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3 years ago

Convert the speed of light, 3.0 x 108 m/s, to km/s.

Physics

1 answer:

zhuklara [117]3 years ago

3 0

Answer:

I don't know

Explanation:

I don't know

Explanation:

Answer:

I don't know

Explanation:

I don't know

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A block starting from rest slides down the length of an 18 plank with an acceleration of 4.0 meters per second. How long does th

Snowcat [4.5K]

Answer:

$\boxed{\text{\sf \Large 3.0 s}}$

Explanation:

Use distance formula

$\displaystyle d=ut+\frac{1}{2} at^2$

$u= \text{\sf initial velocity}\\d= \text{\sf distance}\\a= \text{\sf acceleration}\\t= \text{\sf time taken}$

$\displaystyle 18=0 \times t+\frac{1}{2} \times 4 \times t^2$

$t=3$

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3 years ago

How much of the universe do scientists speculate is composed of dark energy

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Answer: Approximately 65% from what i have learnt.

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2 years ago

Why did you spin counterclockwise rather than clockwise

rodikova [14]

<span>Because of the orbit of the earth and the sun and the moon. </span>

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3 years ago

Read 2 more answers

A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit

lapo4ka [179]

Answer:

The angle is $\theta = 15.48^o$

Explanation:

From the question we are told that

The distance of the dartboard from the dart is $d = 3.66 \ m$

The time taken is $t = 0.455 \ s$

The horizontal component of the speed of the dart is mathematically represented as

$u_x = ucos \theta$

where u is the the velocity at dart is lunched

$distance = velocity \ in \ the\ x-direction * time$

substituting values

$3.66 = ucos \theta * (0.455)$

=> $ucos \theta = 8.04 \ m/s$

From projectile kinematics the time taken by the dart can be mathematically represented as

$t = \frac{2usin \theta }{g}$

=> $usin \theta = \frac{g * t}{2 }$

$usin \theta = \frac{9.8 * 0.455}{2 }$

$usin \theta = 2.23$

=> $tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}$

$\theta = tan^{-1} [0.277]$

$\theta = 15.48^o$

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3 years ago

an aircraft landing on an air craft carrier is brought to a complete stop from an inital velocity of 215km/hr in 2.7 seconds. wh

worty [1.4K]

u= 215 km/hr = 215 * 1000/ 3600 = aprx 60m/s
v=0
t=2.7sec
v= u - at
u= at
60/2.7 = 22.23 m/s^2

Hope it helps

8 0

3 years ago