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Anika [276]
2 years ago
5

A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward

with a force FS whose direction makes an angle of 30.0° with the ramp (Fig. E4.4). (a) How large a force FS is necessary for the component Fx parallel to the ramp to be 90.0 N?(b) How large will the component Fy perpendicular to the ramp be then?

Physics
1 answer:
AleksandrR [38]2 years ago
5 0

Answer:

(a)  104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.  

The component of F perpendicular to the ramp is Fy.

(a)  

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.  

Resolve F into its x-component from Pythagorean theorem:  

Fx=Fcos30°

Solve for F:  

F= Fx/cos30°  

Substitute for Fx from given data:  

Fx=90 N/cos30°

   =104 N

(b) Resolve r into its y-component from Pythagorean theorem:

     Fy = Fsin 30°

   Substitute for F from part (a):

     Fy = (104 N) (sin 30°)  

          = 52 N  

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GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

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Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

v=\sqrt{v_1^2+v_2^2}

v=\sqrt{5^2+8^2}=9.43\ m/s

Let E is the kinetic energy of the plane. It is given by :

E=\dfrac{1}{2}mv^2

E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2

E = 133.38 Joules

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2 years ago
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16) A wheel of moment of inertia of 5.00 kg-m2 starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s.
KiRa [710]

Answer:

57.6Joules

Explanation:

Rotational kinetic energy of a body can be determined using the expression

Rotational kinetic energy = 1/2Iω²where;

I is the moment of inertia around axis of rotation. = 5kgm/s²

ω is the angular velocity = ?

Note that torque (T) = I¶ where;

¶ is the angular acceleration.

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¶ = 3.0/5.0

¶ = 0.6rad/s²

Angular acceleration (¶) = ∆ω/∆t

∆ω = ¶∆t

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ω = 4.8rad/s

Therefore, rotational kinetic energy = 1/2×5×4.8²

= 5×4.8×2.4

= 57.6Joules

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Dao makes a table to identify the variables used in the equations for centripetal acceleration. A 2 column 5 rows. The first col
Zanzabum

Answer:

Column X. Tangential Speed

Column Y. radius  

Explanation:

The equation for centripetal acceleration is

           a_{c} = v² / r

Where v is the tangential velocity of the body and the radius of curvature.

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Let's check the answers

Column X. Tangential Speed

Column Y. radius  

This is the correct answer.

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The young tree was bent and has been brought into a vertical position by the three guy cables. If tension at AB = 0, AC = 10 lb,
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Answer:

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You drop your keys in a high-speed elevator going up at a constant speed. Part APart complete Do the keys accelerate faster towa
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Answer:

Explained

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b)Yes, keys will accelerate towards the floor faster if it is a constant speed than it is moving downward because if the elevator is accelerating downward, the downward change in velocity of the keys is at least partially matched by a downward change in the velocity of the of the elevator.

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