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likoan [24]
3 years ago
7

If a team pulls with a combined force of 4 newtons on an airplace with a mass of 36 kilograms,what is the acceleration of the ai

rplane
Physics
1 answer:
MatroZZZ [7]3 years ago
5 0

Heya!!

For calculate aceleration, lets applicate second law of Newton:

                                                    \boxed{F=ma}

                                                 <u>Δ   Being   Δ</u>

                                                F = Force = 4 N

                                              m = mass = 36 kg

                                             a = Aceleration = ?

⇒ Let's replace according the formula:

\boxed{4\ N = 36\ kg * \textbf{a}}

⇒ Clear acceleration and resolve it:

\boxed{\textbf{a}=0,111\ m/s^{2}}}

Result:

The aceleration is of <u>0,111 meters per second squared (m/s²)</u>

Good Luck!!

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A flat uniform circular disk (radius = 2.00 m, mass= 100
ExtremeBDS [4]

Answer:

0.5 rad / s

Explanation:

Moment of inertia of the disk I₁ = 1/2 MR²

M is mass of the disc and R is radius

Putting the values in the formula

Moment of inertia of the disc  I₁  = 1/2 x 100 x 2 x 2

= 200 kgm²

Moment of inertia of man about the axis of rotation of disc

mass x( distance from axis )²

I₂  = 40 x 1.25²

= 62.5 kgm²

Let ω₁ and ω₂ be the angular speed of disc and man about the axis

ω₂ = tangential speed / radius of circular path

= 2 /1.25 rad / s

= 1.6 rad /s

ω₁ = ?

Applying conservation of angular moment ( no external torque is acting on the disc )

I₁ω₁ = I₂ω₂

200 X ω₁ = 62.5 X 1.6

ω₁ =  0.5 rad / s

7 0
3 years ago
At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan
Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time t_1, the acceleration vector has direction \theta_1 such that

\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ

which indicates the particle is situated at a point on the lower left half of the circle, while at time t_2 the acceleration has direction \theta_2 such that

\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ

which indicates the particle lies on the upper left half of the circle.

Notice that \theta_1-\theta_2=90^\circ. That is, the measure of the major arc between the particle's positions at t_1 and t_2 is 270 degrees, which means that t_2-t_1 is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}

where R is the radius of the circle and T is the period. We have

t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s

and the magnitude of the particle's acceleration toward the center of the circle is

\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}

So we find that the path has a radius R of

7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}

8 0
3 years ago
What should be done to lift the same load by applying less effort on an inclined plane​
Scorpion4ik [409]

Answer:

Reduce the friction at the surface

Explanation:

If you can reduce the friction between the load and the plane less effort will be required as you are not having to apply effort to overcome friction.

5 0
2 years ago
Calculate the pressure in SI unit when a force of 100 dyne acts on area of 0.02cm²​
Andru [333]

Answer:

500 Pa

Explanation:

Convert given units to SI:

100 dyne = 0.001 N

0.02 cm² = 2×10⁻⁶ m²

Pressure = force / area

P = 0.001 N / (2×10⁻⁶ m²)

P = 500 Pa

7 0
4 years ago
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