As per the third law of Newton, the force exerted by the boat over the student is equal in magnitude to the force that the student exerted on the boat.
So, calculate the force on the student using the second law of Newton, Force = mass * acceleration.
Force on the student = 60 kg * 2.0 m/s^2 = 120 N.
=> horizontal force exerted by the student on the boat = 120 N
Answer: option d. 120 N. toward the back of the boat.
Of course it is toward the back because that is where the student jumped from..
The answer is:
B)They are in the same group because they have similar chemical properties, but they are in different periods because they have very different atomic numbers.
The explanation:
The vertical columns on the periodic table are called groups or families because of their similar chemical behavior. All the members of a family of elements have the same number of valence electrons and similar chemical properties.
A period is a horizontal row of elements on the periodic table. For example, the elements sodium ( Na ) and magnesium ( Mg ) are both in period 3. The elements astatine ( At ) and radon ( Rn ) are both in period 6.
Answer
The answer and procedures of the exercise are attached in the following archives.
Explanation
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
C
Explanation:
To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):
d = d₀ + d₀αT
for the sphere, we were given
D₀ = 4.000 cm
α = 1.1 x 10⁻⁵/degrees celsius
we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T EQN 1
Similarly for the Aluminium ring we have
we were given
d₀ = 3.994 cm
α = 2.4 x 10⁻⁵/degrees celsius
we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T EQN 2
Since @ the temperature T at which the sphere fall through the ring, d=D
Eqn 1 = Eqn 2
4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms
0.006=5.18x10⁻⁵T
T=115.7K
<span>Answer: 1600 J
Explanation:
1) Data:
a) ideal gas: ⇒ pV = nRT and work = ∫ pdV
b) slowly compressed ⇒ constant temperature and not heat exchange
c) pressure: p = 2 atm
d) intitial volume: Vi = 10 liters
e) final volumen: Vf = 2 liters.
f) then the
volume of the gas is held constant ⇒ not work in this stage.
g) calculate the work done on the gas: W = ?
2) Equation
W = ∫pdV
3) Solution:
Since p = constant, W = p ∫dV = p ΔV = p (Vf - Vi)
p = 2 atm × 1.0 ×10⁵ Pa / atm = 200.000 Pa
Vi = 10 liter × 0.001 m³ ./ liter = 0.01 m³
Vf = 2 liter × 0.001 m³ / liter = 0.002 m³
W = 200.000 Pa × (0.002 m³ - 0.01m³) = - 1.600 J.
The negative sign means the work is done over the system.
That is all the work in the system because at the second stage the volume is held constant.
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