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beks73 [17]
2 years ago
10

The curved section of a speedway is a circular arc having a radius of 190 m. this curve is properly banked for racecars moving a

t 34 m/s. at what angle with the horizontal is the curved part of the speedway banked?
Physics
1 answer:
Anestetic [448]2 years ago
8 0

The banking angle of the curved part of the speedway is determined as 32⁰.

<h3>Banking angle of the curved road</h3>

The banking angle of the curved part of the speedway is calculated as follows;

V(max) = √(rg tanθ)

where;

  • r is radius of the path
  • g is acceleration due to gravity

V² = rg tanθ

tanθ = V²/rg

tanθ = (34²)/(190 x 9.8)

tanθ = 0.62

θ = arc tan(0.62)

θ = 31.8

θ ≈ 32⁰

Learn more about banking angle here: brainly.com/question/8169892

#SPJ1

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Infrared radiation has wavelengths ranging from about 800 nm to 1 mm. what is the frequency of radiation of wavelength 960 nm? a
Xelga [282]
The correct answer to your question and how to solve it is
 
The relation between wavelength (λ)and the frequency of electromagnetic oscillation (f) is described by the following expression: λ=c/f, where c–is the speed of light in vacuum = 3*10^8 m/s
Derive f from above: f = c/λ.How to Calculate: λ=890nm = 890*10^-9m = 8.9*10^-7m
f =3*10^8m/s Divided by 8.9*10^-7m = 0.34*10^15 s-1=3.4*10^14 s-1
So your Answer is: The frequency of radiation of wavelength 890 nm is 3.4*10^14s-1
3 0
3 years ago
A fluid is a substance that can _____________ and takes the _____________ of its container.
mezya [45]
Evaporate
Shape

I wish there were provided answer choices. Hope this helped
7 0
3 years ago
A hot-air balloon consists of a basket hanging beneath a large envelope filled with hot air. A typical hot-air balloon has a tot
Deffense [45]

Answer:

1.05045 kg/m³

Explanation:

\rho_a = Density of air = 1.26 kg/m³

\rho_{ha} = Density of hot air

m_a = Mass of balloon = 539 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Volume of air in balloon = 2.62\times 10^3\ m^3

The net force on the balloon will be

F_n=(\rho_a-\rho_{ha})vg

Also

F_n=m_ag

\\\Rightarrow 549g=(\rho_a-\rho_{ha})vg\\\Rightarrow 549=(1.26-\rho_{ha})2.62\times 10^3\\\Rightarrow -\rho_{ha}=\frac{549}{2.62\times 10^3}-1.26\\\Rightarrow \rho_{ha}=1.05045\ kg/m^3

The density of hot air inside the envelope is 1.05045 kg/m³

4 0
3 years ago
At the end of a race a runner decelerates from a velocity of 11.00 m/s at a rate of 0.300 m/s2. (a) how far does she travel in t
inessss [21]

given that initial velocity is

v_i = 11 m/s

deceleration is given as

a = -0.3 m/s^2

now we have to find the distance covered in 16 s

d = v_i * t + \frac{1}{2} at^2

d = 11*16 - \frac{1}{2}*0.3*16^2

d = 137.6 m

so it will cover 137.6 m distance

part b)

in order to find the final speed

v_f = v_i + at

v_f = 11 - 0.3*16

v_f = 6.2 m/s

so its speed will be 6.2 m/s

8 0
3 years ago
There are two identical small metal spheres with charges 38.9 µC and −27.6399 µC. Thedistance between them is 6 cm. The spheres
Greeley [361]

Answer:

2683.3N

Explanation:

According to coulombs law which states that "the force of attraction existing between two charge q1 and q2 is directly proportional to the product of the charges and inversely proportional to the square of the distance (d) between them. Mathematically |F|= k|q1| |q2| /d² where;

F is the force of attraction between the charges

q1 and q2 are the charges

d is the distance between them

k is the coulombs constant

Given |q1|= 38.9 × 10^-6C and |q2| = 27.6399×10^-6C d = 6cm = 0.06m

k = 8.98755 × 109 Nm² /C²

Substituting the given data's in the equation we have;

|F| = 8.98755 × 10^9×38.9×10^-6×27.6399×10^-6/0.06²

|F| = 9.66/0.06²

|F| = 9.66/0.0036

|F| = 2683.3N

The magnitude of the force will be 2683.3N

Note that the modulus of the charges changes negative value of q2 to positive value. The opposite signs of the charges doesn't affect the final calculation, it only tells the force of attraction or repulsion between the charges. Since they are unlike charges, they will attract each other in the field.

4 0
3 years ago
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