Answer:
Hello your question is incomplete below is the complete question
Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000
answer : V = 1.624* 10^-5 m/s
Explanation:
First we have to calculate the value of a
a = 93 * 10^6 mile/m * 1609.344 m
= 149.668 * 10^8 m
next we will express the distance between the earth and the sun
--------- (1)
a = 149.668 * 10^8
E (eccentricity ) = ( 1/60 )^2
= 90°
input the given values into equation 1 above
r = 149.626 * 10^9 m
next calculate the Earths velocity of approach towards the sun using this equation
------ (2)
Note :
Rc = 149.626 * 10^9 m
equation 2 becomes
(
therefore : V = 1.624* 10^-5 m/s
Gravitational potential energy<span> is </span>energy<span> an object possesses because of its position in a </span>gravitational<span> field. The most common use of </span>gravitational potential energy<span> is for an object near the surface of the Earth where the </span>gravitational<span> acceleration can be assumed to be constant at about 9.8 m/s</span>2<span>.</span>
Answer:
3.6 KJ
Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
The workdone = the energy.
There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )
P.E = mgh
P.E = 70 × 9.8 × 1.6
P.E = 1097.6 J
P.E = 1.098 KJ
K.E = 1/2mv^2
K.E = 1/2 × 70 × 8.5^2
K.E = 2528.75 J
K.E = 2.529 KJ
The non conservative workdone = K.E + P.E
Work done = 1.098 + 2.529
Work done = 3.63 KJ
Therefore, the non conservative workdone is 3.6 KJ approximately
Answer:
74.86°C
Explanation:
P₂ = Vapour pressure of water at sea level = 760 mmHg
P₁ = Pressure at base camp = 296 mmHg
T₂ = Temperature of water = 373 K
ΔH°vap for H2O = 40.7 kJ/mol = 40700 J/mol
R = Gas constant = 8.314 J/mol K
From Claussius Clapeyron equation
T₁ = 347.996 K = 74.86°C
∴Water will boil at 74.86°C
Answer:
c. 12,500
Explanation:
Original number of atoms = 100,000 atoms
Half- life = 10min
Unknown:
The number of atoms that will remain after 10min = ?
Solution:
The half - life is the time taken for half of a radioactive substance to decay by half.
Time taken Number of atom half life
10min 100000 _
20min 50000 1
30min 25000 2
40min 12500 3
