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1 year ago

The free-body diagram of a crate is shown.

2 answers:

4 0

The net force acting on the crate is the sum of all the forces acting in the different directions on the crate, which is 176N to the left.

Upward force = 440 N
Downward force = - 440N
Rightward force = 176 N
Leftward Force = - 352N

The net vertical force acting on the crate :

Upward force + Downward force
440N + (-440N) = 0 N

The net horizontal force acting on the crate :

Rightward force + Leftward force
176 N + (- 352 N) = - 176N

The total net force acting on the crate :

Vertical + horizontal force
0 + (-176 N) = - 176 N

Therefore, the net force acting on the crate is - 176 N, which is 176 N to the left.

Learn more :brainly.com/question/24119881

Angelina_Jolie [31]1 year ago

3 0

Answer:

B. is correct

Explanation:

EDGE2022

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Answer:

Explanation:

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2 years ago

21. A toy car starts from rest and begins to

Bond [772]

Answer:

v = 66 m/s

Explanation:

Given that,

The initial velocity of a car, u = 0

Acceleration of the car, a = 11 m/s²

We need to find the final velocity of the toy after 6 seconds.

Let v is the final velocity. It can be calculated using first equation of motion. It is given by :

v = u +at

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3 years ago

4. Una cuerda de acero de piano mide 1.60 m de longitud y 0.20 cm de diámetro. ¿Cuál es la tensión en la cuerda si se estira 0.2

bazaltina [42]

Answer:

$1030.83\ \text{N}$

Explanation:

$\Delta L$ = Cambio en la longitud de la cuerda = 0.25 cm

T = tensión en cuerda

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d = Diámetro de la cuerda = 0.2 cm

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El cambio de longitud de una cuerda viene dado por

$\Delta L=\dfrac{TL}{AE}\\\Rightarrow T=\dfrac{\Delta LAE}{L}\\\Rightarrow T=\dfrac{0.25\times 10^{-2}\times \dfrac{\pi}{4}(0.2\times 10^{-2})^2\times 210\times 10^9}{1.6}\\\Rightarrow T=1030.84\ \text{N}$

La tensión en la cuerda es $1030.84\ \text{N}$ .

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