Answer:
a. slow down
Explanation:
To properly answer this question, you must understand what hydroplaning is. Hydroplaning is essentially when a car begins to slide on a wet road and the driver cannot control the vehicle. This happens because of an overflow in water on the road which causes the tires to begin to skim on the surface of the water. Having a proper amount of tire tread, typically 5/23 inches deep or above, will reduce chances of a car hydroplaning on a wet surface.
* note:
if you would like a better understanding of what hydroplaning looks like, there are videos on the internet you could watch that show real life footage of what it looks like when cars hydroplane. I recommend searching up 'car hydroplaning.'
Now that we fully understand the question, let's examine each option individually and determine what is the best thing to do to prevent hydroplaning.
option a
slow down
When a persons car begins to hydroplane, it may be going so fast that it becomes difficult to control. Therefore, slowing down to reduce the speed of the car would be your best bet to try and regain control.
option b
speed up
Stepping on the accelerator will increase the speed of the car which will cause it continue to skim on the surface of the water. This may cause the driver to lose even more control of the vehicle.
option c
deflate your tires
Deflating your tires would not be a very reliable option, as it can make you lose control even more. It is also not a guarantee that you would even be able to deflate your tires.
option d
use cruise control
If you put your car on cruise control it can punctually cause you to lose more control and cause your car to increase more speed.
Now that we have gone through each option individually, we can determine that option a 'slow down' is the best thing to do to prevent hydroplaning.
To prevent hydroplaning, slow down. Thus, option A is correct.
What is hydroplaning?When ever a vehicle's tyres hydroplane, they lose traction on the road and move along a thin layer of water that is on the surface of the ground.
A person's car may very well be moving so quickly when it starts to hydroplane that it is challenging to control. In the interest of attempting or regain control, slows to lower the car's speed would be their best choice.
One not only has improved sight but driving more slowly will decrease your risk of hydroplaning. This is so because tyres have more stability at normal revs on wet pavement.
The amount of rain, the type of road, and the state of your tyres will all affect how much slower you should travel. Therefore, option A is the correct option.
Learn more about hydroplaning, here:
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When x = 10 ft, the crate has a speed of 20 ft/s which is increasing at 6 ft/s^2. Determine the direction of the crate's velocity and the magnitude of the crate's acceleration at this instant.
Answer:
The direction will be "39.8°". The further explanation is given below.
Explanation:
The equation will be:
⇒ [tex]y=\frac{x^2}{24}[/tex]
On differentiating the above, we get
⇒ [tex]\frac{dy}{dx}=\frac{2x}{24}[/tex]
[tex]=\frac{x}{12}[/tex]
On differentiating again, we get
⇒ [tex]\frac{d^2y}{dx^2}=\frac{1}{12}[/tex]
Demonstrate the radius of the path curvature .
⇒ [tex]\rho=\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\left | \frac{d^2y}{dx^2} \right |}[/tex]
[tex]=\frac{[1+(\frac{x}{12} )^2]^{\frac{3}{2}}}{\left |\frac{1}{12} \right |}[/tex]
[tex]=\frac{[1+(\frac{10}{12})^2]^{\frac{3}{2}}}{\frac{1}{12} }[/tex]
[tex]=26.4 \ ft[/tex]
On calculating the acceleration's normal component, we get
⇒ [tex]a_{n}=\frac{v^2}{\rho}[/tex]
[tex]=\frac{20}{26.4}[/tex]
[tex]=15.15 \ ft/s^2[/tex]
Magnitude,
⇒ [tex]a=\sqrt{a_{n}^2+a_{t}^2}[/tex]
[tex]=\sqrt{(15.15)^2+(6)^2}[/tex]
[tex]=16.29 \ ft/s^2[/tex]
The direction of crate velocity will be:
⇒ [tex]\phi=tan^{-1}(\frac{dy}{dx} )[/tex]
On putting the values, we get
[tex]=tan^{-1}(\frac{x}{12})[/tex]
[tex]=tan^{-1}(\frac{10}{12} )[/tex]
[tex]=39.8^{\circ}[/tex]
How a single force is resolved along the perpendicular axis acting an angle theta with horizontal?
Answer:
A single force, which is acting at angle θ from a horizontal axis, can be resolved into components which act along the perpendicular axis.
Consider the perpendicular axis x and y, where x represents the horizontal axis and y represents vertical axis.
The Force is resolved into 2 parts, one acts along x-axis and is represent by X. The other acts along y-axis and is represented by Y.
From the diagram we can see that the Force and its components X and Y makes up a right angles triangle, where θ is the angle from the x-axis
Find X:We know that:
cosθ = Base/Hypotenuse
cosθ = X/F
X = Fcosθ
Find Y:
We know that:
sinθ = Perpendicular/Hypotenuse
sinθ = Y/F
Y = Fsinθ
Relation of Force and its Components:
Force F can be represent by:
F = Fcosθ (along x-axis) + Fsinθ (along y-axis)
As they form a right angled triangle, we can use Pythagoras Theorem to show the relation between Force and its components.
Hypotenuse² = Base² + Perpendicular²
F² = X² + Y²
F² = (Fcosθ)² + (Fsinθ)²
[tex]F = \sqrt{(F\cos\theta)^2+(Fsin\theta)^2}[/tex]
Where θ can be found by using any of the trignometric functions.
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Engineering Design:Question 10
Which option describes the engineering design process?
Select one:
1. A series of steps engineers use to identify constraints.
2. A series of steps engineers use to decide which solution is best.
3. A series of steps engineers use to test what they build.
4. A series of steps engineers use to solve problems.
Answer:
4. A series of steps engineers use to solve problems.
Explanation:
The process of engineering design is a sequence of procedures that engineers pursue to arrive at a solution to a specific problem. Most times the solution includes creating a product such as a computer code, which fulfills certain conditions or performs a function. If the project in-hand includes designing, constructing, and testing it, then engineers probably adopt the design process. Steps of the process include defining the problem, doing background research, specifying requirements, brainstorming solutions, etc.
BE-40 What is a characteristic of a catamaran hull?
Planes at low speed
Tends to pitch in waves
A stable ride
Only recommended on calm inland waters
Answer:
A stable ride
Explanation:
A Catamaran hull is a form of sea craft invented by the Austronesian peoples, the invention of the Catamaran hull enabled these people to sail across the sea in their expansion to the islands of the Indian and Pacific Oceans. Catamaran has multiple hulls, usually two parallel hulls of equal size. This geometric feature gives the craft an increased stability because, it derives extra stability from its wide beam, in the place of a ballasted keel employed in a regular monohull sailboat. A Catamaran hull will require four times the force needed to capsize it, when compared to an equivalent monohull.
. A series RLC circuit containing a resistance of 12Ω, an inductive reactance of 47.13 Ω and a capacitive reactance of 31.83 Ω are connected in series across a 100V, 50Hz supply. Calculate the
Answer:
a) 19.44 ohm
b) 5.14 A
c) 51.9° lagging
Explanation:
A series RLC circuit containing a resistance of 12Ω, an inductive reactance of 47.13 Ω and a capacitive reactance of 31.83 Ω are connected in series across a 100V, 50Hz supply. Calculate the total circuit impedance, the circuits current, power factor
Given that:
R = 12 Ω, [tex]X_L=47.13\ ohm,\ X_C=31.83 \ ohm[/tex], f = 50 Hz,
A) Total circuit impedance (Z) is given by:
[tex]Z=\sqrt{R^2+(X_L-X_C)^2} =\sqrt{12^2+(47.13^2-31.83)^2} =\sqrt{378.09} =19.44\ ohm[/tex]
B) the circuits current (I) is given by:
[tex]I=\frac{V}{Z}=\frac{100}{19.44}=5.14\ A[/tex]
The voltage across the resistor ([tex]V_R[/tex]) = IR= 5.14 × 12 = 61.68 V
The voltage across the inductor ([tex]V_L[/tex]) = [tex]IX_L[/tex] = 5.14 × 47.13 = 242.25 V
The voltage across the capacitor ([tex]V_c[/tex]) = [tex]IX_C[/tex] = 5.14 × 31.83 = 163.5 V
C) The power factor (Θ) is calculated as:
[tex]cos(\theta)=\frac{R}{Z}\\cos(\theta)=\frac{12}{19.44} =0.619\\\theta=cos^{-1}(0.6172)\\\theta=51.9^o\ laggin[/tex]