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morpeh [17]
3 years ago
15

A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate

r at the bottom of the well 5.00 seconds after being dropped from the tower. Calculate the depth of the well. Given: g = -9.81 meters/second2
Physics
1 answer:
lilavasa [31]3 years ago
7 0

Take the stone's position at ground level to be the origin, and the downward direction to be negative. Then its position in the air y at time t is given by

y=100\,\mathrm m-\dfrac g2t^2

Let d be the depth of the well. The stone hits the bottom of the well after 5.00 s, so that

-d=100\,\mathrm m-\dfrac g2(5.00\,\mathrm s)^2\implies d=\boxed{22.6\,\mathrm m}

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Find the radius of the circle formed by the intersection of a sphere of diameter 26 units and a plane that is 5 units away from
horrorfan [7]

Answer:

12 units

Explanation:

This problem can be solved if we take into account the equation for a sphere

x^{2}+y^{2}+z^{2}=r^{2}\\x^{2}+y^{2}+z^{2}=(\frac{26}{2})^{2}=13^{2}

where we took that the radius is 13 units. If we take z=5 and we replace this value in the equation of the sphere we have

x^{2}+y^{2}+(5)^{5}=13^{2}\\x^{2}+y^{2}=144=(12)^{2}

where we have taken x2 +y2 because if the equation of a circunference.

In this case the intersection is made when we take z=5, for this value the sphere and the plane coincides in values.

Hence, the radius is 12 units

I hope this is useful for you

regards

8 0
2 years ago
Rearrange the equation to solve for the permeability of free space (μ0). Remember that the slope is the ratio of magnetic field
IRINA_888 [86]

Complete Question

The complete question is shown on the first uploaded image

Compare the percent error between your value and the accepted value of 1.26 times 10-6 T · m/A. (Use the accepted value given to three significant figures in your calculation.). 100% % error = %

Answer:

The permeability of free space is \mu_0 = 1.32*10^{-6} \ Tm/A

The percentage error is  % error = 5.25%

Explanation:

From the question we are told that

            The slope is  s= 3.9\  G/A , and

Generally  1 \ gauss =  10^{-4 } tesla

               So  s = 3.9 *10^ {-4} T /A

From the relation given in the question

                        \mu_0 = \frac{2R}{N} [\frac{B}{I} ]

 Where R is the radius of the coil  =\frac{Diameter \ of \ coil }{2} = 0.017m

             N is the number of loops of the coil = 10

Now from the question we are told that

                  s = \frac{B}{I}

substituting into the equation above

               \mu_0 = \frac{2R }{N} s

Substituting values

              \mu_0= \frac{2 * 0.017}{10 } * 3.9*10^{-4}

                   = 1.326 *10^ {-6} \ Tm/A

Generally the % error is mathematically represented as

                    %error = \frac{Measured \  value - accepted \ value}{accepted \  value } *100

Given that the accepted value is \mu_o_ {acc} = 1.26 *10 ^{-6} \ T \cdot m /A

      Hence substituting values

                      %error  = \frac{(1.326-1.26)*10^{-6}}{1.26 *10 ^ {-6}} *100

                                  = 5.24

6 0
3 years ago
Which of these is not an even locomotor movement? Question 3 options:
antiseptic1488 [7]

Answer:

skipping

Explanation:

your welcome :)

7 0
3 years ago
Read 2 more answers
Find the useful power output (in W) of an elevator motor that lifts a 2600 kg load a height of 30.0 m in 12.0 s, if it also incr
Annette [7]

Answer:

P = 251,916.667 W

Cost = 2,267.25 cents

Explanation:

To solve this question we will use the Work Energy Theorem, which is

W = dP + dK\\

Where

dP = Change in Potential Energy

dK = Change in Kinetic Energy

Change in Potential Energy

P_{i} = mgh_{i}\\  P_{f} = mgh_{f}

Where

P_{i} = Initial Potential Energy

P_{f} = Final Potential Energy

m = Mass of System = 10,000 kg

g = Acceleration due to gravity = 9.81 m/s

h_{i} = Initial Height = 0

h_{f} = Final Height = 30 m

Inputting the values we get the answer for dP

dP = P_{f} - P_{i}\\dP= mgh_{f} - mgh_{i}\\ dP= 10000(9.81)(30) - 0\\ dP= 2943000

Change in Kinetic Energy

K_{i} = \frac{1}{2} mv_{i} ^2\\ K_{f} = \frac{1}{2} mv_{f} ^2

Where

K_{i} = Initial Kinetic Energy

K_{f} = Final Kinetic Energy

m = Mass of System = 10,000 kg

g = Acceleration due to gravity = 9.81 m/s

v_{i} = Initial Velocity = 0 m/2

v_{f} = Final Velocity = 4 m/s

Inputting the values we get the answer for dK

dK = K_{f} - K_{i}\\ dK = \frac{1}{2} mv_{f} ^2 - \frac{1}{2} mv_{i} ^2\\ dK = \frac{1}{2} (10000)(4)^2 - 0 \\ dK = 80000

Total Work

W = dP + dK\\

Inputting the values

W = 2943000 + 80000

W = 3,023,000

a) Finding the useful Power Output

P = \frac{W}{t}

Where

P = Power Output

W = Work Done = 3,023,000J

t = Time = 12s

Inputting the values

P = \frac{3,023,000}{12}\\ P = 251,916.667

P = 251,916.667 W

b) Finding the Total Cost

Cost = $0.0900 x P/1000

Cost = $0.0900 x (251,916.667/1000)

Cost = $22.67 or 2,267.25 cents

4 0
3 years ago
Pls help with this problem, it's not too hard
ehidna [41]

Answer:

.25 Hz

Explanation:

Period = 4 sec

freq = 1/p = 1/4 hz  = .25 Hz

4 0
2 years ago
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