Answer:
The percent error in his estimation is 30%
Explanation:
Percentage error = |UA- UE| \ UE × 100
UA= Actual value = 7 pounds
UE = Expected value = 10 pounds
Percent error of his estimate = ∣7-10∣ / 10 ×100
= ∣-3∣ / 10 × 100
= 30%
From sea level to the top of the atmosphere, a column of air with a 1-m2 cross section weighs 101,000 N.
To find the answer we have to know about the pressure.
<h3>
What is Pressure?</h3>
- The pressure at a point on a surface is the thrust acting per unit area around that point.
- A particular mass of air is contained in a column that rises from the ocean to the top of the atmosphere. Then,
Pa is the atmospheric pressure at the sea level.
- By combining both the equations, we get the weight of air,
Thus, we can conclude that, the weight of air in a column 1-m2 in cross section that extends from sea level to the top of the atmosphere is 101300N.
Learn more about the pressure here:
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ANSWER:
1
a. Kinematics.
b. Waves/Acoustics.
c. Optics/Polarization.
d. Thermal/Conservation of Energy.
e. Electricity and magnetism.
f. Renewable energy
The electric force on the proton is:
F = Eq
F = electric force, E = electric field strength, q = proton charge
The gravitational force on the proton is:
F = mg
F = gravitational force, m = proton mass, g = gravitational acceleration
Since the electric force and gravitational force balance each other out, set their magnitudes equal to each other:
Eq = mg
Given values:
q = 1.60×10⁻¹⁹C, m = 1.67×10⁻²⁷kg, g = 9.81m/s²
Plug in and solve for E:
E(1.60×10⁻¹⁹) = 1.67×10⁻²⁷(9.81)
E = 1.02×10⁻⁷N/C
Answer: 43.01 m
Explanation:
Given that :
Time ball uses in the air = 5.93, time it takes to reach maximum height and return = time of flight(T) = 5.93 s
TIME of flight (T) = 2 * time taken(t)
Where g = a = acceleration due to gravity = 9.8m/s²
S = 0.5at²
S = maximum height
Tjme taken (t) = time of flight / 2 = 5.93/2 = 2.965 s
Hence,
S = 0.5at²
S = 2 × 0.5 × 9.8 × 2.965²
S = 43.0770025
S = 43.01 m
