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2 years ago

A stationary body is acted upon by a number of forces. State the two conditions which must apply for the body to remain at rest.

Physics

1 answer:

ratelena [41]2 years ago

8 0

Answer:

1. All the externally applied forces were cancelled out.

2. The externally applied force is not exceeded the limiting friction

Explanation:

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Estás cuentas son ingresos fijos o variables o egresos fijos o variables?

Brut [27]

¿El salario es un costo fijo o variable?
Los salarios anuales son costos fijos, pero otros tipos de compensación, como comisiones o horas extraordinarias, son costos variables.

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3 years ago

A 35 N force makes a 10 degree angle with the positive x-axis. What is the magnitude of the vertical component of the force?

Snowcat [4.5K]

Answer:

6.07 N

Explanation:

Given that,

Force, F = 35 N

It makes 10 degree angle with the positive x-axis.

We need to find the magnitude of the vertical component of the force. It can be given by :

$F_y=F\sin\theta\\\\=35\times \sin(10)\\\\=6.07\ N$

So, the magnitude of the vertical component of the force is 6.07 N.

5 0

3 years ago

Read 2 more answers

Please help!

Trava [24]

Answer:

3 N to the right

Explanation:

There are two forces acting on the car:

- A force of 10 N towards the right

- A force of 7 N towards the left

Therefore, the net force is given by the difference between the two, since they are in opposite directions:

$F=10 N-7 N=3 N$

And the direction is to the right, since the force to the right has greater magnitude than the force to the left.

7 0

3 years ago

Read 2 more answers

how much force is needed to cause a 15 kilogram bike to accelerate at a rate of 10 meters per second?

egoroff_w [7]

F = m*a, mass times acceleration.

F = 15*10 = 150 N

8 0

3 years ago

Read 2 more answers

A thin hoop is hung on a wall, supported by a horizontal nail. The hoop's mass is M=2.0 kg and its radius is R=0.6 m. What is th

boyakko [2]

Answer:

Explanation:

Given that,

Mass of the thin hoop

M = 2kg

Radius of the hoop

R = 0.6m

Moment of inertial of a hoop is

I = MR²

I = 2 × 0.6²

I = 0.72 kgm²

Period of a physical pendulum of small amplitude is given by

T = 2π √(I / Mgd)

Where,

T is the period in seconds

I is the moment of inertia in kgm²

I = 0.72 kgm²

M is the mass of the hoop

M = 2kg

g is the acceleration due to gravity

g = 9.8m/s²

d is the distance from rotational axis to center of of gravity

Therefore, d = r = 0.6m

Then, applying the formula

T = 2π √ (I / MgR)

T = 2π √ (0.72 / (2 × 9.8× 0.6)

T = 2π √ ( 0.72 / 11.76)

T = 2π √0.06122

T = 2π × 0.2474

T = 1.5547 seconds

T ≈ 1.55 seconds to 2d•p

Then, the period of oscillation is 1.55seconds

6 0

3 years ago