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seropon [69]
3 years ago
15

A car of mass 487 kg travels around a flat, circular race track of radius 53.3 m. The coefficient of static friction between the

wheels and the track is 0.19. The acceleration of gravity is 9.8 m/s 2 . What is the maximum speed v that the car can go without flying off the track? Answer in units of m/s.
Physics
1 answer:
aleksklad [387]3 years ago
8 0

Answer:

9.96 m/s

Explanation:

mass of car, m = 487 kg

radius of track, R = 53.3 m

coefficient of static friction, μ = 0.19

acceleration due to gravity, g = 9.8 m/s^2

let v be the maximum speed so that the car can go without flying off the track.

The formula for the maximum speed is given by

v_{max}=\sqrt{\mu Rg}

v_{max}=\sqrt{0.19\times53.3\times9.8

vmax = 9.96 m/s

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There is a species of bamboo that can grow 36 inches per day. If a plant grew at this rate and was measured at 40 inches initial
LekaFEV [45]

Answer:

It will take the plant 4\frac{4}{9} days or 4.44 days to grow to a height of 200 inches tall.

Explanation:

From the question, the rate at which the species of the bamboo tree grows is 36 inches per day.

To determine how long it would take a plant 40 inches tall initially to grow at this rate (that is, 36 inches per day) to a height of 200 inches.

This means we will calculate the number of days it will take the plant to grow additional 160 inches ( 200 inches - 40 inches) at this rate.

Now,

If the plant grows 36 inches in 1 day

then it will grow 160 inches in x days

x = (160 inches × 1 day) / 36 inches

x = 160 / 36

x = 4\frac{4}{9} days or 4.44 days

Hence, it will take the plant 4\frac{4}{9} days or 4.44 days to grow to a height of 200 inches tall.

8 0
3 years ago
Which of the following is true of high clouds?
Wewaii [24]
I’m pretty sure it’s B
3 0
2 years ago
a tire with inner volume of 0.0250m^3 is filled with air at a gauge pressure of 36.0 psi. If the tire valve is opened to the atm
enyata [817]

Answer: Escaped volume = 0.0612m^3

Explanation:

According to Boyle's law

P1V1 = P2V2

P1 = initial pressure in the tire = 36.0psi + 14.696psi = 50.696psi (guage + atmospheric pressure)

P2 = atmospheric pressure= 14.696psi

V1 = volume of tire =0.025m^3

V2 = escaped volume + V1 ( since air still remain in the tire)

V2 = P1V1/P2

V2 = 50.696×0.025/14.696

V2 = 0.0862m^3

Escaped volume = 0.0862 - 0.025 = 0.0612m^3

5 0
3 years ago
A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force of friction on the cart
Ber [7]
Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:
\sum F = ma
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.

We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force F_f acting in the opposite direction. So Newton's second law can be rewritten as
F-F_a = ma
or
F=ma + F_f

since the frictional force is 15 N and we want to achieve an acceleration of a=1.50 m/s^2, we can substitute these values to find what is the force the man needs:
F=(30 kg)(1.5 m/s^2)+15 N=60 N
8 0
4 years ago
With a force of 5 newtons, Amanda pushes the stack of books to the right. At the same time, Jeremy, her little brother, pushes t
Vaselesa [24]
Its letter C. 5N to the left. Since Jeremy's force in Newtons are higher than Amanda's (in newtons), and since Jeremy's force directs to the left, then the direction of the force will be to the LEFT. Then subtract the higher one to the lower one so that would be: 10N-5N=5N. So it is C. 5N to the left.
5 0
3 years ago
Read 2 more answers
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