where:

ΔF: force that the men puts in the car, in this case, the weight.

Δx: the sinking of the car, which is 2 cm or 0.02 m.

With this data, and knowing that there are four mens, replace the data in the above formula:

W = 80 * 10 = 800 N

This is the weight for 1 man, so the 4 men together would be:

W = 800 * 4 = 3200 N

So, replacing this data in the formula:

k = 3200 / 0.02 = 160,000 N/m

This means that one spring will be:

k' = 160,000 / 4 = 40,000 N/m

b) An axle and two wheels has a mass of 50 kg, so we can assume they have a parallel connection to the car. If this is true, then:

k^n = 2k

To get the frequency, we need to know the angular speed of the car with the following expression:

wo = √k^n / M

M: mass of the wheel and axle, which is 50 kg

k = 40,000 N/m

Replacing the data:

wo = √2 * 40,000 / 50 = 40 rad/s

And the frequency:

f = wo/2π

f = 40 / 2π = 6.37 Hz

c) finally for the speed, we have the time and the distance, so:

V = x * t

The only way to hit bumps at this frequency, is covering the gaps of bumping, about 6 times per second so:

x: distance of 80 cm or 0.8 m

V = 0.8 * 6 =

V = 4.8 m/s

Final answer:

The speed at which these oscillations go into resonance is approximately 12.6 m/s.

Explanation:

(a) To estimate the spring constant k of each of the four springs, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. Mathematically, this can be expressed as:

F = -k * x

where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

When four 80-kg men climb into the car, the body sinks by a couple of centimeters. Let's assume this displacement is 2 cm (0.02 m). The force exerted by each spring can be calculated using the weight of the men and Hooke's Law:

F = -k * x

mg = -k * x

k = -mg / x

Substituting the given values, we get:

k = -(80 kg * 9.8 m/s^2) / 0.02 m

k ≈ -39200 N/m

Since the spring constant k is a positive value, we can take the magnitude of the spring constant as:

|k| = 39200 N/m

(b) The natural frequency of the axle assembly oscillating on its two springs can be calculated using the formula for the natural frequency of a simple harmonic oscillator:

f = (1 / (2 * pi)) * sqrt(k / m)

where f is the natural frequency, k is the spring constant, and m is the mass of the axle assembly.

Substituting the given values, we get:

f = (1 / (2 * pi)) * sqrt(39200 N/m / 50 kg)

f ≈ 7.89 Hz

(c) The speed at which these oscillations go into resonance can be calculated using the formula for the resonance frequency of a simple harmonic oscillator:

f_resonance = v / (2 * L)

where f_resonance is the resonance frequency, v is the speed of the car, and L is the distance between the bumps on the road.

Substituting the given values, we get:

7.89 Hz = v / (2 * 0.8 m)

v ≈ 12.6 m/s

So, the speed at which these oscillations go into resonance is approximately 12.6 m/s.