The bending of a wave as it moves from one medium to another is called

Vinvika [58]

The answer to the question is choice 2

7 0

3 years ago

A 2.2-m-long steel rod must not stretch more than 1.2 mm when it is subjected to a 8.5-kN tension force. Knowing that E = 200 GP

kodGreya [7K]

Answer:

a) 0.00996 m

b) 109090909 Pa

Explanation:

Unit conversions:

$E = 200GPa = 200\times10^9 Pa$

1.2 mm = 0.0012 m

8.5 kN = 8500 N

If the 2.2m rod cannot stretch more than 0.0012 m, its maximum strain is

$\epsilon = \frac{\Delta L}{L} = \frac{0.0012}{2.2} = 0.000545455$

With elastic modulus being E = 200 GPa, then its maximum stress must be

$\sigma = E\epsilon = 200\times10^9*0.000545455 = 109090909 Pa$

Knowing the tension force being F = 8500 N, we can calculate the appropriate cross section area

$A = \frac{F}{\sigma} = \frac{8500}{109090909} = 7.79\times10^{-5}m^2$

And its corresponding diameter is

$A = \pi d^2/4$

$7.79\times10^{-5} = \pi d^2/4$

$d^2 = \frac{4*7.79\times10^{-5}}{\pi} = 9.92\times10^{-5}$

$d = \sqrt{9.92\times10^{-5}} = 0.00996 m \approx 1 cm$

7 0

3 years ago

A mechanic jacks up the front of a car to an angle of 8.0° with the horizontal in order to change the front tires. the car is 3.

Zina [86]

First draw a free-body diagram. Torque T = Force F x Distance d where force is the component of gravitational force g and d is the lever arm distance to the pivot point. Since the pivot point is at the back tire we subtract that from the length of the car resulting in d = 1.12 - 0.40 = 0.72 meters = d. We are interested in the perpendicular component of the force exerted on the car jack so use sin 8 degrees then T=1130 kg x 9.81 m/s^2 x sin(8 degrees) x0.72 m = 1,110.80 Newton-meters

5 0

4 years ago

Displacement is used to measure

GenaCL600 [577]

Answer:

D. Volume

Hope this helped <3

6 0

3 years ago

Read 2 more answers

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly

Mnenie [13.5K]

Answer:

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The acceleration of the tortoise is 0.28 m/s²

Explanation:

The equations that describe the position and velocity of the hare and the tortoise are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

To find the acceleration of the hare once it begins to slow down, we have to find how much time the hare traveled during the deceleration and what was its initial speed.

First, the hare moves with constant acceleration for 4.7 s. Then, its velocity at t = 4.7 s will be:

v = v0 + a · t (v0 = 0 because the hare starts form rest)

v = a · t = 0.9 m/s² · 4.7 s = 4.2 m/s

The distance traveled by the hare while accelerating can be calculated using the equation of the position:

x = x0 + v0 · t + 1/2 · a · t² (x0 = 0 and v0 = 0)

x = 1/2 · a · t² = 1/2 · 0.9 m/s² · (4.7)² = 9.9 m

Then, the hare runs at a constant speed of 4.2 m/s for 11.7 s. The distance traveled at constant speed will be:

x = v · t

x = 4.2 m/s · 11.7 s = 49.1 m

Then, the distance traveled by the hare while slowing down was:

Distance traveled while slowing down = 72 m - 49.1 m - 9.9 m = 13 m

Let´s find how much time it took the hare to come to stop, so we can calculate the acceleration. We know that when the position is 13 m, the velocity is 0.

v = v0 + a · t

0 = 4.2 m/s + a · t

-4.2 m/s / t = a

Replacing in the equation of the position:

x = v0 · t + 1/2 · a · t² (considering x0 as the point at which the hare started to slow down)

13 m = 4.2 m/s · t - 1/2 · 4.2 m/s / t · t²

13 m = 4.2 m/s · t - 2.1 m/s · t

13 m = 2.1 m/s · t

t = 13 m / 2.1 m/s

t = 6.2 s

Then, the acceleration of the hare while slowing down will be:

-v0/t = a

-4.2 m/s / 6.2 s = a

a = -0.68 m/s²

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The hare traveled 72 m in (6.2 s + 11.7 s + 4.7 s) 22.6 s. The tortoise reaches the final position of the hare at the same time, so, using the equation of the position we can calculate the acceleration of the tortoise:

x = x0 + v0 · t + 1/2 · a · t² (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

72 m = 1/2 · a · (22.6 s)²

144 m / (22.6 s)² = a

a = 0.28 m/s²

The acceleration of the tortoise is 0.28 m/s²

6 0

3 years ago