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IRISSAK [1]
3 years ago
15

A forklift raises a 1020 N crate 3.50 m up to a shelf l. How much work is done by the forklift on the crate? The forklift does J

work on the crate.
Physics
2 answers:
Rom4ik [11]3 years ago
7 0

A forklift raises a 1,020 N crate 3.50 m up to a shelf. How much work is done by the forklift on the crate?

The forklift does  

⇒ 3,570 J of work on the crate. the answer is 3570

lord [1]3 years ago
5 0

Answer : Work done on the crate is 3570 J

Explanation :

Force acting on forklift due to its mass is 1020 N

Distance covered in lifting it, d = 3.5 m

Mathematically, the work done is defined as :

W=F\times d

So, W=1020\ N\times 3.5\ m

W=3570\ J

The forklift does 3570 J work on the crate.      

Hence, this is the required solution.                      

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3 years ago
Que capacidade física consiste em deslocar o corpo no espaço o mais rápido possível, mudando o centro de gravidade de posição, s
grin007 [14]

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6 0
3 years ago
What's the average speed of an object moving 3790 meters in 249 s.
NISA [10]

Heya!!

For calculate velocity, lets applicate formula

                                                 \boxed{d = v * t}

                                               <u>Δ   Being   Δ</u>

                                         d = Distance = 3790 m

                                             t = Time = 249 s

                                             v = Velocity = ?

⇒ Let's replace according the formula and clear "v":

\boxed{v= 3790\ m / 249\ s}

⇒ Resolving

\boxed{ v = 15,22 \ m/s }

Result:

The velocity is <u>15,22 meters per second.</u>

Good Luck!!

4 0
2 years ago
I need help with question 4 please​
Yuki888 [10]

Answer:

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V = -10 j km / hr   if one were to use i, j, k as unit vectors with the usual orientation

4 0
3 years ago
8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.
mart [117]

Answer:

The resultant tension of the two ropes is approximately 42.4 N

The length of the line representing the resultant tension is approximately 8.48 cm

Please find included  with the answer the scale drawing created with Microsoft Word

Explanation:

The given parameters are;

The tension in rope P, T_P = 30 N

The tension in rope Q, T_Q = 30 N

The angle the rope, 'P', makes with the horizontal = 45°

The angle the rope, 'Q', makes with the horizontal = 45°

The scale factor of the scale diagram, S.F. = 5.0 N/cm

By the resolution of forces at equilibrium, we have;

The sum of the vertical forces, \Sigma F_y = T_P_y + T_Q_y + W = 0

∴ W = -(T_P_y + T_Q_y)

W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712

The weight of the heavy ball, W ≈ 42.4 N acting downwards

The sum of the horizontal forces, \Sigma F_x = T_P_x + T_Q_x  = 0

The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm

The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included

3 0
3 years ago
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