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seropon [69]
3 years ago
10

Which scientist was the first to accurately measure the speed of light?

Physics
2 answers:
oksano4ka [1.4K]3 years ago
6 0

Answer:

Olaus Roemer was the first scientist to accurately measure the speed of  light.

Explanation:

nekit [7.7K]3 years ago
5 0
D. Olaus Roemer was the 1st
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When a ball is launched from the ground at a 45° angle to the horizontal, it falls back to the ground 50 m from the launch point
Inessa [10]

Answer:

Explanation:

Given

angle through which ball is launched=45^{\circ}

Range of ball=50 m

Range of projectile is =\frac{u^2sin2\theta }{g}

50=\frac{u^2sin90}{9.8}

u=22.136 m/s

If ball is thrown straight upward

v^2-u^2=2as

0-(22.136)^2=2(-9.8)s

s=\frac{22.136^2}{2\times 9.8}

s=25 m

(b)For Projectile time of flight is

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

t=\frac{2\times 22.136\times sin45}{9.8}

t=3.19 s

7 0
3 years ago
FLUKE
Thepotemich [5.8K]

Answer:

cant understand anything

Explanation:

3 0
3 years ago
Can someone explain what is loss of seismic energy ?
xxTIMURxx [149]

Answer:

Seismic attenuation describes the energy loss experienced by seismic waves as they propagate. It is controlled by the temperature, composition, melt content, and volatile content of the rocks through which the waves travel.

Explanation:

5 0
3 years ago
Four seconds after being launched, what is the height of a ball that starts from a height of 12 m with an initial upward velocit
MrMuchimi

Answer:

15.24 m/s in the downward direction

Explanation:

Given that the initial upward velocity of the ball is 24 m/s.

Assuming that the upward direction is positive.

As gravitational force acts in the downward direction and the direction of acceleration is the same as the direction of force, so the acceleration due to gravity will be negative.

Now, from the equation of motion, when an object is launched with initial velocity u, the final velocity, v, of an object after time t is v=u+at.

Given that u=24 m/s, t=4 seconds, g=-9.81 m/s^2.

So, the final velocity is

v= 24 + (-9.81)\times 4 \\\\\Rightarrow v= 24-9.81\times 4

\Rightarrow v=-15.24 m/s

Here, the negative sign means the final velocity is in the downward direction.

Hence, the velocity after 4 seconds is 15.24 m/s in the downward direction.

8 0
3 years ago
A test of the prototype of a new automobile shows that the minimum distance for a controlled stop from 95 km/h to zero is 55 m.
iris [78.8K]

Answer:

-0.64525g

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 95 km/h

v = Final velocity = 0 km/h

s = Displacement

a = Acceleration

Equation of motion

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-95^2}{2\times 0.055}\\\Rightarrow a=-82045.45\ km/h^2

Converting to m/s²

a=82045.45=\frac{82045.45\times 1000}{3600\times 3600}=-6.33\ m/s^2

g = Acceleration due to gravity = 9.81 m/s²

Dividing both the accelerations, we get

\frac{a}{g}=\frac{-6.33}{9.81}=-0.64525\\\Rightarrow a=-0.64525g

Hence, acceleration of the car is -0.64525g

8 0
3 years ago
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