Explain how you would determine how much air there is between a vector addition in the displacement of the actual trip taken. Why aren’t vectors true straight-line paths in the real world? Would you expect a vector addition to be larger or smaller than the actual trip taken? Why?

Answer:

Non-zero digits

Explanation:

umm (ꏿ﹏ꏿ;)

Answer:

There's three rules on determining how many important quantities are in a number: Non-zero digits are always predominant. Any zeros between two significant digits are always major. The final zero or dangling zeros in the decimal segment are really only important.

I hope this answered your question, have a nice day!

Answer:

density of the sediment, size of the sediment, and velocity of the agent of erosion.

Explanation:

Have a nice day!

Answer: b., c., e.

Explanation:

Answer:

This question is incomplete

Explanation:

This question is incomplete because there are some parameters that are missing. However, when calculating the average emf and speed/velocity is provided, then

average emf = Brvl

where Br is the vertical component of the earth's magnetic field (in weber per square meter, Wb/m²)

v is velocity (in m/s)

l is the length (in meter) of the wing span

Answer:

Shape and position

Explanation:

Hope this helps! :)

Answer:

The answer is "[tex]5.06 \times 10^{-6} \ kg \ m^2[/tex]"

Explanation:

[tex]\to E_1=0..............(i)\\\\\to E_2= \frac{mV^2}{2} +\frac{Iw^2}{2} - mgh.............(ii)\\\\ \Delta E=0\\\\\to mgh= \frac{mV^2}{2} +\frac{Iw^2}{2} \\\\ \to 2 \ mgh= mV^2 +Iw^2\\\\ \to 2 \ mgh- mV^2 =Iw^2\\\\ \to m(2gh- V^2) =Iw^2\\\\ \to I= \frac{m(2gh- V^2)}{w^2}[/tex]

       [tex]= 5.06 \times 10^{-6} \ kg \ m^2[/tex]

Answer:

The same as

Explanation:

First of all, they are in a merry go round, which is a circular one. And thus, we would be dealing with circular motions.

The angular velocity of a merry go round, in this instance, is given by

w = Δ/Δt

The formula has nothing whatsoever to do with distance, and as such, both Rasheed and Sofia would have the same angular velocity.

Transferring this further, the angular acceleration is given as

α = Δw/Δt

Remember I said the velocity has nothing to do with distance, well, so does the acceleration as we can see from the formula stated. And therefore, both Rasheed and Sofia would have the same angular acceleration

Answer:

F' = 100 F/r²

Explanation:

The gravitational force of attraction between two objects is given by the Newton's Gravitational Formula. The Newton's Gravitational Formula is as follows:

F = Gm₁m₂/r²

where,

F = Force between objects

G = Universal Gravitational Constant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects = 10 m

Therefore,

F = Gm₁m₂/10²

Gm₁m₂ = 100F   --------------------- equation (1)

Now, we consider these objects at any distance r apart. So, the force becomes:

F' = Gm₁m₂/r²

using equation (1), we get:

F' = 100 F/r²

So, if the force (F) between objects 10 m apart is known, we can find it at any distance from the above formula.

The force between objects that are any distance apart is expressed as [tex]P'=\frac{100P}{r^2}[/tex]

According to the gravitational law, the force acting on an object is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. Mathematically,

[tex]P=\frac{GMm}{r^2}[/tex]

M and m are the masses

r is the distance between the masses

If the force between objects that are 10 meters apart, hence;

[tex]P=\frac{GMm}{10^2}\\P=\frac{GMm}{100}\\GMm = 100P[/tex]

To find the force between objects that are any distance apart, we will use the same formula above to have;

[tex]P'=\frac{GMm}{r^2}\\[/tex]

Substitute the result above into the expression to have:

[tex]P'=\frac{100P}{r^2}[/tex]

Hence the force between objects that are any distance apart is expressed as [tex]P'=\frac{100P}{r^2}[/tex]

Learn more on gravitational law here: https://brainly.com/question/11760568

Answer:

The answer is the 4th one

Explanation:

Good luck man

Answer:

The combined force of the catapult and engines is 44545.455 newtons.

Explanation:

Let suppose that platform of the aircraft carrier is horizontal, such that changes in gravitational potential energy can be neglected. In addition, effects of non-conservative forces are neglected. By using Principle of Energy Conservation and Work-Energy Theorem we find that steam catapult-jet aircraft is represented by:

[tex]K_{1}+W_{1\rightarrow 2} = K_{2}[/tex] (1)

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energies of the jet fighter, measured in joules.

[tex]W_{1\rightarrow 2}[/tex] - Work done on the jet fighter due to the combined force of the catapult and engines, measured in joules.

By applying definitions of translational kinetic energy and work, we expand and simplify the equation above as follows:

[tex]F\cdot \Delta s = \frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2})[/tex]

[tex]F = \frac{m\cdot (v_{2}^{2}-v_{1}^{2})}{2\cdot \Delta s}[/tex] (2)

Where:

[tex]F[/tex] - Combined force of the catapult and engines, measured in newtons.

[tex]m[/tex] - Mass of the jet fighter, measured in kilograms.

[tex]\Delta s[/tex] - Length of the catapult, measured in meters.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the jet aircraft, measured in meters per second.

If we know that [tex]m = 4000\,kg[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]v_{2} = 70\,\frac{m}{s}[/tex] and [tex]\Delta s = 220\,m[/tex], then the combined force of the catapult and engines is:

[tex]F = \frac{(4000\,kg)\cdot \left[\left(70\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (220\,m)}[/tex]

[tex]F = 44545.455\,N[/tex]

The combined force of the catapult and engines is 44545.455 newtons.

The combined force of the catapult and engines is 44,545.5 N.

The given parameters;

The combined force of the catapult and engines is determined by applying work energy theorem as shown below;

[tex]W = \Delta E\\\\Fd = \frac{1}{2}m(v^2 -u^2) \\\\220F = \frac{1}{2}(4000)(70^2 - 0^2) \ \\\\220 F = 9,800,000\\\\F = \frac{9,800,000}{220} \\\\F = 44,545.5 \ N[/tex]

Thus, the combined force of the catapult and engines is 44,545.5 N.

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Answer:

the war of currents is

Explanation:

The war of currents was a series of events surrounding the introduction of competing electric power transmission systems in the late 1880s and early 1890s.

Answer:

a series of events surrounding the introduction of competing electric power transmission systems in the late 1880s and early 1890s

Explanation:

Answer:

a) letting in-flowing and/or out-flowing tide flow through turbines in a dam.

Explanation:

Tidal power makes use of the energy from the tidal force and wave action in order to generate electricity. it is a predictable source of energy.

Tidal Barrages

The system allows tides to enter, seawater flows via the dam and is trapped into the basin when the tides subside and system’s gates close.

When the tides start to move out, the gates in the dam are opened up this consist of turbines and water begins to flow out, hitting the turbines this eventually turn to produce energy.

Power is produced when the tidal range, which is the difference between low and high tide, has to be more than 5 meters.

Answer: Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s.

Explanation:

From the graph, it is clear that, the velocity is at a time of 1 s is highest. The velocity at 1 second corresponds to 1250 km/hr. Then it decreases with time.

The velocity - time graph shows the change in velocity with respect to time. The velocity is placed in y -axis and time is given in x - axis. The slope of the curve in velocity - time graph gives the acceleration of the object.

Similarly, the position of the object in meter after a t seconds can be determined from the velocity - time graph. It is the rate of change in velocity of the object.

From the graph, it is clear that, the curve has its peak at 1 second. After that the peak descends down. Hence, the maximum velocity of the car is at a time of 1 second at which the velocity is 1250 km/hr.

Find more on velocity - time graph :

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Answer:

i didnt know ther was differnt types

Explanation:

Answer:

2.51 m/s

Explanation:

Given that,

Angular speed of a fan, [tex]\omega=0.5\ rev/s = 3.14\ rad/s[/tex]

Length of the fan arm, r = 0.8 m

We need to find the speed of a point on the end of the fan arm at this time. Let v is the speed. It is given in terms of angular speed is given by :

[tex]v=r\omega\\\\v=0.8\times 3.14\\\\v=2.51\ m/s[/tex]

So, the required speed is 2.51 m/s.

Answer:

I'm not 100% sure, but I think the answer would be the first one because there's a force pushing the object in every direction, so they would cancel eachother out and make the object stay in the same place.

Explanation:

pls vote brainliest

Answer:

32.67km

Explanation:

~= round off

11.2km/h~3.11m/s

175minutes=10500s

10500*3.11~32, 666.67m

=32.66667km

~ 32.67km

Answer:

f= ma

Explanation:

v-u/ t

30-0/10

3m/s

maas 1500

1500*3

4500

Answer:

A

Explanation:

The image formed by the concave mirror is always virtual. An enlarged image is caused when the concave mirror is positioned too near to the object.

When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.

A concave mirror is a name for this sort of mirror. An enlarged and fake image is caused when the concave mirror is positioned too near to the object.

When the distance between the item and the mirror is increased, however, the size of the image decreases, and a true picture is generated.

The picture created by the concave mirror may be manipulated. The image formed by the concave mirror is always virtual.

Hence, option A is correct.

To learn more about the concave mirror, refer to the link;

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Answer:

T = 12732 [N]

Explanation:

To solve this problem we must use newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

By means of the attached free body diagram, we can better understand the solution to this problem.

As we see in the diagram the tension Force T lifts the car upwards so the movement is upwards. Therefore we have two forces one upward (positive) of the force T and the other negative downward due to the weight of the vehicle.

As the movement is up the acceleration is also up (with a positive sign).

ΣF = m*a

where:

F = forces [N] (units of Newtons)

m = mass = 1200 [kg]

a = acceleration = 0.8 [m/s²]

T - 1200*9.81 = 1200*(0.8)

T = 12732 [N]

Answer:

12,720

Explanation:

Answer:

.....

Explanation:

Answer:

a. Please see the attached graph and drawing combined created with Microsoft Excel

b. Approximately 3.2217 m/s

c. Approximately 14.0 m/s  

d. Approximately 11.05 meters

Explanation:

a. The given parameters are;

The height of the wall, h = 10.0 m

The distance from the base of the wall the car lands = 4.60 m

The time, t, it takes the car to land is given by the equation for free fall as follows;

h = 1/2·g·t²

Where;

g = The acceleration due to gravity = 9.81 m/s²

From the equation for free fall, we have;

h/(1/2·g) = t²

∴ t = √(h/(1/2·g) ) = √(10/(1/2·9.81) ) ≈ 1.4278

The time it takes the car to land, t ≈ 1.4278 seconds

b. The horizontal speed of the car = Horizontal distance/(Time) = 4.6/1.4278 ≈ 3.2217 m/s

The horizontal speed of the car before it drives off the wall  ≈ 3.2217 m/s

c. The car's vertical speed just before impact is given by the following equation;

v = u + gt

Where;

u = The initial vertical speed = 0

t = The time it takes before impact ≈ 1.4278 seconds

∴ v = 9.81 × 1.4278 ≈ 14.0 m/s  

The car's vertical speed just before impact, v ≈ 14.0 m/s  

d. Whereby the car is subject to the gravitational field of the moon, we have;

Gravitational force per kilogram = 1.7 N/kg

∴ Gravitational acceleration = 1.7 m/s².

The time it takes the car to land whilst subject to the gravitational field of the moon is therefore;

[tex]t_{Moon}[/tex] = √(h/(1/2·g)) = √(10/(1/2 × 1.7)) ≈ 3.43

[tex]t_{Moon}[/tex] ≈ 3.43 seconds

The horizontal distance covered, at the car's horizontal speed in the time of free fall ≈ 3.2217 m/s ×  3.43 seconds ≈ 11.05 meters

The horizontal distance covered, at the car's horizontal speed in the time of free fall  = The distance the car will land from the base of the wall ≈ 11.05 meters

The distance the car will land from the base of the wall ≈ 11.05 meters.

Explanation:

hope this answer was helpful

Answer:

The value is  [tex]H = 0.036 \ kg/m^3[/tex]

Explanation:

From the question we are told that

   The saturation density of air is [tex]\rho = 0.04 \ kg/m^3[/tex]

   The temperature for this saturation density to occur  [tex]T_s = 96^o F[/tex]

    The relative humidity at this temperature is  [tex]R_h = 90\% = 0.90[/tex]

Generally the relative humidity is mathematically represented as

      [tex]R_H = \frac{H}{\rho}[/tex]

Here H represents humidity

=> [tex]0.90 = \frac{H}{0.04}[/tex]

=>  [tex]H = 0.90 * 0.04[/tex]

=>  [tex]H = 0.036 \ kg/m^3[/tex]

AnswerAmontons's law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. ... If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas (Figure 1).:

Explanation:

Answer:

A mixture is when two or more substances combine physically together.

However, in water, two hydrogen atoms combine with one oxygen atom chemically, forming a new substance that has properties different from hydrogen alone or oxygen alone.

Explanation:

We are given:

Mass of the Paint bucket (with paint) = 7000 grams

Mass of the paint bucket (without paint) = 500 grams

Volume of Paint in the Bucket = 5000 cm³

Mass of Paint in the Bucket:

To get the mass of the paint in the bucket, we will subtract the mass of the bucket from the mass of the paint bucket (with paint)

Mass of Paint = Mass of Paint bucket (with paint) - Mass of the paint Bucket (without paint)

Mass of Paint = 7000 - 500

Mass of Paint = 6500 grams

Density of the Paint:

We know that density = Mass / Volume

Density of Paint  = Mass of Paint / Volume occupied by Paint

Density of Paint = 6500/5000

Density of Paint = 1.3 grams / cm³

Answer:

The answer is D. In the direction of the wave

Explanation:

I took the test, hope this helps! :)