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rjkz [21]
3 years ago
11

Jay fills a wagon with sand (about 20 kg) and pulls it with a rope 40 m along the beach. He holds the rope 25â above the horizon

tal. The rope exerts a 20-N tension force on the wagon. How much work does the rope do on the wagon?
Physics
1 answer:
never [62]3 years ago
8 0

Answer:

W = 725 J

Explanation:

given,

mass of wagon = 20 kg

distance of pull = 40 m

angle made with the horizontal = 25°

tension force on the wagon = 20 N

Work done = ?

Horizontal component of the force will help in movement of the wagon.

Horizontal component of force=

F_x = F cos θ

Work done is equal to force into displacement

W = F.s

W =  F cos θ.s

W = 20 cos 25° x 40

W = 725 J

hence, work done on pulling the wagon is equal to W = 725 J

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3 years ago
A mechanic test driving a car that she has just given a tune-up accelerates from rest to 50.0 m/s in 9.8 s. How far (in meters)
Stels [109]

Answer:

245 m

Explanation:

v = at + v₀

50.0 m/s = a (9.8 s) + 0 m/s

a = 5.10 m/s²

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (9.8 s) + ½ (5.10 m/s²) (9.8 s)²

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3 years ago
A 5c charge experiences a force of 40n when put at a certain location in space. What is the electric field at that location?
stepladder [879]
The 'strength' of the electric field is the force on 1C of charge at that point.

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3 0
3 years ago
A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu
dalvyx [7]

Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 =  0.341 \ kg

   The mass attached to the other end of the string is  m_2 =  0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 =  m_1 *  g

substituting values

      T_1 =  0.341 * 9.8

      T_1 =  3.342 \ N

At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 =  m_2 *  g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

     \tau  = 0.2505 \ N \cdot m

5 0
3 years ago
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