The equation for power is P=w/t, where w=work done in joules and t=time in seconds.
in this case:
p=w/t
p=3500/25
p=140
so, the boat was using 140 watts of power.
Answer:
2.5 kg/m³
Explanation:
Absolute pressure = gauge pressure + atmospheric pressure
P = Pg + Pa
The gauge pressure caused by the weight of a fluid is called the static pressure. It is equal to the density of the fluid × acceleration due to gravity × depth of the fluid,
Pg = ρgh
Therefore:
P = ρgh + Pa
300 Pa = ρ (10 m/s²) (8.00 m) + 100 Pa
ρ = 2.5 kg/m³
The statement Gravity exerts zero joules of energy on the object is false.
<h3>What is gravity?</h3>
- The force of attraction between any two objects in the universe is known as gravity or gravitational force.
- The mass of the object and the square of the distance between them determine the force of attraction.
- The weakest known force in nature, it is by far.
- The force that pulls items toward the center of a planet or other entity .
- All of the planets are kept in orbit around the sun by gravity.
<h3>What is work?</h3>
- Work is defined as the energy that is applied to or removed from an object by applying force along a displacement.
- It is frequently described in its most basic form as the result of force and displacement.
Learn more about work here:
brainly.com/question/18094932
#SPJ4
Answer:
Miller Indices are [2, 4, 3]
Solution:
As per the question:
Lattice Constant, C =
Intercepts along the three axes:
Now,
Miller Indices gives the vector representation of the atomic plane orientation in the lattice and are found by taking the reciprocal of the intercepts.
Now, for the Miller Indices along the three axes:
a =
b =
c =
To find the Miller indices, we divide a, b and c by reciprocal of lattice constant 'C' respectively:
a' =
b' =
c' =
If the atom of carbon has lost 4 electrons, it has now an excess of charge equal to +4e:
where
Its charge is concentrated in the nucleus, so we can treat it as a single-point charge, whose electric field is given by:
where k is the Coulomb's constant and r is the distance from the charge. In our problem,
therefore the electric field at this distance is