What is the magnitude of the tangential acceleration of a bug on the rim of a 13.0-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 79.0 rev/min in 3.80 s

Answer:

.....

Explanation:

Answer:

The answer is D. In the direction of the wave

Explanation:

I took the test, hope this helps! :)

Answer:

f= ma

Explanation:

v-u/ t

30-0/10

3m/s

maas 1500

1500*3

4500

Answer:

Students need a safe environment to take risks and struggle. It’s uncomfortable to struggle, but struggling—falling down and getting back up—is an important facet to learning. Productive struggle is not about being in pain or becoming frustrated.

Please mark branliest!

Explanation: it shows that your really trying your best and when you mess up its telling you that your actually learning from your mistakes.

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 answer is the 4th one

Explanation:

Good luck man

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:

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;

https://brainly.com/question/25937699

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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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#SPJ7

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:

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.

Answer:

[tex]g_2=\dfrac{7}{8}g[/tex]

Explanation:

G = Gravitational constant

M = Mass of planet

R = Radius of planet

Acceleration due to gravity on first planet

[tex]g=\dfrac{GM}{R}[/tex]

Assuming that the planets have the same mass density [tex]\rho[/tex]

Density of first planet

[tex]\rho=\dfrac{M}{V}\\\Rightarrow \rho=\dfrac{M}{\dfrac{4}{3}\pi R^3}\\\Rightarrow M=\rho \dfrac{4}{3}\pi R^3[/tex]

Density of second planet

[tex]\rho=\dfrac{M_2}{V_2}=\dfrac{M_2}{\dfrac{4}{3}\pi R^3-\dfrac{4}{3}\pi (0.5R)^3}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(1-\dfrac{1}{2^3})}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(1-\dfrac{1}{8})}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(\dfrac{7}{8})}\\\Rightarrow M_2=\rho\dfrac{4}{3}\pi R^3(\dfrac{7}{8})\\\Rightarrow M_2=M\dfrac{7}{8}[/tex]

Acceleration due to gravity on the second planet

[tex]g_2=\dfrac{GM_2}{R}\\\Rightarrow g_2=\dfrac{GM\dfrac{7}{8}}{R}\\\Rightarrow g_2=\dfrac{7}{8}g[/tex]

The acceleration due to gravity of the planet would be [tex]\dfrac{7}{8}[/tex] times the acceleration due to gravity on the first planet.

Explanation:

doesn't get the question and the answer to

Answer:

(this is the sample answer:After the launch, the fireball moves upward, than it falls back down to EarthEartEarthEarthEarthEarth

Explanation:

Answer:

i think the answer is solar,

Explanation:

because it is a natural resource im sorry if im wrong

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:

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:

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:

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]

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. moving the magnet farther away and an iron object.

Explanation:

Answer:A

Explanation:

A p e x

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:

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:

Shape and position

Explanation:

Hope this helps! :)

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:

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:

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]