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MatroZZZ [7]
3 years ago
15

What is the potential energy at point e when the total mechanical energy is 4900 for the 100 kg moving at 2 m/ s

Physics
1 answer:
Leto [7]3 years ago
4 0

Answer:

B. 4700 J

Explanation:

Given the following data;

Mechanical energy = 4900J

Mass = 100kg

Velocity = 2m/s

To find the potential energy;

Mechanical energy = kinetic energy + potential energy

First of all, we would determine the kinetic energy of the object;

K.E = ½mv²

K.E = ½*100*2²

K.E = 50*4

K.E = 200 J

Substituting into the equation, we have;

4900 = 200 + P.E

P.E = 4900 - 200

P.E = 4700 Joules

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A box is pulled up a rough ramp that makes an angle of 22 degrees with the horizontal surface. The surface of the ramp is the x-
kifflom [539]

Magnitude of the force  of tension: 139 N

Explanation:

The surface of the ramp here is assumed to be the positive x-direction.

To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.

There are three forces acting along the x-direction:

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  • The force of friction, F_f=14.8 N, acting down along the plane
  • The component of the weight in the x-direction, F_{gx}, acting down along the plane

We know that the magnitude of the weight is

F_g=70.0 N

So its x-component is

F_{gx}=F_g sin \theta =(70.0)(sin 22^{\circ})=26.2 N

The net force along the x-direction can be written as

F_x = F_T-F_f-F_{gx}

And therefore, since the net force is 98 N, we can find the magnitude of the force of tension:

F_T=F_x+F_f+F_{gx}=98+14.8+26.2=139 N

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"A short-wave radio antenna is supported by two guy wires, 150 ft and 170 ft long. Each wire is attached to the top of the anten
Nonamiya [84]

Answer:

The anchor points are 78.37 ft and 111.99 ft

Explanation:

If you look at the attached (Fig 1) you will find that the union of antenna and its guy wires forms two right triangles. To solve problems that involve this kind of triangles, you can apply trigonometric functions (sine, cosine, etc) and Pythagoras Theorem. Trigonometric functions states the relation between angles, sides and hypotenuse of a right triangle. If you look Fig2, considering α angle, "b" is the opposite side, "a" the adjacent side and "c" the hypotenuse. Then

a) Sine (α) = b/c it means opposite side/hypotenuse

b) Cosine (α)= a/c, it means adjacent side/hypotenuse

c) Tangent (α) = b/a opposite side/adjacent side.

Pythagoras theorem states that if you called "a" and "b", the sides of the right triangle, and "c" the hypotenuse, then:

                                      a² + b² = c²    

As the problem states the lengths 150 ft and 170 ft represents the value of the hypotenuse of each triangle and 65° is one of the angles of the triangle with 150 ft hypotenuse. So you can solve this using sin (65°) to find the height of the antenna (h) and then the two distances (x and y,).

Sine (65°) = h/ 150 ft ⇒ Sine (65°) x 150ft = h ⇒ h = 127.9 ft.

To find x : Cosine (65°) = x/ 150 ft ⇒ Cosine (65°) x 150 ft = x

⇒ x = 78.37 ft.

And finally, to find y we can apply Pythagoras theorem

(170 ft)² = (127.9 ft)² + y² ⇒ y² = (170 ft)² - (127.9 ft)² ⇒ y = 111.99 ft

Summarizing, the anchor points are 78.37 ft and 111.99 ft

4 0
3 years ago
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