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musickatia [10]

4 years ago

15

Prof. Kopp is working on an experiment located in an abandoned mine near Duluth, MN. The experiment is located approximately 713

m below the surface. He went to visit the site this last summer and took a harrowing ride in a bumpy, claustrophobic elevator down to the cavern. He timed the ride to last about 90 s. Compute the average speed of the elevator (in mph) as it plummeted to the bottom of the shaft.

1 answer:

MA_775_DIABLO [31]4 years ago

8 0

Answer:

The average speed of the elevator going down in the abandoned mine is 17.722mph.

Explanation:

If the elevator takes 90 seconds to descend a height of 713m, the average speed of the elevator is:

$v_{av}=x_T/t_T=713m/90s=7.922m/s$

And if 1m/s is 2.23694mph, the average speed is:

$v_{av}=7.922m/s=17.722mph$ .

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A girl swings a 0.250 kg rock attached to a taut string in a circle around her head. Her hand holds the end of the string above

nydimaria [60]

Complete Question

The diagram of with this question is shown on the first uploaded image

Answer:

The value is $v = -6.543 \^ i + 9.47 \^ j + 0 \^ k$

Explanation:

From the question we are told that

The mass of the rock is $m = 0.250 \ kg$

The length of the string is $L = 0.75 \ m$

The angle the string makes horizontal is $\theta = 11.9^o$

The angle which the projection of the string onto the xy -plane makes with the positive x-axis is $\phi = 34.6^o$

The angular velocity of the rock is $w = 2.50 rev/s = 2.50 * 2\pi =15.7 \ rad/s$

Generally the radius of the circle made by the length of the string is mathematically represented as

$r = L cos(\theta )$

=> $r = 0.75 cos(11.9 )$

=> $r = 0.734 \ m$

Generally the resultant tangential velocity is mathematically represented as

$v__{R}} = w * r$

=> $v__{R}} = 15.7 *0.734$

=> $v__{R}} = 11.5 \ m/s$

Generally the tangential velocity along the x-axis is

$v_x = -v__{R}} * sin(\phi)$

=> $v_x =- 11.5 * sin(34.6)$

=> $v_x = -6.543 \ m/s$

The negative sign show that the velocity is directed toward the negative x-axis

Generally the tangential velocity along the y-axis is

$v_y = v__{R}} * cos(\phi)$

=> $v_y = 11.5 * cos(34.6)$

=> $v_y = 9.47 \ m/s$

Generally the tangential velocity along the y-axis is

$v_z = v__{R}} * cos(90)$

=> $v_z = 0 \ m/s$

Generally the tangential velocity at that instant is mathematically represented as

$v = -6.543 \^ i + 9.47 \^ j + 0 \^ k$

8 0

3 years ago

A book that weighs 5 newtons is sitting on a table 1.5 meters above the floor. How much gravitational potential energy does the

valina [46]

Answer:7.5J

Explanation:7.5

J

Explanation:

P

E

=

m

g

h

Since

F

=

m

g

in this case

P

E

=

F

h

P

E

=

5

N

⋅

1.5

m

=

7.5

5 0

3 years ago

An automobile is traveling on a long, straight highway at a steady 76.0 mi/h when the driver sees a wreck 180 m ahead. at that i

masha68 [24]

a) The car’s speed just after leaving the icy portion of the road is the first part

We have s = ut + $\frac{1}{2} at^2$

Here s = 130 m, a = -1.2 $m/s^2$ , u = 76 mile/hr = 33.975 m/s

$130 = 0*t - 0.5*(-1.2)*t^2\\ \\ \\ t=14.72 seconds$

After 14.72 seconds cars speed is given by v =u +at

u = 33.975m/s, a = -1.2 $m/s^2$

v = 33.975-1.2*14.72 = 16.31 m/s

So velocity of car on leaving icy surface = 16.31 m/s

b) Time taken after icy surface

0 = 16.31 - 7.10*t

t = 2.23 seconds

Distance traveled during this time

$v^2 = u^2+2as\\ \\ 0 = 16.31^2 - 2*7.10*s\\ \\ s = 18.74 m$

Total distance traveled = 130+18.74 = 148.74 m

c) Total time taken = 14.72+2.23=16.95 seconds

3 0

3 years ago

the velocity of a woman running increases by 5 m/s in 15 seconds. what is the acceleration of the woman

zhannawk [14.2K]

0.33m\s2
hope this helps

4 0

3 years ago

Please help me with my Science homework!! 6-15 (30 points)

andriy [413]

Hello!

6. Answer: B) Apparent magnitude.

7. Answer: A) Absolute magnitude.

8. Answer: C) Fusion begins.

11. Answer: D) Cool, bright stars.

12. Answer: D) Prominences.

13. Answer: A) Clusters.

14. Answer: B) Elliptical.

15. Answer: A) Spectrum.

I really hope this helped you! c:

5 0

3 years ago