A photographer in a helicopter ascending vertically at a constant rate of accidentally drops a camera out the window when the helicopter is 60.0 m above the ground. (a) How long will the camera take to reach the ground?

Answer:

D=7.385km

Explanation:

First we calculate how much it moves vertically (north and south), we will call this distance Y,.

taking into account that the car moves to the southwest and that it has a slope of 45 degrees, we can say that the component that moved south can be subtracted from the distance that moved north.

Y=6.71-1.33sen45=5.77km

then the distance that moves in the east we will call it X

we can add the component of the distance that moves to the southwest with the distance that moves to the east

X=3.67+1.33cos45=4.61Km

then we use the distance equation

[tex]D=\sqrt{x^{2} +y^{2} }[/tex]

[tex]D=\sqrt{4.61^{2}+5.77^{2} }[/tex]

D=7.385km

Answer:

Option c and d

Explanation:

The two mechanical devices which are used in laboratories for accurate measurement of small distances or small objects are calipers and micrometer.

Micrometer is a mechanical device used in laboratories like that of a screw gauge. It is used for the measurement of thickness of objects, length and the depth of the small objects which can be measured by holding the object in between the spindle and anvil of the micrometer.

Calipers is another mechanical device like that of vernier calipers used in laboratories for measurement of small distances, usually the distance between the opposing faces of the object. The measurement is usually taken on a digital display, a dial or a scale that is ruled.

The distance is measured by adjusting the tips of the caliper holding the object on a ruler.

Answer:

The volume of water evaporated is 199mL

Explanation:

Concentration is calculated with the following formula

[tex]C=\frac{n}{V}[/tex]

where n is the number of moles of solute and V is the volume of the solution (in this case is the same as the solvent volume) in liters.

So we isolate the variable n to know the amount of moles, using the volume given in liters

[tex]230mL=0.23L[/tex]

[tex]n=C*V=0.240 M*0.23L=0.055 mol[/tex]

Now, we isolate the variable V to know the new volume with the new concentration given.

[tex]V=\frac{n}{C} =0.055mol/1.75M=0.031L=31mL[/tex]

Finally, the volume of water evaporated is the difference between initial and final volume.

[tex]V_{ev}= V_{i} -V_{f} =230mL-31mL=199mL[/tex]

Answer:

a)y=-131.47 m :if the coordinate system is taken at the pinnacle of the building with positive direction taken to be upward direction.

b)v=-50.764m/s:The minus sign indicates the direction of the speed that is down

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

y = v₀*t +½ g*t² Equation 1

v=v₀+g*t  Equation 2

y: The vertical distance the ball moves at time t  

v₀: Initial speed  in m/s

g= acceleration due to gravity  in m/s²

v= Speed the ball moves at time t  

Known information

v₀=0

t=5.18 s

g=9.8 m/s²

Development of problem

a)We replace t in the equations (1) to y(5.18s) :

y =o+½ *9.8*5.18² =131.47m

y=-131.47 m :if the coordinate system is taken at the pinnacle of the building with positive direction taken to be upward direction

b)We replace t in the equations (2) to v(5.18s) :

v=o+9.8*5.18=50.764m/s

v=-50.764m/s:The minus sign indicates the direction of the speed that is down

Answer:

a)(2×40+2×80)/(2+2)=240/4=60

b)(240)/(120/40+120/80)=240/(3+1.5)

240/4.5=53.33333

no

Answer:

D) He

Explanation:

For adiabatic change

[tex]PV^\gamma = constant[/tex]

\frac{P_2}{P_1} =(\frac{V_1}{V_2} )^\gamma

[tex]\frac{P_2}{P_1} =(\frac{3}{1} )^\gamma[/tex]

[tex]P_2 = P_1 3^\gamma[/tex]

As the value of  \gamma increases, P_2 increases.

\gamma for He is 1.67 and \gamma for O₂ IS 1.4

Naturally The value  of P_2 is greater  for He. Increase in pressure for He is greater.

Answer:

C.) First grade through grades 12 in all states

Explanation:

Free public education is mandated from Kindergarten through grade 12 for all children. The correct option is option (c).

Free public education is mandated for all children in the United States from kindergarten through grade 12. This means that public schools are required to provide education to children starting from kindergarten up to the completion of 12th grade. This mandate ensures that all children have access to education without financial barriers during these years.

While preschool education is not universally mandated, some states or districts may offer free or subsidized preschool programs for certain age groups or based on specific criteria.

Special needs children, as mentioned in option E, also fall under the mandate for free public education from preschool through grade 12. Special education services are provided to ensure that children with disabilities receive an appropriate education tailored to their individual needs.

Therefore, option C, "Kindergarten through grade 12 for all children," best represents the mandate for free public education in the United States.

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The electron configurations correspond to a halogen, transition metals, an inert gas, an alkali metal, and an alkaline earth metal, identified by their completed electron orbitals and the presence of electrons in specific orbitals such as d or p.

Answer:

Part a)

[tex]v_f = 4 m/s[/tex]

Part b)

[tex]t = 0.001 s[/tex]

Part c)

[tex]d = 0.815 m[/tex]

Explanation:

Part a)

As we know that initially the grass hopper is at rest at the ground position

Now the acceleration is given as

[tex]a = 4000 m/s^2[/tex]

distance of the legs that it stretched is given as

[tex]s = 2.0 mm[/tex]

so we have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]v_f^2 - 0 = 2(4000)(0.002)[/tex]

[tex]v_f = 4 m/s[/tex]

Part b)

time taken to reach this speed is given as

[tex]v_f - v_i = at[/tex]

[tex]4 - 0 = 4000 t[/tex]

[tex]t = 0.001 s[/tex]

Part c)

as the grass hopper reach the maximum height its final speed would be zero

so we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 4^2 = 2(-9.81) d[/tex]

[tex]d = 0.815 m[/tex]

Final answer:

The froghopper Philaenus spumarius takes off at an upward velocity of 4 m/s and reaches this velocity in 0.001 seconds. In the absence of air resistance, it can potentially jump to a height of 0.815 m, while air resistance reduces its actual jump to about 70 cm.

Explanation:

The froghopper Philaenus spumarius, known for its powerful jumping ability, can accelerate at 4.00 km/s² over a distance of 2.0 mm. To find the upward velocity (a) with which the insect takes off, the following kinematic equation can be used:v² = u² + 2as

where v is the final velocity, u is the initial velocity (0 m/s in this case), a is the acceleration, and s is the distance.

Firstly, convert the given units to make them consistent. The acceleration becomes 4000 m/s² and the distance becomes 0.002 m. Plugging the values into the equation gives:

v² = 0 + 2(4000)(0.002) = 16

The takeoff velocity, v, is √16, which equals 4 m/s.

For the time interval (b), we use the equation:

v = u + at

So, t = v / a = 4 / 4000 = 0.001 s.

To calculate how high the insect would jump if air resistance were negligible (c), we can use the equation:

s = ut + ½at²

Since initial velocity u is 0 when coming back down, and final velocity v is 0 at the peak of the jump, we set v = 0 and solve for s:

0 = (4)2 - 2(9.81)s

s = 16 / 19.62 = 0.815 m (or 81.5 cm).

Comparing this to the actual jump height of 70 cm indicates a significant effect of air resistance on the insect's jump.

Answer:

0.982 N

Explanation:

mass of ball, m = 52 g = 0.052 kg

initial velocity, u = - 20 m/s (downward)

final velocity, v = 14 m/s (upward)

time of contact, t = 1.8 s

According to Newton's second law, the rate of change of momentum of the body is equal to the forced exerted on that body.

initial momentum , pi  = mass x initial velocity = 0.052 x (-20) = - 1.04 kg m/s

final momentum, pf = mass x final velocity = 0.052 x 14 = 0.728 kg m/s

Change in momentum = final momentum - initial momentum

                                      = 0.728 - (- 1.04)= 1.768 kg m/s

So, force = change in momentum / time

Force = 1.768 / 1.8 = 0.982 N

The magnitude of the average force exerted on the superball by the sidewalk is 1414.4 N. This is calculated by determining the impulse based on the change in velocity and mass of the superball, and dividing the impulse by the contact time.

The question asks to calculate the magnitude of the average force exerted on a superball by the sidewalk during its bounce. The superball's mass is 52 g (0.052 kg), and it experiences a velocity change from 20 m/s downward to 14 m/s upward. The contact time with the sidewalk is given as 1/800 s. First, we must calculate the change in velocity (impulse) and then use it to find the average force.

To calculate the impulse, we add the magnitudes of both components of velocity together since the ball changes direction, the total change in velocity (Δv) is 20 m/s (downward) + 14 m/s (upward) = 34 m/s. The impulse (J) is then given by Δp = mΔv, where m is the mass of the ball. Therefore, J = 0.052 kg * 34 m/s = 1.768 kg·m/s.

The average force (Favg) can be found by dividing the impulse by the contact time (Δt): Favg = J / Δt. Thus, Favg = 1.768 kg·m/s / (1/800 s), which equals an average force of 1414.4 N.

8 am to 3 pm is 7 hours.

(675 km) / (7 hrs) = 96.4 km/hr .

Her average speed for the whole 7 hours has to be not less than 96.4 km/hr. Any less, she's not home by 3:00.

We don't technically know the speed limit at every point on her trip, because you technically haven't told us.

But in 7 hours, she MUST stop for gas, she MUST get some rest, she MUST make a pit stop, and she most likely encounters some traffic somewhere. So in order to average 96.4 for the whole trip, she MUST exceed it for PART of the trip ... possibly by a lot.

Whatever the speed limit may be, I think it's likely that she'll exceed it SOMEwhere, at least for SOME time.

The student must maintain a minimum average speed of 96.43 km/h.

Whether this exceeds the speed limit depends on the highway's speed limit.

To determine the minimum average speed the student needs to maintain to complete the 675-km trip by 3:00 pm, starting at 8:00 am,

We first calculate the total travel time allowed, the difference between 3:00 pm and 8:00 am is 7 hours.

Therefore, to find the minimum average speed, we divide the total distance by the total time:

Average Speed = Total Distance / Total Time

Average Speed = 675 km / 7 hours

= 96.43 km/h

The student must maintain a minimum average speed of 96.43 km/h to reach the destination on time.

Answer:

Given the equation of the particle, we know that:

[tex]x=(2.9\frac{m}{s^{3} } )*t^{3} -(2\frac{m}{s^{6} } )*t^{6}[/tex]

a) In [tex]t=0.0s[/tex] we evaluate [tex]t[/tex] in the former equation:

[tex]x_{1} =(2.9\frac{m}{s^{3} } )*(0)^{3} -(2\frac{m}{s^{6} } )*(0)^{6}=0m[/tex]

In [tex]t=1.3s[/tex]

[tex]x_{2} =(2.9\frac{m}{s^{3} } )*(1.3s)^{3} -(2\frac{m}{s^{6} } )*(1.3s)^{6}=-3,28232m[/tex]

a) So, we know that the displacement of te particle is given by:

[tex]x=x_{2}-x_{1} =-3.28232m-0m=-3.28232m[/tex]

To find it's velocity, we need to derivate the equation of position by the formula:

[tex]v_{x} =\frac{dx}{dt} =3ct^{2}-6bt^{5} =(8.7\frac{m}{s^{3} })t^{2} -(12\frac{m}{s^{6} } )t^{5}[/tex]

And evaluate this expression at each specified t:

b) [tex]v(1s)_{x} =(8.7\frac{m}{s^{3} })(1s)^{2} -(12\frac{m}{s^{6} } )(1s)^{5}=-3.3\frac{m}{s}[/tex]

c) [tex]v(2s)_{x} =(8.7\frac{m}{s^{3} })(2s)^{2} -(12\frac{m}{s^{6} } )(2s)^{5}=-349.2\frac{m}{s}[/tex]

d) [tex]v(3s)_{x} =(8.7\frac{m}{s^{3} })(3s)^{2} -(12\frac{m}{s^{6} } )(3s)^{5}=-2837.7\frac{m}{s}[/tex]

e) [tex]v(4s)_{x} =(8.7\frac{m}{s^{3} })(4s)^{2} -(12\frac{m}{s^{6} } )(4s)^{5}=-12148.8\frac{m}{s}[/tex]

To find it's acceleration, we need to derivate the equation of velocity by the formula:

[tex]a_{x} =\frac{dv}{dt} =(17.4\frac{m}{s^{3} } )t-(60\frac{m}{s^{6} } )t^{4}[/tex]

And evaluate this expression at each specified t:

f) [tex]a(1s)_{x} =(17.4\frac{m}{s^{3} } )(1s)-(60\frac{m}{s^{6} } )(1s)^{4}=-42.6\frac{m}{s^{2} }[/tex]

g) [tex]a(2s)_{x} =(17.4\frac{m}{s^{3} } )(2s)-(60\frac{m}{s^{6} } )(2s)^{4}=-925.2\frac{m}{s^{2} }[/tex]

h) [tex]a(3s)_{x} =(17.4\frac{m}{s^{3} } )(3s)-(60\frac{m}{s^{6} } )(3s)^{4}=-4807.8\frac{m}{s^{2} }[/tex]

i) [tex]a(4s)_{x} =(17.4\frac{m}{s^{3} } )(4s)-(60\frac{m}{s^{6} } )(4s)^{4}=-15290.4\frac{m}{s^{2} }[/tex]

Answer:

Speed of plane is 300.5 m/s at angle of 6.22 degree South of West

Explanation:

Air speed of the plane is given as

v = 264 m/s in direction 5 degree South of West

So we have

[tex]v_1 = 264 cos5 \hat i + 264 sin5 \hat j[/tex]

[tex]v_1 = 263 \hat i + 23 \hat j[/tex]

Also we have speed of air is given as

v = 37 m/s at 15 degree South of West

so it is

[tex]v_2 = 37 cos15\hat i + 37 sin15 \hat j[/tex]

[tex]v_2 = 35.74 \hat i + 9.58 \hat j[/tex]

So the net speed of plane with respect to ground is given as

[tex]v_p = v_1 + v_2[/tex]

[tex]v_p = (263 \hat i + 23 \hat j) + (35.74 \hat i + 9.58 \hat j)[/tex]

[tex]v_p = 298.74 \hat i + 32.58\hat j[/tex]

so it is

[tex]v_p = \sqrt{298.74^2 + 32.58^2}[/tex]

[tex]v_p = 300.5 m/s[/tex]

direction is given as

[tex]\theta =tan^{-1} \frac{v_y}{v_x}[/tex]

[tex]\theta = tan^{-1} \frac{32.58}{298.74}[/tex]

[tex]\theta = 6.22 degree[/tex]

Answer:

The answer to your question is: b. 2.7m/s²

Explanation:

Data

Force = 10 N

a = 5.3 m/s²

Moon

F = 5 N

a = ?

Formula

F = m x a

Process

Find the mass of the table

              m = F / a

              m = 10 / 5.3

              m = 1.887 kg

Now, find the acceleration

            a = F / m

            a = 5 / 1.887

            a = 2.65 m/s² ≈ 2.7 m/s²

Answer:

a = F / m = 5N / 1.887kg = 2.65 m/s (answer choice B)

Explanation:

rounded the answer

Answer:

[tex]\Delta S = 36.55 J/K[/tex]

Explanation:

As we know that entropy change for phase conversion is given as

[tex]\Delta S = \frac{\Delta Q}{T}[/tex]

Here we know that heat required to change the phase of the ice is given as

[tex]\Delta Q = mL[/tex]

here we have

m = 29.8 g = 0.0298 kg

L = 335000 J/kg

now we have

[tex]\Delta Q = 0.0298\times 335000[/tex]

[tex]\Delta Q = 9979.4 J[/tex]

also we know that temperature is approximately same as freezing temperature

so we have

[tex]T = 273 k[/tex]

so here we have

[tex]\Delta S = \frac{9979.4}{273}[/tex]

[tex]\Delta S = 36.55 J/K[/tex]

Answer:

A) E = 3.70*10^{4} N/C

B)  E = 2.281*10^3 N/C

Explanation:

given data:

charge density [tex] \lambda = 130*10^{-9} C/m[/tex]

length of wire = 9.50 cm

a) at x  = 4.5 m above midpoint, electric field is calculated as

[tex]E = \frac{1}{ 2\pi \epsilon} * \frac{ \lambda}{x\sqrt{(x^2/a^2)+1}}[/tex]

x = 4.5 cm

midpoint a = 4.5 cm = 0.0475 m

[tex]E =2{\frac{1}{ 8.99*10^9} * \frac{130*10^{-9} }{0.045\sqrt{(4.5^2/4.75^2)+1}}[/tex]

E = 3.70*10^{4} N/C

B) when wire is in circle form

[tex]Q = \lambda * L[/tex]

[tex]= 130*10^{-9} *9.5*10^{-2}[/tex]

   = 1.235*10^{-8} C

Radius of circle

[tex]r = \frac{L}{2\pi}[/tex]

[tex]r = \frac{9.5*10^{-2}}{2\pi}[/tex]

r = 1.511*10^{-2} m

[tex]E = \frac{1}{ 2\pi \epsilon} * \frac{Qx}{(x^2+r^2)^{3/2}}[/tex]

[tex]E =8.99*10^{9} * \frac{1.23*10^{-8}*4.5*10^{-2}}{((4.5*10^{-2})^2+(1.511*10^{-2})^2)^{3/2}}[/tex]

E = 2.281*10^3 N/C

Answer:

7,14545 mph and 3,1936 m/s

Explanation:

The average speed is calculated by dividing the displacement over time, then it is 26,2 miles/(3 2/3 hours), here 3 (2/3) hours is a mixed number, that represents 11/3 hours or 3,66 hours. Then the average speed is 7,14545 mph, now to turn this into meters per second, we notice as mentioned that 1 mile =1609 meters and 1 hour=3600 seconds. Then 7,14545 miles/hour* (1 hour/3600 seconds) * (1609 meters/1 mile)=3,1936 m/s

The tank containing water filled to a height of 4 meters, the pressure at any point on bottom of tank is 39.2 kPa

Answer: Option C

Solution:

[tex]\begin{aligned} \text {Pressure}=& \text {Density of water} \times \text {Acceleration due to gravity} \\ & \times \text {Height to which water is filled in tank} \end{aligned}[/tex]

Density of water is 1000 [tex]\mathrm{kg} / m^{3}[/tex]

Acceleration due to gravity is 9.8 m/s  

The water is filled to a height of 4 meters in the tank  

[tex]\text { Pressure }=4 \times 1000 \times 9.8=4000 \times 9.8=39200[/tex]

Pressure at any point at the bottom of tank = 39200 Pascal or 39.2 kPa

Answer:

The wavelength of this wave is 1.01 meters.

Explanation:

The variation in the pressure of helium gas, measured from its equilibrium value, is given by :

[tex]\Delta P=2.9\times 10^{-5}\ cos(6.2x-3000t)[/tex]..............(1)

The general equation is given by :

[tex]\Delat P=P_o\ cos(kx-\omega t)[/tex]...........(2)

On comparing equation (1) and (2) :

[tex]k=6.2[/tex]

Since, [tex]k=\dfrac{2\pi}{\lambda}[/tex]

[tex]\dfrac{2\pi}{\lambda}=6.2[/tex]

[tex]\lambda=1.01\ m[/tex]

So, the wavelength of this wave is 1.01 meters. Hence, this is the required solution.

We can find the wavelength of the wave represented by the pressure variation equation by noting the wave number term in the cosine function, recognizing it as 2π divided by the wavelength, and solving for the wavelength. The wavelength in this case is approximately 1.01 meters.

To solve this problem, we must understand that the equation given, ΔP = 2.9 × 10−5 cos (6.20x − 3 000t), is a representation of a wave, where the term inside the cosine function represents the wave number (k). The wavenumber is the spatial frequency of the wave, measured in radians per unit distance, and in wave equations is often given as k = 2π/λ, where λ is the wavelength. Here, we have that k = 6.20, so we can solve for the wavelength (λ). Rearranging our equation, we find λ = 2π/6.20 ≈ 1.01 m. So, the wavelength of the wave is approximately 1.01 meters. This seems reasonable as the units we have used are all compatible — the wave number is in units of per meter and the wavelength we found is in meters.

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Answer:

Explanation:

There are various examples around us that states " internal hold back property " and is different from frictional force or the force of Earth's gravity.

The plaster which is the mixture of cement and water has an adhesive nature and molecules inside them has a internal hold back property. They stick to the walls and ceiling and do not fall.

Though gravity is applied on it still its molecules tends to hold the ceiling firmly and do not fall upon.

Explanation:

3000 to 1,000,000 nm for IR-C

700 to 1400 nm for IR-A

Now, we are working with light, so we can use that the energy of a photon with frequency f is : E(f) = h*f

where h is the plank constant h = 6.62607015×10−34 j*s

and f is the frequency that f= c/λ, where λ is the wavelenght.

so, for IR-C the energy lies between h*c/3000 and h*c/1000000

where c is the light speed.

For IR-A the energy lies between h*c/700 and h*c/1400.

so here you can se that IR-A has a lot more energy than IR-C, you can se that the minimal energy of IR-A is h*c/1400 and the maximal of IR-C is h*c/3000 so the minimal energy of IR-A is almost twice times the maximum energy of IR-C

Answer:

C.

Explanation:

A meter is 8.56 centimeters longer than a yard. Something to keep in mind is that a meter is about 10% longer than a yard.

Hope this helps :)

Answer:

C. 8.56cm

Explanation:

Since we are to get the difference of cord length in centimeter, there fore we will have to convert the one yard and one metre length of wires to centimeters.

For 1m length of cord

1m = 100cm

For 1yard length of cord

1yard = 91.44cm

Difference in cord length = 100cm - 91.44cm

= 8.56cm

Answer:

second medium is less optical denser than first medium

Explanation:

As per Snell's law we know that

[tex]\mu_1 sin\theta_1 = \mu_2 sin\theta_2[/tex]

so here we know that

[tex]\theta_1 [/tex] = angle of incidence

[tex]\theta_2[/tex] = angle of refraction

As we know that light ray goes away from the normal when it move from medium 1 to medium 2

so here angle of refraction will me more than the angle of incidence

so we have

[tex]\frac{sin\theta_1}{sin\theta_2} = \frac{\mu_2}{\mu_1}[/tex]

since we know that

[tex]\theta_1 < \theta_2[/tex]

so we have

[tex]\frac{sin\theta_1}{sin\theta_2} < 1[/tex]

[tex]\frac{\mu_2}{\mu_1} <1[/tex]

so we can say that

[tex]\mu_2 < \mu_1[/tex]

so we can say that second medium is less optical denser than first medium

To open the hatch of a submerged vehicle, the crew needs to overcome the difference in pressure applied on both sides of the hatch. By calculating the external pressure at the depth, the force outside is found. Subtracting the force inside (internal pressure plus hatch weight) gives the requisite force (~1.95 x 10⁵ N).

The question is about determining the force the crew must exert to open the hatch of a submerged vehicle. In physics, this problem involves understanding the principles of fluid statics and the pressure difference between the interior and exterior of the hatch, contributing to the net force on the hatch.

Given that the vehicle is submerged 28 meters in saltwater, with an internal pressure of 1 atm, the external pressure at that depth will be greater than the internal pressure. The pressure a diver experiences under water increases by 1 ATA every 33 feet of salt water. Hence, at 28 meters (approximately 92 feet), the external pressure is about 1 (atmospheric pressure) + 92/33 (pressure due to ocean) = ~3.8 ATA. This is the pressure acting against the crew from outside.

The force acting on the hatch from outside (pressurized water) can be obtained by multiplying the pressure with the area (Force = Pressure x Area). So the Force outside = 3.8 atm x 0.70 m² = ~2.66 x 10⁵ N (considering that 1 atm is approximately 1.013 x 10⁵ N/m²).

Now, the force acting from inside consists of the atmospheric pressure (1 atm) and the weight of the hatch. Force inside = (1 atm x 0.70 m²) + weight of the hatch = 0.71 x 10⁵ N + 200 N = ~0.71 x 10⁵ N

Therefore, the resultant force (force required to open the hatch) is the difference between the outside and inside forces. That becomes ~2.66 x 10⁵ N - 0.71 x 10⁵ N = ~1.95 x 10⁵ N.

This is the downward force the crew must exert on the hatch to open it and escape.

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The downward force that the crew must exert on the hatch to open it at a depth of 28 m below the surface of the ocean is [tex]\( {199567.2 \, \text{N}} \)[/tex].

To find the downward force (F) that the crew must exert on the hatch to open it at a depth of 28 m below the surface of the ocean, we need to consider the hydrostatic pressure acting on the hatch.

Step 1: Calculate the pressure at the depth

The hydrostatic pressure [tex]\( P_{\text{water}} \)[/tex] at a depth h below the ocean surface is given by:

[tex]\[P_{\text{water}} = \rho g h\][/tex]

Calculate [tex]\( P_{\text{water}} \):[/tex]

[tex]\[P_{\text{water}} = 1025 \times 9.81 \times 28\][/tex]

[tex]\[P_{\text{water}} = 285396 \, \text{Pa}\][/tex]

Step 2: Calculate the total pressure inside the submersible

The total pressure inside the submersible is the sum of the atmospheric pressure and the hydrostatic pressure at depth:

[tex]\[P_{\text{total}} = P_{\text{inside}} + P_{\text{water}}\][/tex]

[tex]\[P_{\text{total}} = 1.0 \times 1.013 \times 10^5 + 285396\][/tex]

[tex]\[P_{\text{total}} = 1.013 \times 10^5 + 285396\][/tex]

[tex]\[P_{\text{total}} = 386396 \, \text{Pa}\][/tex]

Step 3: Calculate the force required to open the hatch

The force F required to open the hatch against the pressure difference is:

[tex]\[F = (P_{\text{total}} - P_{\text{atm}}) \times A\][/tex]

Substitute the values:

[tex]\[F = (386396 - 1.013 \times 10^5) \times 0.70\][/tex]

[tex]\[F = (386396 - 101300) \times 0.70\][/tex]

[tex]\[F = 285096 \times 0.70\][/tex]

[tex]\[F = 199567.2 \, \text{N}\][/tex]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

[tex]v = \frac{2 \pi r}{T}[/tex]  

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

[tex]T^{2} = r^{3}[/tex]

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

[tex]T^{2} = r^{3}[/tex]

[tex]T = \sqrt{(133370 Km)^{3}}[/tex]

[tex]T = \sqrt{(2.372x10^{15} Km)}[/tex]

[tex]T = 4.870x10^{7} Km[/tex]

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( [tex]1.50x10^{8} Km[/tex] )

1 AU is defined as the distance between the earth and the sun.

[tex]\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU[/tex]

[tex]T = 0.324 AU[/tex]

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

[tex]T = \frac{0.324 AU}{1 AU} . 1 year[/tex]

[tex]T = 0.324 year[/tex]

That can be expressed in units of days

[tex]T = \frac{0.324 year}{1 year} . 365.25 days[/tex]  

[tex]T = 118.60 days[/tex]

Circular velocity for the particle in the Encke Division:

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

[tex]v = \frac{2 \pi (133370 Km)}{(118.60 days)}[/tex]

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

[tex]118.60 days .\frac{86400 s}{1 day}[/tex] ⇒ 10247040 s

[tex]133370 Km .\frac{1000 m}{1 Km}[/tex] ⇒ 133370000 m

[tex]v = \frac{2 \pi (133370000 m)}{(10247040 s)}[/tex]

[tex]v = 81.778 m/s[/tex]

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

[tex]T^{2} = r^{3}[/tex]

[tex]T = \sqrt{(69000 Km)^{3}}[/tex]

[tex]T = \sqrt{(3.285x10^{14} Km)}[/tex]

[tex]T = 1.812x10^{7} Km[/tex]

[tex]\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU[/tex]

[tex]T = 0.120 AU[/tex]

[tex]T = \frac{0.120 AU}{1 AU} . 1 year[/tex]

[tex]T = 0.120 year[/tex]

[tex]T = \frac{0.120 year}{1 year} . 365.25 days[/tex]  

[tex]T = 43.83 days[/tex]

Circular velocity for the particle in D Ring:

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

[tex]v = \frac{2 \pi (69000 Km)}{(43.83 days)}[/tex]

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

[tex]43.83 days . \frac{86400 s}{1 day}[/tex] ⇒ 3786912 s

[tex]69000 Km . \frac{1000 m}{ 1 Km}[/tex] ⇒ 69000000 m

[tex]v = \frac{2 \pi (69000000 m)}{(3786912 s)}[/tex]

[tex]v = 114.483 m/s[/tex]

[tex]\frac{114.483 m/s}{81.778 m/s} = 1.399[/tex]            

The particle in the D ring is 1399 times faster than the particle in the Encke Division.  

Answer:

a) F = 84.64N

b) The force has to increase.

Explanation:

First of all, let's convert everything to the same unit system:

Vo = 0 m/s    Vf = 98 mi/h * 1609.34 m / 1mi * 1h / 3600s = 43.8m/s

d = 1.7m         m = 0.15kg

We can calculate force as:

F = m*a    where a is the acceleration experiencd by the ball and m is its mass.

In order to calculate acceleration, we can use this formula:

[tex]Vf^2 = V^2 +s*a*d[/tex]    Solving for a:

[tex]a = \frac{Vf^2-Vo^2}{2*d} = 564.25m/s^2[/tex]

Now, the force will be:

F = m * a = 84.6N

As we can see in that previous equation, the force is directly proportional to the mass of the ball, so, assuming the final speed of the ball is the same as before (that is, the acceleration is the same), the force will increase in the same proportion as the mass does.

a) The average force exerted on the ball during the pitch F = 84.64N

b)If the mass of the ball has increased the force Will increase

First of all, let's convert everything to the same unit system:

Vo = 0 m/s    

[tex]V_f =\dfrac{98 \frac{mile}{h} \times 1609.34 }{3600} =43.8 \frac{m}{s}[/tex]

d = 1.7m        

m = 0.15kg

The force is from newtons second law

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

Now for finding acceleration we have

[tex]V_{f}^2 =V^2+ 2ad[/tex]

[tex]a=\dfrac{V_{f}^2 -V^2}{2d}[/tex]

[tex]a= \dfrac{43.8^2}{1.7} =564.25\dfrac{m}{s^2}[/tex]

Now, the force will be:

[tex]F=m\times a = 0.15\times 564=84.6 N[/tex]

As we can see in that previous equation, the force is directly proportional to the mass of the ball,

so, assuming the final speed of the ball is the same as before (that is, the acceleration is the same)

The force will increase in the same proportion as the mass does.

Thus

a) The average force exerted on the ball during the pitch F = 84.64N

b)If the mass of the ball has increased the force Will increase

To know more about the acceleration of anybody follow

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Answer:

(a) the approximate mass of a column of air is: 1225 (Kg), (b) the weight of this amount of air is: 12017.25 (N) and (c) the pressure at the bottom is: 120172.5 (KPa)

Explanation:

We need to remember that the equation that related mass and volume is the density as:[tex]p=\frac{m}{V}[/tex], where p is the density, m is the mass and V is the volume and assuming that air density is 1.225 (Kg/m3) and knowing that exosphere is the upper atmosphere and its height is 10000 (Km). We now can get the volume as:[tex]Volume=A*h_{upper atmosphere} =0.0001*10000000=1000(metre^{3})[/tex], so that the approximate mass of a column of air is: [tex]m=p*V=1.225*1000=1225 (Kg)[/tex]. Then the weight of this amount of air is:[tex]W=m*g[/tex], where W is the weight, m is the mass and g the gravity acceleration, so:[tex]W=1225*9.81=12017.25 (N)[/tex]. Finally to find the pressure we know that gauge pressure equation [tex]P=p*g*h[/tex] where P is the pressure, p is the density of the column and h is the height of the column, so [tex]P=1.225*9.81*10000000=120172.5 (KPa)[/tex].

The approximate mass of the air column can be calculated using the density of air and the volume of the column. The weight of the air column can be calculated by multiplying the mass by the acceleration due to gravity. The pressure at the bottom can be calculated by dividing the weight by the area.

The approximate mass of a column of air with an area of 1cm² that extends from sea level to the upper atmosphere can be calculated using the density of air. The density of air near the Earth's surface is about 1.29 kg/m³. Using this density, we can calculate the mass by multiplying the density with the volume of the column of air. The volume can be calculated by multiplying the area of 1 cm² with the height of the column, which is the distance from sea level to the upper atmosphere.

To calculate the weight of this amount of air, we need to multiply the mass of the air column by the acceleration due to gravity. The acceleration due to gravity is approximately 9.81 m/s². Multiplying the mass by the acceleration due to gravity will give us the weight of the air column in newtons.

The pressure at the bottom, or at sea level, can be calculated using the weight of the air column and the area of 1 cm². The pressure is equal to the weight divided by the area.

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Answer:

a = 252 m / s²

Explanation:

The initial speed Vο = 0

The speed of baseball with witch it is thrown  Vf = 42 m/s

Displacement covered = 3.5 m

Hence acceleration will be constant at this moment, so as per the equation of constant acceleration : ( Vf² - Vο² ) / 2d = a

                                                            a = 252 m / s²

Answer:

None of them is 100% correct.

Technician A is correct when he says that the higher compression rate helps the car to use less fuel. A higher compression ratio gives the engine a higher thermal efficiency which in turn translates to higher fuel efficiency but it does not produce more power than a gasoline engine.

Technician B is correct when he says the diesel engine produces more torque, but he also says that it is the only thing that is better with the diesel engine which is wrong because we already know that it also helps on fuel efficiency.

A ship at 0 degrees latitude and 27 degrees west longitude is on the equator and in the Atlantic Ocean. The equator indicates a latitude of 0 degrees, and a longitude of 27 degrees west places the ship in the Atlantic Ocean, west of the Prime Meridian.

We define a ship’s position using latitude (north-south position) and longitude (east-west position). A latitude of 0 degrees signifies the ship is on the equator. Longitude, on the other hand, is measured in degrees east or west of the Prime Meridian, that passes through Greenwich, England and is set at 0 degrees. The longitude of 27 degrees west implies the ship is to the west of the Prime Meridian.

Therefore, when a ship's position is given as 0 degrees latitude and 27 degrees west longitude, it means the ship is located on the equator and in the Atlantic Ocean. The Atlantic Ocean lies to the west of the Prime Meridian and between the Prime Meridian and the International Date Line which is roughly along the 180° meridian of longitude.

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