the density of dry air at 20 degrees celsius is 1.20 g/L. What is the mass of air, in kilograms, of a room that measures 24.0m b
1 answer:
<h2>Mass of air in a room that measures 24.0 m by 15.0 m by 4.0 m is 1728 kg.</h2>
Explanation:
Density of air = 1.20 g/L =
Size of room is 24.0m by 15.0 m by 4.0 m
Volume of room = 24 x 15 x 4 = 1440 m³
We know the equation
Mass = Volume x Density
Mass = 1440 x 1.2
Mass = 1728 kg
Mass of air in a room that measures 24.0 m by 15.0 m by 4.0 m is 1728 kg.
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Answer:
14m/s
Explanation:
Given parameters:
Radius of the curve = 50m
Centripetal acceleration = 3.92m/s²
Unknown:
Speed needed to keep the car on the curve = ?
Solution:
The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.
It is given as;
a =
a is the centripetal acceleration
v is the speed
r is the radius
Now insert the parameters and find v;
v² = ar
v² = 3.92 x 50 = 196
v = √196 = 14m/s
Answer:
t = 6.09 seconds
Explanation:
Given that,
Speed, v = 44.1 cm/s
Distance, d = 269 cm
We need to find the time interval of the marble. Speed is distance per unit time.
Hence, the time interval of the marble is 6.09 seconds.
Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;
E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C