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1 year ago

PLEASE HELP WILL MARK BRAINLIEST IF YOU EXPLAIN HOW YOU GOT IT!!!

Physics

1 answer:

allochka39001 [22]1 year ago

6 0

For an object at equilibrium :

.option D: It might be stationary.

.option E: Fnetx must be zero, Fnetx can be anything.

What is equilibrium?

This refers to a state in which an object is at rest or balanced to the equal action of opposing forces.

At equilibrium, in the presence of external forces, the opposite forces normally have a balanced effect on the object in concern. Objects at equilibrium are known for their stable nature in terms of motion and energy.

Conditions for equilibrium

.The object must have no acceleration.

.The total force acting on the body i.e the net force and the net torque on the object must be zero.

Learn more about equilibrium on

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Suppose that she pushes on the sphere tangent to its surface with a steady force of F = 60 N and that the pressured water provid

sergey [27]

Answer:

25.06s

Explanation:

Remaining part of the question.

(A large stone sphere has a mass of 8200 kg and a radius of 90 cm and floats with nearly zero friction on a thin layer of pressurized water.)

Solution:

F = 60N

r = 90cm = 0.9m

M = 8200kg

Moment of inertia for a sphere (I) = ⅖mr²

I = ⅖ * m * r²

I = ⅖ * 8200 * (0.9)²

I = 0.4 * 8200 * 0.81

I = 2656.8 kgm²

Torque (T) = Iα

but T = Fr

Equating both equations,

Iα = Fr

α = Fr / I

α = (60 * 0.9) / 2656.8

α = 0.020rad/s²

The time it will take her to rotate the sphere,

Θ = w₀t + ½αt²

Angular displacement for one revolution is 2Π rads..

θ = 2π rads

2π = 0 + ½ * 0.02 * t²

(w₀ is equal to zero since sphere is at rest)

2π = ½ * 0.02 * t²

6.284 = 0.01 t²

t² =6.284 / 0.01

t² = 628.4

t = √(628.4)

t = 25.06s

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Will humans ever grow wings?

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Explanation:

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4 years ago

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A 2 000-kg car is slowed down uniformly from 20.0 m/s to 5.00 m/s in 4.00 s. (a) What average force acted on the car during that

slava [35]

Answer:

the answer is c

Explanation:

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3 years ago

Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the

Svet_ta [14]

Answer:

b) $x_{cm}$ = 4.88 cm , c) $x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃) and d)

$x_{cm}$ ’= 1.88 cm

Explanation:

The definition of mass center is

$x_{cm}$ = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

D₂ = d₁ + d₂

D₂ = 1.1 + 1.9

D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

D₃ = D₂ + d₂

D₃ = 3.0 + 3.9

D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

M = m₁ + m₂ + m₃

M = 24 + 12 + 56

M = 92 g

$x_{cm}$ = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

$x_{cm}$ = 1/92 (448.8)

$x_{cm}$ = 4,878 cm

$x_{cm}$ = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

$x_{cm}$ ’= 1/M (m₁ d₁ + m₂ d₂ + m₃ d₃)

$x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃)