Answer:
between 10 and 15 percent
Explanation:
How to put your load
- First load the heavy
The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.
In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.
- Hold your load
After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.
- Trailer weight
To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.
GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.
Answer:
A λ = 97.23 nm
, B) λ = 486.2 nm
, C) λ = 53326 nm
Explanation:
With that problem let's use the Bohr model equation for the hydrogen atom
= -k e² /2a₀ 1/n²
For a transition between two states we have
- = -k e² /2a₀ (1/ ² - 1 / n₀²)
Now this energy is given by the Planck equation
E = h f
And the speed of light is
c = λ f
Let's replace
h c / λ = - k e² /2a₀ (1 / ² - 1 / no₀²)
1 / λ = - k e² /2a₀ hc (1 / ² -1 / n₀²)
Where the constants are the Rydberg constant = 1.097 10⁷ m⁻¹
1 / λ = (1 / n₀² - 1 / nf²)
Now we can substitute the given values
Part A
Initial state n₀ = 1 to the final state = 4
1 / λ = 1.097 10⁷ (1/1 - 1/4²)
1 / λ = 1.0284 10⁷ m⁻¹
λ = 9.723 10⁻⁸ m
We reduce to nm
λ = 9.723 10⁻⁸ m (10⁹ nm / 1m)
λ = 97.23 nm
Part B
Initial state n₀ = 2 final state = 4
1 / λ = 1.097 10⁷ (1/2² - 1/4²)
1 / λ = 0.2056 10⁻⁷ m
λ = 486.2 nm
Part C
Initial state n₀ = 3
1 / λ = 1,097 10⁷ (1/3² - 1/4²)
1 / λ = 5.3326 10⁵ m⁻¹
λ = 5.3326 10-5 m
λ = 53326 nm
Answer:
The no of revolutions is 2.032 revolution.
Explanation:
Given that,
Moment of inertia = 0.85 Kgm²
Radius = 170 mm
Force = 32 N
Time = 2s
We need to calculate the angular acceleration
Using formula of torque
Where, F = force
r = radius
I = moment of inertia
Put the value into the formula
We need to calculate the rotational speed
Using equation of angular motion
We need to calculate the angular position
Using equation of angular motion
We need to calculate no of revolutions
Hence, The no of revolutions is 2.032 revolution.
It looks like the diagram is telling you that there is a net force of 8 N + 4 N = 12 N pointed to the right. By Newton's second law, the magnitude of the net force <em>F</em> is equal to the mass <em>m</em> times the acceleration <em>a</em> :
<em>F</em> = <em>m</em> <em>a</em>
12 N = (4 kg) <em>a</em>
<em>a</em> = (12 N) / (4 kg)
<em>a</em> = 3 m/s²
Answer:
The acceleration is - 3 m/s2.
Explanation:
initial speed, u = 54 km/h = 15 m/s
final speed, v = 0
time, t = 5 s
Let the acceleration is a.
use first equation of motion
v = u + at
0 = 15 + a x 5
a = - 3 m/s2