While in empty space, an astronaut throws a ball at a velocity of 11 m/s. What will the velocity of the ball be after it has traveled 7 meters? A. 0 m/s B. 4 m/s C. 18 m/s D. 11 m/s\

A rear-wheel-drive car has the given specifications and parameters: Car weight = 2600 lb;Wheelbase = 84 inches;Center of gravity = 20 inches above the roadway surface and 30 inches behind the front axle; Drivetrain efficiency = 85%; Wheel radius = 14 inches; Overall gear reduction = 7 to 1;

Torque/engine speed curve = Me = 6ne - 0.045ne2. The car is on a paved, level roadway surface with a coefficient of adhesion of 0.75. We have to determine the car's maximum acceleration from rest. To find the maximum acceleration from rest of the rear-wheel-drive car, we have to follow the below steps:

Step 1: Find out the static weight transfer when the car is accelerating from rest.

Step 2: Calculate the force of rolling resistance

Step 3: Find out the force of air resistance.

Step 4: Calculate the force to accelerate the car from rest.

Step 5: Find out the maximum acceleration of the car from rest.

The car would be going 60.5 mph if it could achieve its maximum speed when its engine is producing maximum power.

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The percentage of time the operator of the single entrance booth will be free is approximately 38.462%.

To determine the percentage of time that the operator of the single entrance booth will be free, we need to calculate the service rate and the arrival rate. The service rate is given as 400 vehicles per hour, and the arrival rate is 250 vehicles per hour. The percentage of time the operator will be free can be calculated using the formula:

Free time percentage = (Service rate - Arrival rate) / Service rate * 100

Substituting the given values into the formula:

Free time percentage = (400 - 250) / 400 * 100

= 150 / 400 * 100

= 0.375 * 100

= 37.5%

Rounding off the answer to the nearest thousandths, the percentage of time the operator of the single entrance booth will be free is approximately 38.462%.

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Answer:

Fc = 4340,93 Newton

Explanation:

Dados los siguientes datos;

Masa = 900 kg

Velocidad, V = 50 km/h a metros por segundo = (50 * 1000)/(60 * 60) = 50000/3600 = 13,89 m/s

Radio, r = 40 m

Para encontrar la fuerza centrípeta;

Fc = mv² / r

Fc = (900 * 13,89²)/40

Fc = (900 * 192,93)/40

Fc = 173637/40

Fc = 4340,93 Newton

The light from Galaxy HD1 is believed to have traversed 1.273 10 26 metres.

To convert the distance traveled by the light from Galaxy HD1 from light-years to meters, we can use the following conversion factor:

1 light-year = 9.461×10^15 meters

Therefore, the distance traveled by the light from Galaxy HD1 is:

13.463 billion light-years × 9.461×10^15 meters/light-year = 1.273×10^26 meters

So, the distance traveled by the light from Galaxy HD1 is approximately 1.273×10^26 meters, which is an incredibly large distance. It is important to note that this distance represents the comoving distance, which takes into account the expansion of the universe over time.

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Answer:

24

Explanation:

If it took 3 hours to clean 6 houses than you can figure that is 2 houses per hours. so if there are 12 hours and 2 houses an hour, you multiply 12x2 and get 24. Another way to do it is to use proportions (aka two fractions that equal eachother):

  3 hours                12 hours

----------------    =    ---------------

 6 houses          number of houses

To calculate this, we can think about how we go from 3 to 12. Well, 3x4 is 12, so we got to 12 by doing multiplying by 4. So we can multiply 6 by 4 and you get 24.

Hope it helps! have a great day :)

The diameter of the aorta can be determined by using the given velocity and flow rate. The velocity in branch arteries can be calculated using the given flow rate and the total area of the arteries. The first branch off the aorta can be identified based on its location. The valve area can be determined by utilizing the given velocities and the area of the LVOT.

To calculate the diameter of the aorta, we need to relate the velocity and the flow rate. Velocity is given as 33 cm/sec, and the flow rate is 4.4 L/min. We can convert the flow rate to cm^3/sec (1 L = 1000 cm^3), which gives us a flow rate of 73.33 cm^3/sec. The flow rate is proportional to the cross-sectional area of the vessel multiplied by its velocity. Using the formula Q = A * V, where Q is the flow rate, A is the cross-sectional area, and V is the velocity, we can rearrange the formula to solve for the diameter of the aorta.

To determine the velocity in each branch artery, we can use the given flow rate of 5.1 L/min and the total area of the branch arteries, which is 50 cm^2. By dividing the flow rate by the total area, we can find the velocity in the branch arteries.

Identifying the first branch off the aorta requires knowledge of the anatomy. Typically, the first branch off the aorta is the brachiocephalic trunk, which divides into the right subclavian artery and the right common carotid artery.

To calculate the valve area, we can utilize the velocities and the area of the left ventricular outflow tract (LVOT). The velocity in the aortic valve is given as 4.4 m/sec, and the velocity in the LVOT is 88 cm/sec. The valve area can be determined using the continuity equation, which states that the product of the cross-sectional area and velocity at one point is equal to the product of the cross-sectional area and velocity at another point along the flow path.

In summary, the diameter of the aorta can be determined using the given velocity and flow rate. The velocity in branch arteries can be calculated using the flow rate and the total area of the arteries. The first branch off the aorta is typically the brachiocephalic trunk. The valve area can be calculated using the velocities and the area of the LVOT, utilizing the continuity equation.

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The ratio of the pressure of solar radiation on sphere 2 to that on sphere 1 can be calculated using the equation P = 2I/c which is 1:4.

The pressure of solar radiation on a spherical object can be calculated using the equation P = 2I/c, where P is the pressure, I is the intensity of the radiation, and c is the speed of light.

The intensity of solar radiation at a distance r from the sun is proportional to 1/r². Therefore, the intensity of solar radiation on sphere 1 is 1/1² = 1, and the intensity on sphere 2 is 1/2² = 1/4.

Thus, the pressure of solar radiation on sphere 1 is 2/c, and the pressure on sphere 2 is 2/(4c) = 1/2c. Therefore, the ratio of the pressure of solar radiation on sphere 2 to that on sphere 1 is (1/2c) / (2/c) = 1/4.

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Answer:

125,000

Explanation:

The mass of the block 2 is 5.04 kg when Two blocks of mass 1 and 2 are connected by a massless string that passes over a massless pulley.

To find the mass of block M for the system to be in equilibrium, we need to consider the forces acting on the blocks and the angles of the inclines.

For block M:

The weight of block M acts vertically downward and can be decomposed into two components: one parallel to the incline and one perpendicular to the incline.

According to the question for the system to remain in equilibrium,

\(M_1sin\theta_1=M_2sin\theta_2\\\\2.75sin73.5=M_2 sin31.5\\\\M_2=5.04 kg\)

The mass of the block 2 is 5.04 kg. when the block connected to the pully.

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Answer:

Explanation:

It would be option C. Ray x appears as though it passed through the focal point in front of the lens, and ray Y passes through the focal point on the other side of the lens.

Answer:

c. Ray X appears as though it passed through the focal point in front of the lens, and ray Y passes through the focal point on the other side of the lens.

Explanation:

Answer:

c

Explanation:

tbh I think it is c because of what the explanation says so

Answer:

D

Explanation:

because i got it right when guessing its d

Answer:

it gravitional pull on earth will increased becauste it is compress to a form of moon which is comperatively smaller so the gravitonal pull on per cm of earth will incrased so we can say that there will be change in acceleration due to gravity

The statement that best illustrates that the light behaves like particles is "light bounces off a white cement sidewalk".

Light is electromagnetic radiation which can be seen by Bare eyes. It is the part of the electromagnetic spectrum.

Light is a particulate matter the reflection of light by striking to a surface is the best example to understand its nature.

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The current flowing through the loop is approximately 1.96 amperes.

Ampere's law states that the magnetic field around a closed loop is directly proportional to the current passing through the loop and inversely proportional to the distance from the center of the loop. By rearranging the formula, we can solve for the current:

B = (μ₀ * I) / (2 * π * r),

where B is the magnetic field, I is the current, μ₀ is the permeability of free space (a constant), and r is the radius of the loop.

In this case, we are given the magnetic field B = 2.5 mT and the diameter of the loop, which is equal to twice the radius (1.0 cm). We can convert the magnetic field to tesla by dividing by 1000 (1 T = 1000 mT). Thus, B becomes 0.0025 T, and the radius r is 0.5 cm.

Substituting these values into the rearranged Ampere's law equation, we have:

\(0.0025 T = (4\pi * 10^{-7} T m/A * I) / (2 * \pi * 0.5 cm).\)

Simplifying the equation, we cancel out the common factors and convert cm to meters:

\(0.0025 = (4\pi * 10^{-7}I) / 0.01.\).

Further simplification yields:

\(I = (0.0025 * 0.01) / (4\pi * 10^{-7}).\)

Evaluating the right side of the equation, we find:

I ≈ 1.96 A.

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Higher high water
Lower high water
Higher low water
Lower low water

In a mixed-tide system, the water levels exhibit two high tides and two low tides within a tidal cycle. The ranking of the water levels according to height is as follows:

Higher high water: This is the highest water level observed during the tidal cycle. It occurs when the gravitational forces of the Moon and the Sun align to produce a stronger tidal bulge. It typically happens around the time of a new moon or a full moon.

Lower high water: This is the second-highest water level observed during the tidal cycle. It occurs when the gravitational forces of the Moon and the Sun are not aligned, resulting in a weaker tidal bulge. It typically happens around the time of the first quarter and third quarter moon phases.

Higher low water: This is the higher of the two low water levels observed during the tidal cycle. It occurs when the gravitational forces of the Moon and the Sun produce a weaker tidal trough. It typically happens between the two high tides.

Lower low water: This is the lowest water level observed during the tidal cycle. It occurs when the gravitational forces of the Moon and the Sun are not aligned, resulting in a stronger tidal trough. It typically happens between the higher low water and the next high tide.

The ranking of water levels in a mixed-tide system, from highest to lowest, is: higher high water, lower high water, higher low water, lower low water

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Answer:

Current = 2 amps

Explanation:

Remark

The voltage of the circuit is 24 volts.

The Resistance Series circuit is

Givens

E = 24 volts

R = 12 ohms

Formula

E = I * R where

Solution

24 volts = 12 * I         Divide by 12

24/12 = I

I = 2 amperes.

The crankshaft's angular acceleration at time zero is thus \(100 rad/s^2\).

Crankshaft is shown as a graph of angular velocity against time. The graph of the crankshaft of a car's angular velocity against time is shown in the image below. The formula for angular acceleration is the product of the angular velocity and the acceleration time. Alternatively, pi () divided by the acceleration time (t) and 30 times driving speed (n).

The radians per second squared unit of measurement for angular acceleration is obtained from this equation. This equation's first term, which is the rod torque adjusted for articulating inertial effects, second term, which is the counterbalance torque, and final term, which is the rotating inertial torque.

\(a = (w_2-w_1) /(t_2-t_1)\\a= (150-50) / (1-0)\\a= 50 m/s^2\)

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Correct Question:

What is the crankshaft's angular acceleration at t = 1 s?

A portion of the paper near the balloon becomes positive when you place the balloon close to a small piece of paper because the negative charge balloon repels the electrons charge in the paper.

Electric charge, a fundamental characteristic of matter carried by some elementary particles, controls how an electric or magnetic field affects the particles. Positive or negative electric charge exists in distinct natural units and cannot be created or destroyed.

Positive and negative charges are the two main categories of electric charges. When two objects are relatively close to one another and one type of charge is present in excess, they repel one another. When they are relatively close to each other, two objects with excess opposite charges—one positively charged and the other negatively charged—attract to one another.

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To determine which subatomic particle has a longer lifespan, we compare the average lifetimes of the tau lepton and the neutral pion. The average lifetime of the tau lepton is 2.906 x 10^-12 seconds, and the average lifetime of the neutral pion is 8.4 x 10^-17 seconds.

Comparing these values, we can see that the average lifetime of the tau lepton (2.906 x 10^-12 seconds) is significantly longer than the average lifetime of the neutral pion (8.4 x 10^-17 seconds).

To calculate how many times longer the lifespan of the tau lepton is compared to the neutral pion, we divide the average lifetime of the tau lepton by the average lifetime of the neutral pion:

(2.906 x 10^-12 seconds) / (8.4 x 10^-17 seconds) = 3.453 x 10^4

Therefore, the tau lepton has a lifespan that is approximately 3.453 x 10^4 (34,530) times longer than the neutral pion.

Answer:

The circuit works the same no matter where you put the switch

Explanation:

Part of the electrical code (for house wiring) says that the switch should always go between the hot conductor and the load. This is for safety

The theoretical density of BCC chromium is , 2.48g/cm³

ρ = \(\frac{N * M }{N_{A} * a^{3} }\)

where

ρ  = density of unit cell

a = edge length of unit cell

M = Atomic mass

Z = no. of atoms in unit cell

\(N_{A}\) = Avogadro's number

(given)

Z = 2 (BCC)

M = 52.0 g/mol

atomic radius = 0.125 nm

edge length of unit cell is a

r = 0.866a (BCC unit cell)

a = 0.125nm/0.866 = 0.144×10⁻⁷cm

1nm = 10⁻⁷cm

using above values

ρ = 2× 52.0 / (6.022×10²³mol⁻¹ )×(0.144×10⁻⁷cm)³

ρ =   2.48g/cm³

The theoretical density of BCC chromium is , 2.48g/cm³

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Answer:

vf = 7 m/s

Explanation:

The final velocity of the ball right before it hits the ground can be found by using the third equation of motion:

\(2gh = v_f^2 - v_i^2\)

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 2.5 m

vf = final velocity = ?

vi = initial velocity =  0 m/s

Therefore,

\(2(9.81\ m/s^2)(2.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{49.05\ m^2/s^2}\\\)

vf = 7 m/s

39°C=1.28L

8°C=? Less

=8°C/39°C×1.28L

=8/39×1.28L

=10.24/39

=512/195

=2.6L

The part of the microscope that the objectives are attached to is called the (C) nosepiece.

The nosepiece is a rotating mechanism located below the microscope's body tube. It holds the objectives, which are the lenses responsible for magnifying the specimen. The nosepiece typically has multiple positions, allowing the user to switch between different objective lenses for varying levels of magnification.

This convenient feature eliminates the need to manually remove and replace objectives when changing magnification. By rotating the nosepiece, different objectives can be brought into position above the specimen. This allows for quick and efficient adjustments in magnification without disrupting the viewing process.

Hence, the nosepiece plays a critical role in the microscope's functionality by providing a convenient way to switch between objectives and adjust the magnification level.

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Here is the complete question:

What is the name of the part of the microscope that the objectives are attached to? (Choose the best answer)

A. Ocular

B. Stage

C. Nosepiece

D. Arm

If a wave has a wavelength of 50 m and is traveling through water that is 30 m deep then this kind of wave is called deep water wave.

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Answer:(:

Explanation:

discovered alpha and beta rays, and proposed the laws of radioactive decay.

What is the numeric value of the prefix "micro"?

a. 0.00001

c. 0.0001

d. 0.001

In this example, there would be 6 stirrups crossing the shear crack.

The number of stirrups crossing a shear crack can be calculated using the formula n = d/s, where n represents the number of stirrups, d represents the crack diagonal, and s represents the spacing between the stirrups.

To calculate the number of stirrups crossing a shear crack, you need to determine the crack diagonal, which is the length of the crack along its diagonal path. This length can be measured or estimated based on the shear crack angle assumption of 45 degrees.

Once you have the crack diagonal length, you need to determine the spacing between the stirrups, which is the distance between each stirrup. This spacing can be specified in the design code or determined based on structural requirements.

Using the formula n = d/s, you can then divide the crack diagonal length by the stirrup spacing to find the number of stirrups crossing the shear crack.

For example, let's say the crack diagonal length is 900 mm and the stirrup spacing is 150 mm. Plugging these values into the formula, we have n = 900 mm / 150 mm = 6 stirrups.

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