Answer:
1225 J
Explanation:
The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by
where
m is the mass
g = 9.8 m/s^2 is the acceleration due to gravity
h is the height at which the object has been lifted
In this problem, we have
m = 250 kg
h = 0.5 m
So, the PE of the object is
An electron has a negative charge
<h2>
Answer: Pulsars</h2>
A <u>pulsar</u> is a neutron star that emits very intense electromagnetic radiation at short and periodic intervals ( rotating really fast) due to its intense magnetic field that induces this emission.
Nevertheless, it is important to note that all pulsars are neutron stars, but not all neutron stars are pulsars.
Let's clarify:
A neutron star, is the name given to the remains of a supernova. In itself it is the result of the gravitational collapse of a massive supergiant star after exhausting the fuel in its core.
Neutron stars have a small size for their very high density and they rotate at a huge speed.
However, the way to know that a pulsar is a neutron star is because of its high rotating speed.
Answer:
physics, escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body. The escape velocity from Earth is about 11.186 km/s (6.951 mi/s; 40,270 km/h; 25,020 mph)[1] at the surface. ... [2] Speeds higher than escape velocity have a positive speed at infinity.
The bouyancy force is:
Since the wood-lead system is completely submerged, the bouyancy force
is FB = ĎwgVl + ĎwgVb, where Ďw is the density of water,Vl
is the volume of
the piece of lead and Vb is the volume of the wooden block. The weight of the
combined lead and wooden block is: W = ĎlgVl + ĎbgVb. Since the system is
in equilibrium, the bouyancy force must be equal to the total weight:
ĎwgVl + ĎwgVb = ĎlgVl + ĎbgVb
now we can solve for the volume of lead:
ĎwgVl â’ ĎlgVl = ĎbgVb â’ ĎwgVb
Vl(Ďw â’ Ďl) = Vb(Ďb â’ Ďw)
Vl =
Ďbâ’Ďw
Ďwâ’Ďl
Vb
Now we substitute the values for the density of lead Ďl = 11.3 Ă— 103kg/m3 ,
the density of the wood and the density of water Ďw = 1000kg/m3
. We get:
Vl =
600â’1000
1000â’11300
(0.6m Ă— 0.25m Ă— 0.08m) = 4.66 Ă— 10â’4m3
