tính các mô men quán tính chính trung tâm của tiết diện, cho biết a = 10cm

From the relationship between acceleration a and g on an inclined plane, the angle of the slope is 13 degrees

Given that a solid sphere starts from rest and rolls down a slope that is 6.4 m long. The speed at the bottom of the slope is 5.3 m/s, the distance travelled is 6.4 m. That is,

Initial velocity U = 0 ( since it starts from rest)

Final velocity V = 5.3 m/s

distance S = 6.4 m

Let us first calculate its acceleration by using  third equation of motion.

[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2aS

[tex]5.3^{2}[/tex] = 0 + 2 x 6.4a

28.09 = 12.8a

a = 28.09 / 12.8

a = 2.2 m / [tex]s^{2}[/tex]

To calculate the angle of the slope, let us use the relationship between acceleration a and g on an inclined plane.

acceleration a = gsin∅

substitute all the relevant parameters

2.2 = 9.8 sin∅

sin∅ = 2.2/9.8

sin∅ = 0.224

∅ = [tex]Sin^{-1}[/tex](0.224)

∅ = 12.97 degrees

∅ = 13 degrees (approximately)

Therefore, the angle of the slope is 13 degrees

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Answer:

thx for the points

Explanation:

no need brainliest

The maximum mass of a load that can be lifted by the jack and the distance covered are:

m = 160.2 Kg

h = 25 cm

Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.

The parameters given are

[tex]F_{1}[/tex] = 250

[tex]A_{1}[/tex] = Area of the small piston = π[tex]r^{2}[/tex]

[tex]A_{1}[/tex] = 22/7 x [tex]0.4^{2}[/tex]

[tex]A_{1}[/tex] = 0.5 [tex]m^{2}[/tex]

[tex]F_{2}[/tex] = ?

[tex]A_{2}[/tex] = Area of the large piston = π[tex]r^{2}[/tex]

[tex]A_{2}[/tex] = π x 1

[tex]A_{2}[/tex] = 3.14 [tex]m^{2}[/tex]

To calculate the force on the large piston, we will use the below formula

[tex]F_{1}[/tex]/ [tex]A_{1}[/tex] = [tex]F_{2}[/tex] / [tex]A_{2}[/tex]

Substitute all the parameters into the equation

250/0.5 =  [tex]F_{2}[/tex]/3.14

[tex]F_{2}[/tex] = 1570 N

To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law

F = mg

1570 = 9.8m

m = 1570/9.8

m = 160.2 Kg

.(take g=9.81ms^-2)​

If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be

[tex]F_{1}[/tex]/ 0.25[tex]A_{1}[/tex] = [tex]F_{2}[/tex] / [tex]A_{2}[/tex]h

250/0.125 = 1570/3.14h

make h the subject of the formula

6280h = 1570

h = 1570/6280

h = 0.25 m

Therefore, the distance through which the load is lifted is 25 cm

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Answer:

D. An area of low pressure might move more quickly on Tuesday and Wednesday than expected.

Explanation:

I took the test and got it wrong :/

Answer:

C. Velocity and Position

Explanation:

The quantities that are increasing for the object represented by line B are velocity and position. The correct option is b.

The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity. In essence, velocity is a vector quantity. It is the speed at which distance changes. It is the displacement change rate.

Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is traveling along a path. In other words, velocity is a vector, whereas speed is a scalar value.

The graph is given which represents the velocity and time with terms A, B, and C. As opposed to the position-time graph, which describes an object's motion over time, the velocity-time graph reveals an object's speed.

Therefore, the correct option is b. velocity and position.

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The question is incomplete. Your most probably complete question is given below:

Velocity and acceleration

velocity and position

velocity only

velocity, position, and acceleration

Answer:

3h/4

Explanation:

At speed v/2 height will be 3/4 h

Equation that describes the motion of point , bodies , and system of bodies without considering the force that cause them to move is called equation of motion in kinematics

When block is at top of first ramp

u=0 ( block was at rest )

a = g ( acceleration due to gravity

using equation of motion                                                                                                                                                                                                                                                                    

2as = v^2 - u^2

2gh = v^2

Then the block continued and reached a speed of v1 = v/2 on second ramp

now , final velocity = v= v1 =[tex]\sqrt{2gh}[/tex] / 2

u= [tex]\sqrt{2gh\\}[/tex]

s= h1

using equation of motion , we get

2as = v^2 - u^2

2(-g)h1 =( [tex]\sqrt{2gh}[/tex]/2)^2 - [tex]\sqrt{2gh}[/tex]

2(-g)h1 = (g h - 4 g h) / 2

h1 = 3/4 h

At speed v/2 height will be 3/4 h

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Answer:

sorry for you

The ratio of the man's moment of inertia in the first case (arms extended) to that in the second case (arms drawn in) is 3.2.

The relationship between the rotational frequency [tex](\(\omega\))[/tex] and moment of inertia (I) is given by the equation:

[tex]\[I_1\omega_1 = I_2\omega_2\][/tex]

where [tex]\(I_1\)[/tex]and [tex]\(I_2\)[/tex] are the moments of inertia in the two cases, and [tex]\(\omega_1\) and \(\omega_2\)[/tex] are the corresponding rotational frequencies.

Let's denote the moment of inertia in the first case (arms extended) as [tex]\(I_1\)[/tex] and in the second case (arms drawn in) as [tex]\(I_2\)[/tex]. The given rotational frequencies are [tex]\(\omega_1 = 0.25 \, \text{rev/s}\) and \(\omega_2 = 0.80 \, \text{rev/s}\)[/tex].

Using the equation [tex]\(I_1\omega_1 = I_2\omega_2\)[/tex], we can rearrange it to solve for the ratio of moments of inertia:

[tex]\[\frac{I_1}{I_2} = \frac{\omega_2}{\omega_1}\][/tex]

Substituting the given values, we have:

[tex]\[\frac{I_1}{I_2} = \frac{0.80 \, \text{rev/s}}{0.25 \, \text{rev/s}}\][/tex]

Simplifying the expression, we get:

[tex]\[\frac{I_1}{I_2} = 3.2\][/tex]

Therefore, the ratio of the man's moment of inertia in the first case (arms extended) to that in the second case (arms drawn in) is 3.2.

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Answer:

Moment of a force is the product of the force and the perpendicular distance of force from axis of rotation.

The SI unit of force is newton (N).

Explanation:

Hope it helps you again!

Answer:

the different about the molecules we needed to make protein 3 compared to the molecules we used to make protein 2 is that if we used 2 molecules than it will be smaller than using protein 3.

Hi there!

We know that:

Initial Total Mechanical Energy = Final Total Mechanical Energy

(Ei = Ef)

In this instance:

Ei = Gravitational Potential Energy

Ef = Kinetic Energy

In the absence of friction, ALL of the initial potential energy will be changed into kinetic energy at the bottom of the slide. Thus, the maximum kinetic energy of the child will be 1800 J.

Answer:

I'm pretty sure this is not a complete question. My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.

IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43

It could have any direction of

θ = (225 - 180) ± arcsin(13/30)

θ = 45 ± 25.679...

70.679 ≤ θ ≤ 19.321

components of vector B would be

Bx = |B|cosθ

By = |B|sinθ

My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.

IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43

It could have any direction of

θ = (225 - 180) ± arcsin(13/30)

θ = 45 ± 25.679...

70.679 ≤ θ ≤ 19.321

components of vector B would be

Bx = |B|cosθ

By = |B|sinθ

My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird. A vector has a magnitude of 30m at an angle of 225 degrees with respect to the positive x-axis and has a magnitude of 13m.

Therefore, My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.

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Answer:

Sorry, This Photo is not clear.

The total reactance of the inductor and the capacitor is 470.1 ohms.

The given parameters;

The inductive reactance of the coil is calculated as follows;

[tex]X_l = \omega L\\\\X_l = 2\pi f L\\\\X_l = 2\pi \times 60 \times 250 \times 10^{-3}\\\\X_l = 94.26 \ ohms[/tex]

The capacitive reactance of the capacitor is calculated as follows;

[tex]X_c = \frac{1}{\omega C} \\\\X_c = \frac{1}{2\pi f C} \\\\X_c = \frac{1}{2\pi \times 60 \times 4.7 \times 10^{-6}} \\\\X_c = 564.31 \ ohms[/tex]

The impedance of the circuit is calculated as follows;

[tex]Z = \sqrt{(X_c - X_l)^2} \\\\Z = X_c - X_l\\\\Z = 564.31 - 94.26\\\\Z = 470.1 \ ohms[/tex]

Thus, the total reactance of the inductor and the capacitor is 470.1 ohms.

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this is where the sun and moon line up where you asleep only a tiny bit of the sun it's pretty cool to see

Explanation:

a solar eclipse means when the moon goes infront of the sun and the earth turns dark

Answer:

3.9m/s² is the acceleration

The acceleration of the object is 3.9m/s².

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).

When you fall off a bridge. The car turning at the corner is an example of acceleration because the direction is changing. The quicker the turns, the greater the acceleration.

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The magnet is attracted to the ring since the north pole of the current loop is above the ring and the south pole is below the ring.

Answer:

a

Explanation:

Hi there!

We can begin by solving for the linear acceleration as we are given sufficient values to do so.

We can use the following equation:

vf = vi + at

Plug in given values:

4 = 9.7 + 4.4a

Solve for a:

a = -1.295 m/s²

We can use the following equation to convert from linear to angular acceleration:

a = αr

a/r = α

Thus:

-1.295/0.61 = -2.124 rad/sec² ⇒ 2.124 rad/sec² since counterclockwise is positive.

Now, we can find the angular displacement using the following:

θ = ωit + 1/2αt²

We must convert the initial velocity of the tire (9.7 m/s) to angular velocity:

v = ωr

v/r = ω

9.7/0.61 = 15.9 rad/sec

Plug into the equation:

θ = 15.9(4.4) + 1/2(2.124)(4.4²) = 20.56 rad

The rate of heat energy exhausted to the river is 600.96 MW.

The ratio of usable output to total input can be used to objectively measure efficiency. The efficiency of the device is defined as the ratio of energy converted to a useable form to the original amount of energy supplied.

Given parameters:

Efficiency of the power plant; η = 31 %

Output  electric power; O = 270 MW.

We know that, Efficiency of the power plant;

η  = (Output  electric power/ input power)× 100%

⇒ input power = (Output  electric power × 100)/η

⇒ input power = (270 × 100)/31 MW

= 870.96 MW.

So, the rate of heat energy exhausted to the river that cools the plant =  Input power- output power

= (870.96 - 270) MW

= 600.96 MW.

Hence, heat energy exhausted to the river that cools the plant  is 600.96 MW.

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Answer:

5460000000 J OR 5460000 KJ

Explanation:

GPE = mgh

21000*10*26000

=5460000000J OR 5460000KJ

Answer:

its b it must be charge

Explanation:

Answer:

25.0 cm3

Explanation:

The volume is 25.0 cm3 .

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]

Age of the universe; [tex]t = \ ?[/tex]

We know that, the reciprocal of the Hubble's constant ( [tex]H_0[/tex] ) gives an estimate of the age of the universe ( [tex]t[/tex] ). It is expressed as:

[tex]Age\ of\ Universe; t = \frac{1}{H_0}[/tex]

Now,

Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]

We know that;

[tex]1\ light\ years = 9.46*10^{15}m[/tex]

so

[tex]1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m[/tex]

Therefore;

[tex]H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\[/tex]

Now, we input this Hubble's constant value into our equation;

[tex]Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years[/tex]

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

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Answer:

Correct option is

C

The listener is moving towards the source of sound

The Doppler effect (or the Doppler shift) is the change in frequency or wavelength of a wave (or another periodic event) for an observer moving relative to its source.

So if both listener and source of sound are stationary or are moving with the same velocity that is relative velocity is zero, there won't be any change in frequency.

Thus, in this case, the effect will arise when a listener is moving towards the source of the sound.

From kinematic, the maximum height above ground a projectile of mass 0.79 kg, launched from ground level can achieve is 329 meters approximately.

The parameters given from the question are

mass m = 0.79kg

initial speed U = 80.3 m/s

Maximum height H = ?

The object will be going up under the influence of gravity. Acceleration due to gravity g = -9.8m/[tex]s^{2}[/tex]

At maximum height, final velocity V = 0  

From kinematic formula, the best formula to use is

[tex]V^{2} = U^{2} - 2gH[/tex]

Substitute all the parameters into the equation

0 = [tex]80.3^{2}[/tex] - 2 x 9.8 x H

19.6H = 6448.09

H = 6448.09 / 19.6

H = 328.98 m

Therefore, the maximum height above ground a projectile of mass 0.79 kg, launched from ground level can achieve is 329 meters approximately.

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Answer:

1.904

Explanation:

F= ma

8 = 4.2 a

a = 8/4.2

a = 1.904

Answer:

answer

transverse waves

Answer:

The Estes C6-0 engine is a booster stage engine designed for model rocket flight and has to be used with a standard engine. This engine is for flights in rockets weighing less than 4 ounces, including the engines. Each package includes 3 engines, 4 starters and 4 plugs.