A golf ball with an initial angle of 32° lands exactly 224 m down the range on a level course.

(a) Neglecting air friction, what initial speed would achieve this result? [m/s]
(b) Using the speed determined in item (a), find the maximum height reached by the ball. [m]

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:

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:

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:

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) 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:

A. Statistics addresses gaps in knowledge.

Explanation:

"Statistics addresses gaps in knowledge." does not describe a limitation of statistics

Final answer:

Option A, 'Statistics addresses gaps in knowledge,' does not describe a limitation of statistics, but rather a purpose of using statistics to enhance understanding.

Explanation:

To identify which statement does not describe a limitation of statistics, we need to evaluate the options provided:

The best answer here is A, as this option reflects a benefit of statistics rather than a limitation.

Answer:

The magnitude of force on P is 2.75 N

The magnitude of force on Q is 7.15 N

Solution:

As per the question:

Mass of the object, M = 11.0 kg

Acceleration of the object when the forces are directed leftwards, a = [tex]0.900 m/s^{2}[/tex]

Acceleration when the forces are in opposite direction, a' = [tex]0.400 m/s^{2}[/tex]

Now,

The net force on the object in first case is given by:

[tex]F_{net} = |\vec{F_{P}}| + |\vec{F_{Q}}| = Ma[/tex]       (1)

The net force on the object in second case is given by:

[tex]F_{net} = |\vec{F_{P}}| - |\vec{F_{Q}}| = Ma'[/tex]       (2)

Adding both eqn (1) and (2):

[tex]2|\vec{F_{Q}}| = M(a + a')[/tex]

[tex]|\vec{F_{Q}}| = \frac{11.0(0.900 + 0.400)}{2} = 7.15 N[/tex]

Putting the above value in eqn (1):

[tex]|\vec{F_{P}}| = 11\times 0.900 - |\vec{F_{Q}}|[/tex]

[tex]|\vec{F_{P}}| = 9.900 - 7.15 = 2.75[/tex]

The rate at which the angle is changing is determined as - 0.0175 rad/s.

How to calculate the rate at which the angle is changing?

The rate at which the angle is changing is calculated by applying the following formula as shown below.

From the right triangle attached;

tan θ = y / x

So, cot θ = x / y

cot θ = x / 100

we will take the derivative of both sides of the equation;

- csc²θ (dθ/dt) = 1/100 (dx / dt)

Since we are looking for the rate of change of the angle, we will divide both sides by "- csc²θ".

[tex]\frac{d\theta }{dt} = \frac{\frac{1}{100} \times \frac{dx}{dt} }{-(csc \theta) ^2}[/tex]

But cscθ  = 1/sin θ

sin θ  = 100/200

sin θ = 1/2

cscθ = 2

Also, we are given dx/dt = 7 ft/s

Now, we will calculate the rate at which the angle is decreasing;

[tex]\frac{d\theta }{dt} = \frac{\frac{1}{100} \times \frac{dx}{dt} }{-(csc \theta) ^2}\\\\\frac{d\theta }{dt} = \frac{\frac{1}{100} \times 7 }{-(2) ^2}\\\\\frac{d\theta }{dt} = \frac{7}{-100 \times 4} \\\\\frac{d\theta }{dt} = - \frac{7}{400} \ rad/s\\\\\frac{d\theta }{dt} = - 0.0175 \ rad/s[/tex]

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.

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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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.

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:

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.

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:

Arrival times of P and S waves

Explanation:

Seismological recording station has a seismometer  that senses that motion in the ground, a clock that records time and a data recorded.

The distance between beginning of the first P wave and the first S wave will give you the time the waves are apart.This time value will be used to find the distance between the seismograph and the epicenter of earthquake and you mark it.This is corresponding distance in km to the time in seconds obtained  before.You then find the amplitude of the strongest wave and mark it on the right side of chart.Amplitude is the height on paper of the strongest wave.Using a ruler join the amplitude point and the point where you marked the distance to epicenter.This line will cross the magnitude chart at a point which represents the magnitude of the Earthquake.

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.

To calculate the number of hours of usage required for the LED light bulb to become more cost-effective than the incandescent bulb, we need to compare the total costs of each bulb. The LED bulb uses 80% less energy than the incandescent bulb, resulting in electricity cost savings. By comparing the total costs, we can find the point at which the LED bulb becomes more cost-effective.

To calculate the number of hours of usage required for the LED light bulb to become more cost-effective than the incandescent bulb, we need to compare the total costs of each bulb. The LED bulb uses 80% less energy than the incandescent bulb, resulting in electricity cost savings. We also need to consider the purchase price and lifetime of each bulb. By comparing the total costs, we can find the point at which the LED bulb becomes more cost-effective.

Let's use the given values: The LED bulb uses 20W of electricity and the incandescent bulb uses 100W. Assume the cost of electricity is $0.10 per kilowatt-hour. The LED bulb costs $20.00 and the incandescent bulb costs $0.75.

First, we calculate the energy used during the year for each bulb: E = Pt. For the LED bulb, the energy is (20W)(3 hours/day)(365 days/year) = 21.9 kilowatt-hours. For the incandescent bulb, the energy is (100W)(3 hours/day)(365 days/year) = 109.5 kilowatt-hours.

Next, we can multiply the energy by the cost of electricity to find the cost for each bulb. For the LED bulb, the cost is (21.9 kWh)($0.10/kWh) = $2.19. For the incandescent bulb, the cost is (109.5 kWh)($0.10/kWh) = $10.95.

Now we need to consider the initial purchase price and the lifetime of each bulb. The incandescent bulb lasts for 1.08 years (1200 hours) and the LED bulb lasts for 45.66 years (50,000 hours).

Finally, we can calculate the total cost for each bulb, including the purchase price and the energy cost. For the LED bulb, the total cost is $20.00 + $2.19 = $22.19. For the incandescent bulb, the total cost is $0.75 + $10.95 = $11.70.

Therefore, the LED bulb becomes more cost-effective than the incandescent bulb after approximately 545 hours of usage, since the total cost of the LED bulb is lower.

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.

Answer:

The time it takes for the camera to reach the ground is 5 s.

Explanation:

To solve this problem, we will use the free fall cinematic equation.

Since the helicopter ascends with constant speed, the camera falls to the ground only by the effect of gravity on it.

The speed at which the helicopter ascends is not specified in the statement, but according to a similar problem, we will use 12.5 [tex]\frac{m}{s}[/tex].

First, we must calculate the time and the maximum height at which the camera arrives after leaving the helicopter.

To calculate the maximum height to which it arrives, we will use the formula of vertical shot (since the camera leaves the helicopter with a speed upwards of 12,5 [tex]\frac{m}{s}[/tex]).

[tex]V_{f} ^{2}[/tex]=[tex]V_{0} ^{2}[/tex] - 2 * g * h

Where:

[tex]V_{f}[/tex]: final speed at maximum height.

[tex]V_{0}[/tex]: initial speed when it falls from the helicopter.

g: gravity taken at 9.8 [tex]\frac{m}{s^{2} }[/tex]

h: height reached from 60 m when leaving the helicopter

as [tex]V_{f}[/tex]=0

0=[tex](12,5\frac{m}{s}) ^{2}[/tex] - 2 * [tex]9,8\frac{m}{s^{2} }[/tex] * h

clear h:

h=[tex](12,5 \frac{m}{s^{2} } )^{2}[/tex] / (2 * [tex]9,8 \frac{m}{s^{2} }[/tex])

h=7,97 m

Then we must calculate the time it takes to reach its maximum height:

[tex]V_{f}[/tex]=[tex]V_{0}[/tex] - g * t

t: time it takes to arrive from the moment it leaves the helicopter at its maximum height.

as [tex]V_{f}[/tex]=0

0=12.5 [tex]\frac{m}{s}[/tex] - 9.8 [tex]\frac{m}{s^{2} }[/tex] * t

clearing t

t=12.5 [tex]\frac{m}{s}[/tex] / 9.8 [tex]\frac{m}{s^{2} }[/tex]

t=1.27 s.

Now we can calculate the time it takes to fall from the maximum height of 67.97 m.

The equation we will use is Y=[tex]v_{0}*t+(\frac{g*t^{2} }{2} )[/tex]

where:

t: time it takes for the camera to fall.

Y: height from where the camera falls concerning the ground.

[tex]v_{0}[/tex]: initial speed of the camera at the time of starting the fall.

g: acceleration of gravity, estimated at 9.8 [tex]\frac{m}{s^{2} }[/tex]

Step 1: As the helicopter ascends with constant speed, the initial speed of the camera at the moment of falling is 0.

[tex]v_{0}[/tex]=0

So the first term of our equation is nullified.

Step 2: To calculate the time it takes to fall, we clear "t" of the equation:

Y=[tex]\frac{(g*t^{2})}{2}[/tex]

Y*2=(g*[tex]t^{2}[/tex])

[tex]\frac{Y*2}{g}[/tex]=[tex]t^{2}[/tex]

[tex]\sqrt{\frac{Y*2}{g} }[/tex]=t

Step 3: I replace the values with the incognites and get "t".

t=[tex]\sqrt{\frac{67,97m*2}{9,8\frac{m}{s^{2} } } }[/tex]

t=3,73 s

The total time it takes for the camera to fall from the moment it leaves the helicopter is the sum of the time it takes to reach the maximum point of height and the time it takes to fall to the ground from that height.

t= 1,27 s + 3,73 s = 5 s

Have a nice day!

Final answer:

The camera will take approximately 3.9 seconds to reach the ground.

Explanation:

To find the time it takes for the camera to reach the ground, we can use the equation of motion for an object in free fall:

h = (1/2)gt²

where h is the initial height (60.0 m), g is the acceleration due to gravity (9.8 m/s²), and t is the time. Since the camera is dropped, its initial velocity is zero. Rearranging the equation, we have:

t = [tex]\sqrt{(2h / g)[/tex]

Substituting the given values:

t = [tex]\sqrt{2 * 60.0 / 9.8)[/tex]

t ≈ 3.9 seconds

So, the camera will take approximately 3.9 seconds to reach the ground.

Answer:

through layers of hot air just above a surface,causing it to follow a curved path.

. . . downward (through the air).

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 correct statement about the structure or function of the plasma membrane is: The processes of endocytosis and exocytosis occur here.

The plasma membrane, also known as the cell membrane, is a selective barrier that surrounds the cell and separates its internal contents from the external environment. It is composed of a phospholipid bilayer with embedded proteins, cholesterol, and carbohydrates. Here's the explanation for each statement:

1. Hydrophilic molecules attract the water the cell requires: This statement is true. The phospholipid bilayer has hydrophilic (water-attracting) heads that face outward towards the aqueous environments inside and outside the cell. This helps to maintain the cell's water balance.

2. The double layer prevents anything from entering the cell: This statement is partially true but can be misleading. While the phospholipid bilayer is selectively permeable and does restrict the passage of many substances, it is not an absolute barrier. Small, nonpolar molecules can pass through the lipid bilayer by simple diffusion. Additionally, the membrane contains various proteins that facilitate the transport of ions, nutrients, and waste products across the membrane.

3. The processes of endocytosis and exocytosis occur here: This statement is true. Endocytosis is the process by which cells take in material from the external environment by forming a vesicle from the plasma membrane. Exocytosis is the reverse process, where cells export material by fusing vesicles with the plasma membrane, releasing their contents to the outside.

4. It is made entirely of integral proteins: This statement is false. While integral proteins are an essential component of the plasma membrane, they do not make up the entire structure. The plasma membrane is primarily composed of a phospholipid bilayer, with proteins, cholesterol, and carbohydrates interspersed within it or attached to it. Integral proteins are embedded within the bilayer, but they are just one part of the overall structure.

Therefore, the statement that best describes a function of the plasma membrane is that it is the site of endocytosis and exocytosis, which are critical processes for cellular uptake and secretion.

Answer:

The total electrical power we are using is: 1316 W.

Explanation:

Using the ohm´s law [tex]V=I*R[/tex] and the formula for calculate the electrical power, we can find the total electrical power that we are using. First we need to find each electrical power that is using every single component, so the radio power is:[tex]I=\frac{V}{R}=\frac{110 (v)}{56(ohms)}=1.96(A)[/tex], so the radio power is: [tex]P=I*V=1.96(A)*110(v)=216(W)[/tex], then we find the pop-corn machine power as: [tex]P=I*V=7(A)*110(v)=770(W)[/tex] and finally there are three light bulbs of 110(W) so: P=3*110(W)=330(W) and the total electrical power is the adding up every single power so that: P=330(W)+770(W)+216(W)=1316(W).

Answer: The answer is that they rely on shock waves that are created by earthquakes and explosions,and that is how they find out the structure of the interior of the planet. :)

Answer:

New force, [tex]F'=\dfrac{F}{12}[/tex]

Explanation:

Given that, two point charges attract each other with an electric force of magnitude F. It is given by :

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

If one charge is reduced to one-third its original value and the distance between the charges is doubled such that,

[tex]q_1'=\dfrac{q_1}{3}[/tex], [tex]r'=2r[/tex]

[tex]F'=k\dfrac{q_1'q_2'}{r'^2}[/tex]

[tex]F'=k\dfrac{(q_1/3)q_2}{(2r)^2}[/tex]

[tex]F'=\dfrac{F}{12}[/tex]

So, the electric force between them is reduced to (1/12). Hence, the correct option is (d).

The magnitude of the electric force when one charge is reduced to one-third and the distance is doubled becomes F/12, following Coulomb's Law where force is proportional to the charge and inversely proportional to the square of the distance.

The question is asking how the electric force between two charges changes when one charge is reduced to one-third its original value and the distance between the charges is doubled. According to Coulomb's Law, the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Initially, we have a force of magnitude F. When one charge is reduced to one-third, the force becomes one-third of its original force (since force is directly proportional to the charge). This results in a force of F/3. When the distance is doubled, the force is reduced to one-fourth of its value (since force is inversely proportional to the square of the distance). So, F/3 is further reduced to (F/3) / 4 = F/12. Hence, the resulting magnitude of the electric force is F/12.

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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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 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]

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