Answer:
Explanation:
First we need to determine the distance covered during deceleration. According to the equation of motion.
S = ut+1/2at²
Given:
u = 20m/s
t = 0.50s
a = -10m/s (deceleration is negative acceleration)
S = 20²+1/2(-10)(0.5)²
S = 400-5(0.5)²
S = 400-5(0.25)
S = 400-1.25
S = 398.75m
If the deer steps onto the road 35m in front of you, the distance between you and the deer when you come to a stop will be 398.75-35 = 363.75m
We are aware that weight is the product of applied gravitational force and mass. W = MG thus, where W represents the weight, M the mass, and G the gravitational force. As a result, it might also mean that "an object's weight is directly proportionate to its mass."
<h3>What is mass?</h3>
- Mass is a physical body's total amount of matter.
- It also serves as a gauge for the body's inertia, or resistance to acceleration (change in velocity) in the presence of a net force.
- The strength of an object's gravitational pull to other bodies is also influenced by its mass.
- The kilogram is the primary mass unit in the SI (kg).
- Even though weight is frequently measured using a spring scale rather than a balancing scale and directly compared with known masses, mass is not the same as weight in physics.
<h3>What is weight?</h3>
- The force exerted on an object by gravity is known as the weight of the object in science and engineering.
- Weight is sometimes described as a vector quantity, or the gravitational force exerted on the object, in some common textbooks.
- Others define weight as a scalar quantity, the gravitational force's strength.
- Others define it as the strength of the force applied to a body as a result of systems designed to resist the effects of gravity; the weight is the amount that is determined, for instance, by a spring scale.
Learn more about mass here:
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Answer:
According to Archimedes principle, volume of water displaced = volume of water.
Hence volume of rock is = 1.65L or 1650 cm^3
Explanation:
Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.
