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BabaBlast [244]
2 years ago
9

What happens to the cost of goods when inflation increases?

Physics
1 answer:
Dmitriy789 [7]2 years ago
7 0

Answer:

The answer is B

Explanation:

  • <u><em>Generally speaking, a company selling goods during periods of inflation will see an increase in its cost of goods sold.</em></u>
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In which of the following cases is the torque about the shoulder due to the weight of the arm the greatest? Case 1: A person hol
Alexeev081 [22]

Answer: a) Case 2 b) Case 1

Explanation:

a) By definition, the magnitude of a torque, referred to a given point, is expressed as the product of the force that causes the torque, times the perpendicular distance to the reference point.

If we assume that the only force acting on the arm is the weight of the arm, and that this is concentrated in a point in the center of it (taking the arm as a solid bar with the center of mass at the mid-point), clearly the torque will be the greatest when the force be exactly perpendicular, which is the case of the arm placed straight out parallel to the ground (Case 2).

b)  As the torque and the angular acceleration are directly proportional each other (being the rotational inertia the proportionality constant) the angular acceleration will be maximum when torque be maximum also, which is the case that the arm begins to swim, due to the perpendicular distance to the shoulder is the maximum possible (Case 1).

6 0
3 years ago
Will the April 29th 2020 astroid be the end of humanity as we know it?
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Answer:

No.

Explanation:

If the so called Dinosaurs died from an asteroid but yet the species of DOGS lived, I'm sure humans can live.

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4 years ago
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Yuck! This statement represents which of the folllwoing functions of language
rodikova [14]

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3 years ago
What are 90 percent of the stars in space called?
Naddik [55]
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6 0
4 years ago
In a physics lab experiment, a compressed spring launches a 24 g metal ball at a 35o angle above the horizontal. Compressing the
Levart [38]

Answer:

k = 45.95 N/m

Explanation:

First, we will find the launch speed of the ball by using the formula for the horizontal range of the projectile.

R = \frac{v_{o}^{2}\ Sin\ 2\theta}{g} \\\\v_{o}^{2} = \frac{Rg}{Sin\ 2\theta}\\

where,

Vo = Launch Speed = ?

R = Horizontal Range = 5.3 m

θ = Launch Angle = 35°

Therefore,

v_{o}^{2} = \frac{(5.3\ m)(9.81\ m/s^{2})}{Sin\ 2(35^{o})}\\

v₀² = 55.33 m²/s²

Now, we know that the kinetic energy gain of ball is equal to the potential energy stored by spring:

Kinetic\ Energy\ Gained\ By\ Ball = Elastic\ Potential\ Energy\ Stored\ in \ Spring\\\frac{1}{2}mv_{o}^{2} = \frac{1}{2}kx^{2}\\\\k = \frac{mv_{o}^{2}}{x^2} \\

where,

k = spring constant = ?

x = compression = 17 cm = 0.17 m

m = mass of ball = 24 g = 0.024 kg

Therefore,

k = \frac{(0.024\ kg)(55.33\ m^2/s^2)}{(0.17\ m)^2} \\

<u>k = 45.95 N/m</u>

4 0
3 years ago
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