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elena-s [515]
3 years ago
15

Air rushing over the wings of high-performance race cars generates unwanted horizontal air resistance but also causes a vertical

downforce, which helps cars hug the track more securely. The coefficient of static friction between the track and the tires of a 678-kg car is 0.843. What is the magnitude of the maximum acceleration at which the car can speed up without its tires slipping when a 3620-N downforce and an 1270-N horizontal air resistance force act on it?
Physics
1 answer:
Nookie1986 [14]3 years ago
5 0

Answer:

10.897 m/s²

Explanation:

f_s = Slipping force

F_d = Downforce = 3620 N

\mu = Coefficient of static friction = 0.843

m = Mass of car = 678 kg

f_h = Horizontal force = 1270 N

g = Acceleration due to gravity = 9.81 m/s²

a=\frac{f_s-f_h}{m}\\\Rightarrow a=\frac{\mu(F_D+mg)-f_h}{m}\\\Rightarrow a=\frac{0.843(3620+678\times 9.81)-1270}{678}\\\Rightarrow a=10.897\ m/s^2

Hence, magnitude of the maximum acceleration is 10.897 m/s²

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A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her orig
atroni [7]

Answer:

The plane is 2353.7 mi from the starting position.

Explanation:

Please, see the attached figure for a graphic representation of the problem.

We have 2 displacement vectors "a" and "b" and a vector "c" that is the sum of vectors "a" plus "b" (c = a + b). The module of "c" will be the distance of the plane from the starting point.

vector a = (xa, ya)

vector b = (xb, yb)

where “xa” and “xb” are the horizontal components of the vectors “a” and “b” respectively and “ya” and “yb” are the vertical components of each vector.

Then, the vector c = a + b will be:

c = (xa + xb, ya + yb)

The module of a vector is calculated using the following expression for a vector “v”:

module of v = \sqrt{x^{2} + y^{2} }

Then, the module of c will be:

module of c = \sqrt{(xa + xb)^{2} + (ya + yb)^{2}} = distance from starting point

Then, we have to find the components of vectors “a” and “b”

The distance traveled during the first 1.5 hours of the trip is the module of the vector “a”. Then:

module of a = \sqrt{xa^{2} + ya^{2} } = distance traveled during the first 1.5 hours.

The distance can be calculated using the equation of the position of an object moving in a straight line at constant speed:

x = x0 + v * t

where

x = position at time t

x0 = initial position

v = speed

t = time

Considering x0 as the starting point (x0 = 0)

x = 675 mi/h * 1.5 h = 1012.5 mi

Then:

module of a = \sqrt{xa^{2} + ya^{2} } = 1012. 5 mi

Since the plane moves only on the horizontal (see figure), the "y" component of the vector, "ya", will be 0.

Then:

(1012.5 mi)² = xa²

xa = 1012. 5mi

a = (1012.5 mi, 0)

In the same way, we have fo find the components of the vector “b”. The module of “b” will be the distance traveled during this part of the flight:

module of b = \sqrt{xb^{2} + yb^{2} } = x = x0 + v * t

Considering x0 as the point at which the plane turns (x0 = 0)

x = 675 mi / h * 2 h = 1350 mi

Using trigonometry, we can calculate xb and yb (see figure):

sin angle = opposite / hypotenuse

cos angle = adjacent / hypotenuse

In this case:

opposite = yb

adjacent = xb

hypotenuse = module of “b”

Then:

sin 10° = yb / module of “b”

sin 10° * module of “b” = yb

In the same way:

cos 10° * module of “b” = xb

Since module of “b” = 1350 mi

xb = 1329.5 mi

yb = 234.4 mi

b = (1329.5 mi, 234.4 mi)

The vector c = a+b can now be calculated:

c = (xa + xb, ya + yb)

c =(1012.5 mi + 1329.5 mi, 0 mi + 234.4 mi) = (2342 mi, 234.4 mi)

The module of c will be:

module of c = \sqrt{(2342 mi)^{2} + (234.4 mi)^{2} } = 2353.7 mi

The plane is 2353.7 mi from the starting position.

4 0
3 years ago
The tired of a car support the weight of a stationary. If onetire has a slow leak, the air pressure within the tire will_____wit
Westkost [7]

Answer:

d) Decrease, increase, remain constant

Explanation:

If one tire has a slow leak, the air pressure within the tire will_DECREASE____with time due to outflow of air , the surface area between the tire and the road will__INCREASE__in time,due to flattening of tire.

The net force the tire exerts on the road will_REMAIN CONSTANT____in time. It is so because force does not depend upon area. It is pressure which depends upon area. As there is no change in the weight of the car , force on the road will remain constant.

7 0
3 years ago
804 n of force are applied to a 51.7 kg. What is the acceleration that the object experiences?
Andreyy89

We can use Newton II here  (where F=m*a), that F is the net (or resultant) force on the object, m is the mass of the object and a is the acceleration the object experiences.

This means, in this case there would be no friction and absolutely no other force which gives a component in the plane of motion, only then can you assume that F=804N.

Now using F= m*a

804 = 51.7*a

Therefore a = 804/51.7 = 15.55 m/s²


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X=(150^2•sin(2•42))/9.8
amid [387]

Answer:

2283.3410863

Explanation:

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3 years ago
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