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Oksana_A [137]
4 years ago
8

Tarzan swings on a 30.0m long vine initially inclined at an angle 37 degrees with the vertical. What is his speed at the bottom

of the swing if he starts from rest?
Physics
1 answer:
Nutka1998 [239]4 years ago
7 0
Tarzan is a real n***a fax b
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Find the potential energy associated with a 61-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level. Take the
Natali [406]

Answer:

The potential energy of the hiker is 1.13\times 10^6\ J.

Explanation:

Given that,

Mass of the hiker, m = 61 kg

Height above sea level, h = 1900 m

We need to find the potential energy associated with a 61-kg hiker atop New Hampshire's Mount Washington. The potential energy is given by :

E=mgh

g is the acceleration due to gravity

E=61\ kg\times 9.8\ m/s^2\times 1900\ m\\\\E=1.13\times 10^6\ J

So, the potential energy of the hiker is 1.13\times 10^6\ J. Hence, this is the required solution.

8 0
3 years ago
A solenoid of length 5.0 cm has a cross sectional area of 1.0 cm2 and has 300 uniformly spaced turns. The solenoid is wrapped in
OleMash [197]

The calculated mutual inductance is 8.544 x 10⁻⁵ H.

Two coils have a mutual inductance of 1 henry when emf of 1 volt is induced in coil 1 and when the current flowing through coil 2 is changing at the rate of one ampere per second.

Length of the solenoid= 5.0 cm

Area of cross-section=1.0 cm²

no of spaced turns=300 turns

turns of insulated wire=180 turns

Mutual inductance (M) = μ₀μr N1N2 A/ L

                                     =(4xπx 10⁻⁷) x (6.3 x 10⁻³) x 300 x 180 x 1/ 5

                                     =79.12 x 10⁻¹⁰ x 54000 / 5

                                     =8.544 x 10⁻⁵  H

hence, the mutual inductance is 8.544 x 10⁻⁵ H.

Learn more about Mutual inductance here-

brainly.com/question/14014588

#SPJ4

                                   

4 0
1 year ago
To store stacks of clean plates, a cafeteria uses a closed cart with a spring-loaded shelf inside. Customers can take plates off
ruslelena [56]

The answer for the following problem is mentioned below.

The option for the question is "A" approximately.

  • <u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>

Explanation:

Given:

Spring constant (k) = 240 N/m

amount of the compression (x) = 0.40 m

To calculate:

Elastic potential energy (E)

We know;

<em>According to the formula;</em>

    E = \frac{1}{2} × k × x × x

   <u>E = </u>\frac{1}{2}<u> × k ×(x)²</u>

where;

E represents the elastic potential energy

K represents the spring constant

x represents amount of the compression in the string

So therefore,

Substituting the values in the above formula;

      E = \frac{1}{2} × 240 × (0.40)²

      E =  \frac{1}{2} × 240 × 0.16

      E =  \frac{1}{2} × 38.4

      E = 19.2 J or approximately 20 J

<u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>

5 0
3 years ago
A jogger runs 20 mi West and then 6.0 mi North. Find the magnitude and direction of the resultant displacement.
Black_prince [1.1K]

Answer:

The magnitude of the resultant displacement is 21 mi and its direction is 16.7° north of west

Explanation:

Hi there!

Please see the figure for a better understanding of the problem. The total displacement vector will be the sum of both displacements:

The vector for the first displacement is:

First displacement = (20 mi, 0)

The second displacement:

Second displacement = (0, 6.0 mi)

The resultant displacement will be:

R = (20 mi, 0) + (0, 6.0 mi) = (20 mi + 0, 0 + 6.0 mi) = (20 mi, 6.0 mi)

The magnitude of this vector will be:

|R| = \sqrt{(20 mi)^{2} + (6.0 mi)^{2}} = 21 mi

The magnitude of the vector displacement is 21 mi.

To find the direction of the vector R, we have to apply trigonometry:

In a right triangle the following trigonometric rule applies:

cos θ = adjacent side to the angle/ hypotenuse

In this case:

cos θ = 20 mi / magnitude of R

θ = 16.7°

The direction of the vector is 16.7° north of west.

4 0
3 years ago
HELP ASAP!!! Some bands hate playing in school gyms because sound waves easily reflect off the walls and floor. What could you d
jonny [76]

Answer:

1.) Covering the ceiling and walls with soft perforated boards

2.) Hanging curtains round the hall

3.) Having more opening in the wall.

Explanation:

This is due to what we called Reverberation due to poor acoustic properties.

Reverberation can be reduced by;

1.) Covering the ceiling and walls with soft perforated boards

2.) Hanging curtains round the hall

3.) Having more opening in the wall.

4 0
3 years ago
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