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denis-greek [22]

4 years ago

9

As increases, the humidity of that area will . temperature; also increase dew; also increase rainfall; alwa

ys increase temperature; never increase

1 answer:

Nady [450]4 years ago

4 0

Always increase temperature

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A disk with a rotational inertia of 5.0 kg. m2 and a radius of 0.25 m rotates on a frictionless fixed axis perpendicu-

diamong [38]

Answer:1.6 rad/s

Explanation:

Given

moment of Inertia of disk $I=5 kg-m^2$

radius of disc $r=0.25 m$

Force $F=8 N$

Torque $T=I\alpha =F\cdot r$

$5\times \alpha =8\times 0.25$

$\alpha =0.4 rad/s^2$

using

$\theta =\omega _0\times t+\frac{\alpha t^2}{2}$

$\pi =0+\frac{0.4t^2}{2}$

$2\pi =0.4t^2$

$t^2=5\pi$

$t=\sqrt{5\pi }$

$t=3.96 s$

$\omega =\omega _0+\alpha t$

$\omega =0+0.4\times 3.96$

$\omega =1.58 rad/s\approx 1.6 rad/s$

3 0

4 years ago

PLEASE HELP 40 POINTS

sergij07 [2.7K]

Liar, not even 40 points....

5 0

3 years ago

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons

Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

$H_m=1.65m$

$H_E=1.16307m$

Explanation:

From the question we are told that

Mass of ball $M=2kg$

Length of string $L= 2m$

Wind force $F=13.2N$

Generally the equation for $\angle \theta$ is mathematically given as

$tan\theta=\frac{F}{mg}$

$\theta=tan^-^1\frac{F}{mg}$

$\theta=tan^-^1\frac{13.2}{2*2}$

$\theta=73.14\textdegree$

Max angle = $2*\theta= 2*73.14=>146.28\textdegree$

Generally the equation for max Height $H_m$ is mathematically given as

$H_m=L(1-cos146.28)$

$H_m=0.9(1+0.8318)$

$H_m=1.65m$

Generally the equation for Equilibrium Height $H_E$ is mathematically given as

$H_E=L(1-cos73.14)$

$H_E=0.9(1+0.2923)$

$H_E=1.16307m$

8 0

3 years ago

If you double d, what happens to your force

Tanzania [10]

Answer:

If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases. If the distance is doubled, the force of gravity is one-fourth as strong as before.

3 0

3 years ago

The planet Uranus has a radius of 25,360 km and a surface acceleration due to gravity of 9.0 m/s^2 at its poles. Its moon Mirand

AlexFokin [52]

Answer:

$8.67791\times 10^{25}\ kg$

$0.34589\ m/s^2$

$0.07903\ m/s^2$

Explanation:

M = Mass of Uranus

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Uranus = 25360 km

h = Altitude = 104000 km

$r_m$ = Radius of Miranda = 236 km

m = Mass of Miranda = $6.6\times 10^{19}\ kg$

Acceleration due to gravity is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow M=\dfrac{gr^2}{G}\\\Rightarrow M=\dfrac{9\times 25360000^2}{6.67\times 10^{-11}}\\\Rightarrow M=8.67791\times 10^{25}\ kg$

The mass of Uranus is $8.67791\times 10^{25}\ kg$

Acceleration is given by

$a_m=\dfrac{GM}{(r+h)^2}\\\Rightarrow a_m=\dfrac{6.67\times 10^{-11}\times 8.67791\times 10^{25}}{(25360000+104000000)^2}\\\Rightarrow a_m=0.34589\ m/s^2$

Miranda's acceleration due to its orbital motion about Uranus is $0.34589\ m/s^2$

On Miranda

$g_m=\dfrac{Gm}{r_m^2}\\\Rightarrow g_m=\dfrac{6.67\times 10^{-11}\times 6.6\times 10^{19}}{236000^2}\\\Rightarrow g_m=0.07903\ m/s^2$

Acceleration due to Miranda's gravity at the surface of Miranda is $0.07903\ m/s^2$

No, both the objects will fall towards Uranus. Also, they are not stationary.

6 0

3 years ago