The rate of change of atmospheric pressure with respect to altitude is proportional to the current pressure. Using this information, we can calculate the pressure at different altitudes.
To solve this problem, we can use the fact that the rate of change of atmospheric pressure with respect to altitude is proportional to the current pressure. We can set up a proportion using the given information to find the constant of proportionality. Then, we can use this constant to find the pressure at different altitudes.
(a) Let's use the given information to find the constant of proportionality. We have P = kP, where k is the constant of proportionality. Using the values at sea level and 1000m, we can set up the proportion 102.1/87.8 = k. Solving for k, we find k ≈ 1.16.
Now, we can use this constant to find the pressure at an altitude of 4500m. We set up the proportion 102.1/x = 1.16, where x is the pressure at 4500m. Solving for x, we find x ≈ 122.0 kPa.
(b) We can use the same constant of proportionality to find the pressure at the top of a mountain that is 6165m high. We set up the proportion 102.1/x = 1.16, where x is the pressure at the top of the mountain. Solving for x, we find x ≈ 89.2 kPa.
#SPJ3
Answer:
her initial frequency is 445 Hz
Explanation:
Given;
initial beat frequency, = 5
observed frequency, F = 440 Hz
let the initial frequency = F₁
F₁ = F ± 5 Hz
F₁ = 440 Hz ± 5 Hz
F₁ = 435 or 445 Hz
This result obtained shows that her initial frequency can either be 435 Hz or 445 Hz
The last beat frequency will be used to determine the actual initial frequency.
F = v/λ
Frequency (F) is inversely proportional to wavelength. That is an increase in length will cause a proportional decrease in frequency.
This shows that the final frequency is smaller than the initial frequency because of the increase in length.
Initial frequency - frequency of tuning fork = 5 beat frequency
Reduced initial frequency - frequency of tuning fork = 3 beat frequency
Initial frequency = 5Hz + 440 Hz = 445 Hz
Final frequency (Reduced initial frequency) = 440 + 3 = 443 Hz
Check: 445 Hz - 440 Hz = 5 Hz
443 Hz - 440 Hz = 3 Hz
Answer:
Work done W =1406.25 J
Explanation:
Work done on a body can be calculated using newton's 2nd laws:
F=ma
Hence acceleration of the block is given by:
Displacement of the object is given by:
Substitute the values
Now work done is given by:
W=F.S
W = 25×56.25
W =1406.25 J
Answer:
Time taken for 1 swing = 3.81 second
Explanation:
Given:
Time taken for 1 swing = 2.20 Sec
Find:
Time taken for 1 swing , when triple the length(T2)
Computation:
Time taken for 1 swing = 2π[√l/g]
2.20 = 2π[√l/g].......Eq1
Time taken for 1 swing , when triple the length (3L)
Time taken for 1 swing = 2π[√3l/g].......Eq2
Squaring and dividing the eq(1) by (2)
4.84 / T2² = 1 / 3
T2 = 3.81 second
Time taken for 1 swing = 3.81 second
Answer:
the answer is 3.3 %
Explanation:
Answer:
1838216 J
Explanation:
95 km/h = 26.39 m/s
40 km/h = 11.11 m/s
Initial kinetic energy
= .5 x 1600 x(26.39)²
= 557145.67 J
Final kinetic energy
= .5 x 1600 x ( 11.11)²
= 98745.68 J
Loss of kinetic energy
= 458400 J
Loss of potential energy
= mg x loss of height
= 1600 x 9.8 x 340 sin 15
= 1379816 J
Sum of Loss of potential energy and Loss of kinetic energy
= 1379816 + 458400
= 1838216 J
This is the work done by the friction . So this is heat generated.
To calculate the thermal energy dissipated from the brakes of a car, use the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill. The temperature change of the brakes can then be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.
The thermal energy dissipated from the brakes of a car can be calculated by converting the gravitational potential energy lost by the car into internal energy of the brakes. By using the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill, we can calculate the thermal energy dissipated. From there, the temperature change of the brakes can be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.
#SPJ11