Question
A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for the tangent of the highway's angle of banking. Give your equation in terms of the radius of curvature r, the intended speed of the turn v, and the acceleration due to gravity g
Part (b) what is the angle of banking of the highway? Give your answer in degrees
Answer:
a. Equation of Tangent
tan(θ) = v²/rg
b. Angle of the banking highway
θ = 0.087°
Explanation:
Given
Radius of the curve = r = 330m
Acceleration of gravity = g = 9.8m/s²
Velocity = v = 8km/h = 8 * 1000/3600
v = 2.22 m/s
a . Write an equation for the tangent of the highway's angle of banking
The Angle is calculated by
tan(θ) = v²/rg
θ = tan-1(v²/rg)
b.
Part (b) what is the angle of banking of the highway? Give your answer in degrees
θ = tan-1(v²/rg)
Substituting the values of v,g and r
θ = tan-1(2.22²/(330 * 9.8)
θ = tan-1(0.001523933209647)
θ = 0.087314873580116°
θ = 0.087°
Periods<span> going left to right. The periodic table also has a special name </span>for<span> its vertical columns. Each column </span>is<span> called a </span>group. The elements in eachgroup<span> have the same number </span>of<span> electrons </span>in the<span> outer orbital.</span>
Answer:
Cart A
Explanation:
Momentum can be computed by finding the product of mass and velocity. To solve this, you can use the formula below to find the greatest momentum:
p = mv
where:
p = momentum (kgm/s) m = mass (kg) v = velocity (m/s)
Because carts are moving along with the weights, we need to consider the whole system. This means that you need to add in the masses and the mass of the cart.
<u>Cart A:</u>
m = 200kg + 0 kg = 200 kg
v = 4.8 m/s
p = 200kg x 4.8 m/s = 960 kg-m/s
<u>Cart B:</u>
m = 200kg + 20 kg = 220 kg
v = 4.0 m/s
p = 220kg x 4.0 m/s = 880 kg-m/s
<u>Cart C:</u>
m = 200kg + 40 kg = 240 kg
v = 3.8 m/s
p = 240kg x 3.8 m/s = 912 kg-m/s
<u>Cart D:</u>
m = 200kg + 60 kg = 260 kg
v = 3.5 m/s
p = 260kg x 3.5 m/s = 910 kg-m/s
As you can see, Cart A has the greatest momentum.
Answer:
<u>I had to search the Figure on Google to solve this question.</u>
a) The magnitude of the force F₃ is:
And the direction of F₃:
(with respect to the y-direction, in the third quadrant)
b) P = 4.22 N
Explanation:
<u>I had to search the Figure on Google to solve this question.</u>
a) We can find the force of the third person as follows:
So, in x-direction we have:
In y-direction we have:
The magnitude of the force F₃ is:
To find the direction of F₃ we need to calculate its angle with respect to the y-direction (in the third quadrant):
<em>b) If the third person exerts the force found in part (a) the car will stop, so the only way for the cart to accelerate at 200 m/s² is that the third person does not exert the force found in a. </em>
<u>To find the weight of the cart when it accelerates at 200 m/s², we need to consider: F₃ = 0</u>.
First, we need to find the cart's mass. Since the car is moving in the x-direction we have:
Now, the weight of the cart is:
I hope it helps you!