Answer:
75.6J
Explanation:
Hi!
To solve this problem we must use the first law of thermodynamics that states that the heat required to heat the air is the difference between the energy levels of the air when it enters and when it leaves the body,
Given the above we have the following equation.
Q=(m)(h2)-(m)(h1)
where
m=mass=1.3×10−3kg.
h2= entalpy at 37C
h1= entalpy at -20C
Q=m(h2-h1)
remember that the enthalpy differences for the air can approximate the specific heat multiplied by the temperature difference
Q=mCp(T2-T1)
Cp= specific heat of air = 1020 J/kg⋅K
Q=(1.3×10−3)(1020)(37-(-20))=75.6J
Answer:
No. of laps of Hannah are 7 (approx).
Solution:
According to the question:
The total distance to be covered, D = 5000 m
The distance for each lap, x = 400 m
Time taken by Kara, 
Time taken by Hannah, 
Now, the speed of Kara and Hannah can be calculated respectively as:


Time taken in each lap is given by:



t = 500 s
So, Distance covered by Hannah in 't' sec is given by:


No. of laps taken by Hannah when she passes Kara:

≈ 7 laps
Answer:
: Rocket weight on earth
: Rocket weight on moon
Explanation:
Conceptual analysis
Weight is the force with which a body is attracted due to the action of gravity and is calculated using the following formula:
W = m × g Formula (1)
W: weight
m: mass
g: acceleration due to gravity
The mass of a body on the moon is equal to the mass of a body on the earth
The acceleration due to gravity on a body is different on the moon and on the earth
Equivalences
1 slug = 14.59 kg
Known data



Problem development
To calculate the weight of the rocket on the moon and on earth we replace the data in formula (1):
: Rocket weight on earth
: Rocket weight on moon
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
Answer:
15m
Explanation:
Hello! first to solve this problem we must find that so much that the tree grew in the 15 years this is achieved by dividing the height of the tree before and after

the tree grew 10 times its initial length in 15 years, so to find how tall the nail is, we multiply this factor by 1.5m
X=(1.5m)(10)=15m
the nail is 15 meters above ground level