Answer:
232.641374 mph
Explanation:
A race car has a maximum speed of 0.104km/s
Let X represent the speed in miles per hour
Therefore the speed in miles per hour can be calculated as follows
1 km/s = 2,236.936292 mph
0.104km/s = X
X = 0.104 × 2,236.936292
X = 232.641374
Hence the speed in miles per hour is 232.641374 mph
Answer:
24.71 mm
Explanation:
Distance is proportional to focal length, so
d∝f
which means

Magnification of first lens

and

Similarly, magnification of second lens

and

From the above equations we get

and

which means,

and

So, we get

∴ Focal length should this camera's lens is 24.71 mm
Answer:
h = v₀² / 2g
, h = k/4g x²
Explanation:
In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches
Starting point. Lower compressed spring
Em₀ = K = ½ m v²
Final point. Highest on the path
= U = mg h
As or no friction the energy is conserved
Em₀ = Em_{f}
½ m v₀²² = m g h
h = v₀² / 2g
We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse
½ k x² = 2 g h
h = k/4g x²
Answer: 0.204 s
Explanation:
The speed of sound
is defined as the distance traveled
in a especific time
:
Where:
is the speed of sound in seawater
is the time the sound wave travels from the dolphin and then returns after the reflection
is twice the distance between the dolphin and the object to which the sound waves are reflected
Finding
:
<u>Finally:</u>
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