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zmey [24]
3 years ago
10

Two banked curves have the same radius. Curve A is banked at 12.7 °, and curve B is banked at an angle of 15.1°. A car can trave

l around curve A without relying on friction at a speed of 19.1 m/s. At what speed can this car travel around curve B without relying on friction?
Physics
1 answer:
Elina [12.6K]3 years ago
6 0

Answer:

20.88 m/s

Explanation:

Curve A:

theta = 12.7, vA = 19.1 m/s

Curve B:

Theta = 15.1 degree

Let the speed is v.

By the use of given formula

tanθ = v^2 / rg

For Curve A

tan 12.7 = (19.1)^2 / r g   ...... (1)

For Curve B

tan 15.1 = v^2 / r g       ......(2)

Divide equation (2) by equation (1), we get

tan 15.1 / tan 12.7  = v^2 / (19.1)^2

0.269 / 0.225 = v^2 / 364.81

v = 20.88 m/s

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Sindrei [870]

Answer:

K = 40N/m

Explanation:

According to Hooke’s law

Energy = 1/2 kx^2

K is the spring constant

x = displacement

From the question

Energy =5J

x = 0.5m

K = ?

Using the above formula,

We have

5 = 1/2 x k x (0.5)^2

5 = 1/2 x k x (0.5 x 0.5)

5 = 1/2 x k x 0.25

5 = 0.25k/ 2

Cross multiply

5 x 2 = 0.25k

10 = 0.25k

Divide both sides by 0.25 to get k

10/0.25 = 0.25k/ 0.25

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3 years ago
Which change in an abiotic factor causes coral bleaching? a. increased salinity b. decreased salinity c. decreased turbidity d.
Sedbober [7]

The answer is <u>"d. increased temperature".</u>


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7 0
3 years ago
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A driver drives a car up a hill. The driver keeps pressing the gas so that the engine increases the force it applies to the car.
iris [78.8K]

Answer:

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6 0
3 years ago
1. Her angular speed increases because by pulling in her arms she creates a net torque in the direction of rotation. 2. Her angu
choli [55]

Answer:

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A stone is thrown vertically upward with a speed of 28.0 m/s how much time is required to reach this height
Solnce55 [7]
A stone is thrown vertically upward with a speed of 17.0 m/s. How fast is it moving when it reaches a height of 11.0 m? How long is required to reach this height?

Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½ at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 17.0m/s, a = -g = -9.81m/s^2, s = 11.0m and we want to know v and t, so from equation (2):

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v = √73.18 = 8.55m/s

now from equation (3):

v = u + at

8.55 = 17.0 – 9.81t

t = (8.55 – 17.0)/(-9.81) = 0.86s
6 0
3 years ago
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