Answer:
388.97 nm
Explanation:
The computation of the wavelength of this light in benzene is shown below:
As we know that
n (water) = 1.333
n (benzene) = 1.501
And, the wavelength of water is 438 nm
Now placing these values to the above formula
So,
= 388.97 nm
We simply applied the above formula so that we can easily determine the wavelength of this light in benzene could come
Answer:
a) The linear acceleration of the car is , b) The tires did 7.46 revolutions in 2.50 seconds from rest.
Explanation:
a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:
Where:
- Magnitude of the radial acceleration, measured in meters per square second.
- Magnitude of the tangent acceleration, measured in meters per square second.
Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:
Where:
- Angular acceleration, measured in radians per square second.
- Radius of rotation (Radius of a tire), measured in meters.
Given that and
. The linear acceleration experimented by the car is:
The linear acceleration of the car is .
b) Assuming that angular acceleration is constant, the following kinematic equation is used:
Where:
- Final angular position, measured in radians.
- Initial angular position, measured in radians.
- Initial angular speed, measured in radians per second.
- Angular acceleration, measured in radians per square second.
- Time, measured in seconds.
If ,
,
, the final angular position is:
Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)
The tires did 7.46 revolutions in 2.50 seconds from rest.
Answer: the constant angular velocity of the arms is 86.1883 rad/sec
Explanation:
First we calculate the linear velocity of the single sprinkler;
Area of the nozzle = π/4 × d²
given that d = 8mm = 8 × 10⁻³
Area of the nozzle = π/4 × (8 × 10⁻³)²
A = 5.024 × 10⁻⁵ m²
Now total discharge is dived into 4 jets so discharge for single jet will be;
Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec
So using continuity equation ;
Q_single = A × V_single
V_single = Q_single/A
we substitute
V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)
V_single = 29.8566 m/s
Now resolving the forces as shown in the second image,
Vt = Vcos30°
Vt = 29.8566 × cos30°
Vt = 25.8565 m/s
Finally we calculate the angular velocity;
Vt = rω
ω_single = Vt / r
from the given diagram, radius is 300mm = 0.3m
so we substitute
ω_single = 25.8565 / 0.3
ω_single = 86.1883 rad/sec
Therefore the constant angular velocity of the arms is 86.1883 rad/sec
