Luke investigated the heating of water
Answers
Go Luke !
See Luke investigate !
Water, water, water.
Heat, heat, heat.
Luke rocks !
He probably learned something.
Related Questions
A 2-kg object is moving horizontally with a speed of 4 m/s. How much net force is required to keep the object moving at this speed and in this direction?
Who Can Help ? Much Love If You Can .Select all that apply. A convex lens _____. converges light bulges in the middle diverges light is thinner in the middle
A 70 kilogram mountaineer is standing on the summit of Mt. Everest. The distance between the mountaineer and the center of Earth is 6.39 × 106 meters. What is the magnitude of the force of gravity acting on the mountaineer? (The value of G is 6.673 × 10-11 newton meter2/kilogram2. The mass of Earth is 5.98 × 1024 kilograms.)90.50 newtons111.5 newtons684.1 newtons4041.7 newtons
Which statement best describes the superposition principle?A. If two in-phase waves arrive simultaneously at a point, their amplitudes add up. B. If two out-of-phase waves arrive simultaneously at a point, their amplitudes add up. C. If two in-phase waves arrive at a point one after another, their amplitudes add up. D. If two out-of-phase waves arrive at a point one after the other, their amplitudes adds up
B. miners getting trapped underground
C. oil spills that pollute oceans and beaches
O D. pollutants in the groundwater after drilling
Answers
Answer:
i think it is A
Explanation
Answers
Answer:
26.82m/s
Explanation:
Given
Mass = m= 0.4kg
Initial Velocity = u = 0
Charge = 4.0E-5C
Distance= d = 0.5m
Object Charge = 2E-4C
First, we'll calculate the initial energy (E)
E = Potential Energy
PE = kQq / d
Where k = coulomb constant = 8.99E9Nm²/C²
Energy is then calculated by;
PE = 8.99E9 * 4E-5 * 2E-4 / 0.5
PE = 143.84J
Energy = Potential Energy = Kinetic Energy
K.E = ½mv² = 143.84J
½mv² = ½ * 0.40 * v² = 143.85
0.2v² = 143.85
v² = 143.85/0.2
v² = 719.25
v = √719.25
v = 26.81883666380777
v = 26.82m/s
Hence, the object is 26.82m/s fast when the cart moving is very far (infinity) from the fixed charge
Answers
Explanation:
KE = 1/2 mv²,
where KE is the kinetic energy,
m is the mass,
and v is the velocity.
Given m = 1500 kg and v = 30 m/s:
KE = 1/2 (1500 kg) (30 m/s)²
KE = 675,000 J
KE = 675 kJ
Answers
The speed of electron be 183.73 m/s.
What is potential energy?
The energy held within an object is known as potential energy. There are several different types of potential energy, including gravitational potential energy, elastic potential energy, and electric potential energy. An object's energy as a result of the overall charge it contains is known as its electric potential.
Given parameters:
Distance between proton and electron: d = 0.03 m.
Let, at that point the speed of the electron be v.
So from conservation of energy:
Kinetic energy of electron + potential energy of electron = 0
1/2×mv² - 1/(4πε₀)(e²/d) = 0
⇒v = √(2e²/m(4πε₀)(d))
= √( 2 × (1.6 ×10⁻19)²× 9 × 10⁹/9.1093837 × 10⁻³¹×0.03) m/s
= 183.73 m/s.
The speed of electron be 183.73 m/s.
Learn more about potential energy here:
#SPJ2
Answers
The electric potential difference is the electric potential energy per unit charge
Explanation:
First of all, we define the concept of electric potential. The electric potential is a measure of the gradient of the electric field at a certain point of the space. The electric potential at a distance from a positive charge of magnitude
is given by
where k is the Coulomb's constant.
Now we can define the electric potential energy and the electric potential difference:
- Electric potential energy is the energy possessed by a charge due to the presence of an electric field. For a charge of magnitude
immersed in an electric field, its potential energy is given by
, where V is the electric potential at the location of the charge.
- The electric potential difference is simply the difference in electric potential between two points in the space. For instance, if the potential at point A is V(A) and the potential at point B is V(B), then the potential difference is
The electric potential energy is also defined as the work done on a charge q moved through a potential difference of . Consequently, the potential difference
represents the work per unit charge done, i.e. the work done when moving a unitary charge through a potential difference
.
Learn more about potential difference and current:
#LearnwithBrainly
The pulley increases the force needed to lift the load by moving it over a longer distance. It also changes the direction of the force. As the input force is applied downward, the output force is exerted upward.
B.
The pulley reduces the force needed to lift the load by moving it over a longer distance. It also changes the direction of the force. As the input force is applied downward, the output force is exerted upward.
C.
The pulley increases the force needed to lift the load by moving it over a shorter distance. It also changes the direction of the force. As the input force is applied upward, the output force is exerted downward.
D.
The pulley reduces the force needed to lift the load by moving it over a shorter distance. It also changes the direction of the force. As the input force is applied upward, the output force is exerted downward.
Answers
Answer:
B.
The pulley reduces the force needed to lift the load by moving it over a longer distance. It also changes the direction of the force. As the input force is applied downward, the output force is exerted upward.
Explanation:
When we use pulley system to pull the heavy weight then it is easier to pull the weight upwards because while we use use pulley to pull the objects then the force that is applied on the string is exerted at some angle with the vertical
Due to this we need to apply force at some angle with the vertical and it increase the output force on the object which is to be pulled
So here our effort is in downward direction to pull the object in upward direction
so correct answer will be
B.
The pulley reduces the force needed to lift the load by moving it over a longer distance. It also changes the direction of the force. As the input force is applied downward, the output force is exerted upward.