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tankabanditka [31]
3 years ago
6

Which of Newton’s laws did Nikolas use to have to calculate this force

Physics
1 answer:
devlian [24]3 years ago
6 0

Answer:

the second force

Explanation:

this is the answer i hade for mine

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A waitperson carrying a tray with a platter on it tips the tray at an angle of 12 degrees below the horizontal. If the gravitati
Tasya [4]

Answer:1.04 N

Explanation:

Given

Gravitational Force on the Platter is 5 N

Tray makes an angle of \theta =12^{\circ}

This gravitational Force has components along and Perpendicular to Platter

Perpendicular Force W_p=W\cos \theta

W_p=5\times \cos 12=4.89 N

Along the Tray

W_{along}=W\sin \theta

W_{along}=5\times \sin 12=1.04 N

Thus 1.04 N is the magnitude of force that will cause Platter to slide down  

7 0
3 years ago
An object is suspended by a string from the ceiling of an elevator. If the tension in the string is equal to 25 N at an instant
Phantasy [73]

By Newton's second law, the net force on the object is

∑ <em>F</em> = <em>T</em> - <em>mg</em> = - <em>ma</em>

where

• <em>T</em> = 25 N, the tension in the string

• <em>m</em> is the mass of the object

• <em>g</em> = 9.8 m/s², the acceleration due to gravity

• <em>a</em> = 2.0 m/s², the acceleration of the elevator-object system

Solve for <em>m</em> :

25 N - <em>m</em> (9.8 m/s²) = - <em>m</em> (2.0 m/s²)

==>   <em>m</em> = (25 N) / (9.8 m/s² - 2.0 m/s²) ≈ 3.2 kg

4 0
3 years ago
How does a longshore current affect a beach?
77julia77 [94]
Longshore currents are affected by the velocity and angle of a wave. When a wave breaks at a more acute (steep) angle on a beach, encounters a steeper beach slope, or is very high, longshore currents increase in velocity. ... This process, known as “longshore drift,” can cause significant beach erosion.
8 0
3 years ago
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1) A spring, which has a spring constant k=7.50 N/m, has been stretched 0.40 m from ts equilibrium position . What the potential
cupoosta [38]
<h3>Answer:</h3>

\displaystyle U_s = 0.6 \ J

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Physics</u>

<u>Energy</u>

Elastic Potential Energy: \displaystyle U_s = \frac{1}{2} k \triangle x^2

  • U is energy (in J)
  • k is spring constant (in N/m)
  • Δx is displacement from equilibrium (in m)
<h3>Explanation:</h3>

<u>Step 1: Define</u>

k  = 7.50 N/m

Δx = 0.40 m

<u>Step 2: Find Potential Energy</u>

  1. Substitute in variables [Elastic Potential Energy]:                                        \displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.40 \ m)^2
  2. Evaluate exponents:                                                                                      \displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.16 \ m^2)
  3. Multiply:                                                                                                           \displaystyle U_s = (3.75 \ N/m) (0.16 \ m^2)
  4. Multiply:                                                                                                           \displaystyle U_s = 0.6 \ J
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3 years ago
A car slams on its brakes, coming to a complete stop in 4.0 s. The car was traveling south at 60.0 mph. Calculate the accelerati
emmainna [20.7K]
Acceleration = ms^(-1)
= 60/4
=15 ms with the power of -1
4 0
3 years ago
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