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2 years ago

. Write the following in the scientific notation of 0.0000002400m

Physics

1 answer:

aleksandr82 [10.1K]2 years ago

7 0

Explanation:

<h2>Steps :</h2>

Move decimal from left to right =0 000000240.0
Then count the numbers before decimal and write it like this =240.0x10 power-9
That's all

hope it help

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A shot-putter released the shot at an angle of 41.5 degrees and a height of 1.9 m with an initial velocity of 13.3 m/s. How far

liq [111]

Answer:

x = 17.88[m]

Explanation:

We can find the components of the initial velocity:

$(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]$

We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.

g = - 9.81[m/s^2]

Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.

$y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]$

With this time we can calculate the horizontal distance:

$x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]$

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3 years ago

There is a plate with moment of inertia Ip = 0.0711 kgm2 , rotating around its center of mass with angular speed of 4.53 rad/s.

Anna007 [38]

Answer:

1) their common angular speed = 3.038 rad/s

2) kinetic energy loss = 0.24J

Explanation:

1) This is a case of conservation of angular momentum.

The initial angular momentum must be equal to the final angular momentum

Initial angular momentum = I x w

Where I = moment of inertia, and

w = angular momentum.

Initial angular momentum = 0.0711 x 4.53 = 0.322 kg-m2-rad/s

After addition of ring, moment of inertia becomes,

0.0711 + 0.0353 = 0.106 kg-m2

Therefore, final angular momentum will be

0.106 x Wf

Where Wf = final common angular velocity

Equating the two angular momentum we have

0.322 = 0.106Wf

Wf = 0.322/0.106 = 3.038 rad/s

2) KE = 1/2 x I x w^2

Initial KE = 1/2 x 0.0711 x 4.53^2

= 0.729 J

Final KE = 1/2 x 0.106 x 3.038^2

= 0.489 J

Loss in KE = 0.729 - 0.489 = 0.24 J

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3 years ago

The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.56 m and 1.00 m, respectively. The respective

valentina_108 [34]

Answer:

Explanation:

Given

Diameter of main rotor $d_1=7.56 m$

Tail rotor $d_2=1 m$

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$\omega =\frac{2\pi N}{60}$

$\omega _1=\frac{2\pi 453}{60}=47.44 rad/s$

$\omega _2=\frac{2\pi 4137}{60}=433.28 rad/s$

Speed of the tip of main rotor $=\omega _1\times r_1=47.44\times \frac{7.56}{2}=179.32 m/s$

Speed of tail rotor $=\omega _2\times r_2=433.28\times \frac{1}{2}=216.64 m/s$

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The answer must be a mass x velocity

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What is the SI units of mass and weight?

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SI unit for mas is Kg and the SI unit of weight N for newton

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3 years ago

Read 2 more answers

. Write the following in the scientific notation of 0.0000002400m ​

. Write the following in the scientific notation of 0.0000002400m