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KIM [24]
4 years ago
12

Which particle does the gray ball at position #3 represent?

Physics
2 answers:
Marizza181 [45]4 years ago
6 0

The answer is Neutron


Inessa [10]4 years ago
4 0

Answer:

A. neutron or D. neutron

Explanation:

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Four examples of second class levers​
forsale [732]

Answer:

they are :

Explanation:

  1. wheelbarrow
  2. bottle opener
  3. an oar
5 0
3 years ago
Identify the equation used to calculate the perpendicular force (F⊥) acting on a block on an inclined plane.
pychu [463]
Using geometrical arguments, we can see that the angle of the inclined plane \theta is equal to the angle between Fg and the perpendicular force.

But the perpendicular force is the projection of Fg along the perpendicular axis, and Fg=mg, so the correct answer is
<span>C) F=mg cosΘ </span>
6 0
3 years ago
Read 2 more answers
A uniform stationary ladder of length L and mass M leans against a smooth vertical wall, while its bottom legs rest on a rough h
ikadub [295]

Answer:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

Explanation:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

3 0
4 years ago
An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about _____ years.
Airida [17]

Answer:

An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about  2 years.

Explanation:

Given;

orbital period of 3 years, P = 3 years

To calculate the years of an orbital with a semi-major axis, we apply Kepler's third law.

Kepler's third law;

P² = a³

where;

P is the orbital period

a is the orbital semi-major axis

(3)² = a³

9 = a³

a = a = \sqrt[3]{9} \\\\a = 2.08 \ years

Therefore, An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about  2 years.

5 0
3 years ago
A string is stretched to a length of 376 cm and
sergey [27]

Answer: 53.09Hz

Explanation:

The fundamental frequency of an ideal taut string is:

Fn= n/2L(√T/μ)

Where:

F= frequency per second (Hz)

T= Tension of the string (cm/s sqr)

L= Length of the string (cm)

μ= Linear density or mass per unit length of the string in cm/gm

√T/μ= square root of T divided by μ

It is important to note:

Note: Typically, tension would be in newtons, length in meters and linear density in kg/m, but those units are inconvenient for calculations with strings. Here, the smaller units are used.

F1= 1/2(376cm)(0.01/1) × (√574/(0.036g/cm)(0.1kg/m÷1g/cm)

F1= 0.1329 × 399.30

= 53.09Hz

7 0
3 years ago
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