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Tatiana [17]
3 years ago
10

A race car travels on a circular track at an average rate of 125 mi/h. The radius of the track is 0.320 miles. What is the centr

ipetal acceleration of the car?
Physics
1 answer:
GREYUIT [131]3 years ago
4 0
The centripetal acceleration of the car is 48,800 mi/h2
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A car of mass 1500 kg is negotiating a flat circular curve of radius 50 m with a speed of 20 m/s.
Lilit [14]

Answer:

Explanation:

a. The source of centripetal force on the car is  (3) the static friction force.

b. ac = v²/R = (20²)/50 = 8 m/s²

c.  Fc = m(ac) = 1500(8) = 12 kN

d. μ = Fc/N = Fc/mg = 12000 / 1500(9.8) = 0.8163... ≈ 0.82

6 0
2 years ago
considere que o calor específico de um material presente nas cinzas seja c=0,8j/gc. Supondo que esse material entre na turbina a
drek231 [11]

Answer:

3120J

Explanation:

Given parameters:

C  = Specific heat capacity  = 0.8J/g°C

Initial temperature  = 20°C

Mass given   = 5g

Final temperature  = 800°C

Unknown:

Energy given to the mass  = ?

Solution:

To find the energy given to the mass, let us simply use the expression below:

          H   =   m   c   ΔT

H is the unknown, the energy supplied

m is the mass of the substance

c is the specific heat capacity

ΔT is the change in temperature

Input the variables;

            H    = 5  x   0.8    x    (800 - 20)  = 3120J

7 0
3 years ago
Does air resistence decrease with speed
dybincka [34]

Answer:

yes

Explanation:

because when you slow down, the resistance slows with the speed.

3 0
3 years ago
Read 2 more answers
Jeff is investigating factors that affect the growth rate of potted bean plants. Which of the following experimental variables w
Kipish [7]
It would be C. the color of the pot. its pretty obvious that i would not effect the project.

8 0
2 years ago
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Inserting the formulas you found for Xman(t) and Xbus(t) into the conditionXman(tcatch)=Xbus(tcatch) , you obtain the following-
lara [203]

Answer:

c > √(2ab)

Explanation:

In this exercise we are asked to find the condition for c in such a way that the results have been real

The given equation is

              ½ a t² - c t + b = 0

we can see that this is a quadratic equation whose solution is

             t = [c ±√(c² - 4 (½ a) b)]  / 2

for the results to be real, the square root must be real, so the radicand must be greater than zero

              c² -2a b > 0

              c > √(2ab)

3 0
2 years ago
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