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2 years ago

Which object is in static equilibrium

Physics

2 answers:

Gelneren [198K]2 years ago

6 0

Answer:

a book that has no net force acting on it and is sitting on a table

Explanation:

edge2020

balu736 [363]2 years ago

3 0

Answer:

The body is said to be in static equilibrium if the net force acting on a body at rest is zero.As the net force is zero,the body will not undergo motion.

Explanation:

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Which of the following is the flow of electrons through a wire or a conductor

Hatshy [7]

Wires or silver and copper

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3 years ago

Two identical light springs with spring constant k3 are now individually hung vertically from the ceiling and attached at each e

anyanavicka [17]

Answer:

Keq = 2k₃

Explanation:

We can solve this exercise using Newton's second one

F = m a

Where F is the eleatic force of the spring F = - k x

Since we have two springs, they are parallel or they are stretched the same distance by the object and the response force Fe is the same for the spring age due to having the same displacement

F + F = m a

k₃ x + k₃ x = m a

a = 2k₃ x / m

To find the effective force constant, suppose we change this spring to what creates the cuddly displacement

Keq = 2k₃

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3 years ago

An airplane starts from rest and reaches a takeoff velocity of 60.0m/s due north in 4.0 s. How far has it traveled before takeof

Blizzard [7]

240 meters that's it

7 0

3 years ago

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Fudgin [204]

Idk can u explain to me?

6 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).