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3 years ago

a 3.46 kg box is sitting at rest on a flat floor. a.) what is the weight of the box. b.) what is the normal force on the box

Physics

1 answer:

just olya [345]3 years ago

4 0

A) The weight of the box is given by:

W = mg

W = weight, m = mass, g = gravitational acceleration

Given values:

m = 3.46kg, g = 9.81m/s²

Plug in and solve for W:

W = 3.46(9.81)

W = 33.9N

B) The box is sitting at rest on the floor. This means that the box is in equilibrium; the two forces acting on it, its weight and the normal force that the floor exerts on it, are equal in magnitude and opposite in direction. Therefore the normal force N acting on the box has a magnitude of 33.9N

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You are performing a knee extension exercise. You hold a 20kg weight at full knee extension. The weight is 0.4m from your knee j

dmitriy555 [2]

Answer:

The moment is -78.4 N-m (clockwise).

Explanation:

Given:

Mass of the object (m) = 20 kg

Distance of the object from the knee joint (d) = 0.4 m

Weight of leg is not considered.

Acceleration due to gravity (g) = 9.8 m/s²

Now, weight of the object is equal to the product of its mass and acceleration due to gravity. So,

Weight = Mass × Acceleration due to gravity

= $mg=20\times 9.8 =196\ N$

We know that, moment of a force about a point is defined as the product of force applied and the perpendicular distance between the point and the line of application of force.

Moment of the given weight about the knee joint is given as:

Moment about knee joint = Weight × Distance from knee joint to weight

Moment about knee joint = 196 × 0.4 = 78.4 Nm

Now, from the diagram below, we can observe that, the weight acts vertically down and thus the sense of rotation about the knee joint at point O is clockwise. So, moment is negative.

Therefore, the moment is -78.4 N-m (clockwise).

7 0

3 years ago

The gauge pressure at the bottom of a cylinder of liquid is 0.30atm. The liquid is poured into another cylinder with twice the r

likoan [24]

Answer:

$P_g' = 0.075 atm$

Explanation:

Gauge pressure at the bottom of the cylinder depends on the height of water in the cylinder

So here we can say that

$P_g = \rho g h$

now when liquid is filled to height "h" in base area "A" then gauge pressure of the liquid at the bottom is given as

$P_g = 0.30 atm$

now we put the whole liquid into another cylinder with twice radius of the first cylinder

So area becomes 4 times

now by volume conservation we can say that if area is increased by 4 times then height of liquid will decrease by 4 times

so we have

$h' = \frac{h}{4}$

so gauge pressure is given as

$P_g' = \frac{0.30}{4} = 0.075 atm$

5 0

3 years ago

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g

Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

$k=\sqrt{\frac{I}{M} }$

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration $k=\sqrt{\frac{I}{M} }$

So, moment of rotational inertia (I) of a cylinder about it axis = $\frac{MR^2}{2}$