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TiliK225 [7]
2 years ago
11

"The predictions of Einstein’s Theory of General Relativity were tested on a double pulsar system in January of 2004. His equati

ons proved to be accurate to __________ percent."
Physics
1 answer:
Rasek [7]2 years ago
8 0

Answer:

99.95%

Explanation:

A double pulsar system named PSR J0737-3039A/B  in Puppis constellation was discovered in the year 2003. Pulsars are second most densest object in the universe after black holes and they emit radio waves at regular intervals. This pair presented a great and natural setup to test the Theory of General Relativity presented by Einstein in 1915. In this theory Einstein had presented a set of equations on how the space-time fabric will be curved because of the very dense objects such as Neutron stars. It also predicted how the gravitational waves are created because of stars orbiting each other.

A team of astrophysicists led by Michael Kramer, conducted a study on how these gravitational waves will impact the time in which the radio waves emitted by pulsars will reach Earth. The result of the study proved the theory of General Relativity to be accurate up to 99.95%.

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jesse is swinging miguel in a circle at a tangential speed of 3.50 m/s. if the radius of the circle is 0.600 m and miguel has a
Morgarella [4.7K]
Centripetal acceleration = (speed)² / (radius) .

Force = (mass) · (acceleration)

Centripetal force = (mass) · (speed)² / (radius) .

                             = (11 kg) · (3.5 m/s)² / (0.6 m)

                             = (11 kg) · (12.25 m²/s²) / (0.6 m)

                             =  (11 · 12.25) / 0.6  kg-m/s²

                             =      224.58 newtons.    (about 50.5 pounds)

That's the tension in Miguel's arm or leg or whatever part of his body
Jesse is swinging him by.  It's the centripetal force that's needed in
order to swing 11 kg in a circle with a radius of 0.6 meter, at 3.5
meters/second.  If the force is less than that, then the mass has to
either swing slower or else move out to follow a bigger circle.
6 0
2 years ago
Read 2 more answers
The image shows one complete cycle of a mass on a spring in simple harmonic motion. An illustration of a mass on a vertical spri
Alja [10]

Answer:

D. "The net force is zero, so the acceleration is zero"

Explanation:

edge 2020

6 0
2 years ago
If the period of a clock signal is 500 ps, the frequency is:_____
NNADVOKAT [17]

Answer:

The frequency of the signal is 2 GHz

Explanation:

Given;

period of the clock signal, T = 500 ps = 500 x 10⁻¹² s

the frequency of the signal is given by;

F= \frac{1}{T}\\\\F = \frac{1}{500*10^{-12}}\\\\F = 2*10^{9} \ Hz

F = 2 GHz

Therefore, the frequency of the signal is 2 x 10⁹ Hz or 2 GHz

4 0
2 years ago
an input force of 50 Newtons is applied through a distance of 10 meters to machine with mechanical advantage of 3. If the work o
gladu [14]
The output of the machine is

                                      (output work) =  (output force) x (distance)

                                        450 N-m      =  (output force) x (3 meters)

Divide each side
by  3 meters:                Output force = (450 N-m) / (3 m)

                                                           =    150 newtons .

With all the information given about the output work, we don't need
to know anything about the input work, or even the fact that we're
dealing with a machine.

It's comforting, though, to look back and notice that the output work
(450 N-m) is not more than the input work (500 N-m).  So everything
is nice and hunky-dory.
___________________________________

Well, my goodness !
I didn't even need to go through all of that.

Given:

-- The input force to the machine is 50 newtons.

-- The mechanical advantage of the machine is 3 .

That right there tells us that

-- The output force of the machine is 150 newtons.

We don't need any of the other given information.
5 0
2 years ago
Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally
Bess [88]

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0


east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero


distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m  = 3600m


3 0
2 years ago
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