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alexandr402 [8]
3 years ago
5

Why does the friction force stay the same when the applied force is increasing?

Physics
1 answer:
Rina8888 [55]3 years ago
3 0

Answer:

When we increase the applied force (push harder), the frictional force will also increase until it reaches a maximum value. When the applied force is larger than the maximum force of static friction the object will move. ... For kinetic friction the value remains the same regardless of the magnitude of the applied force.

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Carbon-14 has a half-life of 5,730 years. if the age of an object older than 50,000 years cannot be determined by radiocarbon da
Aleonysh [2.5K]
<h3><u>Answer;</u></h3>

Carbon-14 levels in a sample are undetectable after approximately 9 half lives

<h3><u>Explanation;</u></h3>
  • <em><u>The half life of Carbon-14 is 5,730 years . Half life is the time taken by a radioactive material to decay by half of its original mass.  Therefore, it  would take a time of 5730 years for a sample of 100 g of carbon-14 to decay to 50 grams</u></em>
  • <em><u>A period of 50,000 years, is equivalent to; </u></em>

<em><u>  50,000÷5,730 </u></em>

<em><u>= 8.73 half lives</u></em>

<em>Which is approximately equal to 9 half lives.</em>

  • Therefore, if the age of an object older than 50,000 years cannot be determined by radiocarbon dating, then <em><u>Carbon-14 levels in a sample are undetectable after approximately 9 half lives</u></em>.
6 0
3 years ago
Read 2 more answers
Which of the following actions would decrease the energy stored in a parallel plate capacitor when a constant potential differen
Debora [2.8K]

Answer:

increasing the separation between the plates

Explanation:

The increase in the vacuum/separation between the plates in a parallel plate capacitor connected to a constant potential difference decreases the energy stored in the capacitor. the increase in the separation of the plates of a parallel plate capacitor reduces the capacitance of the capacitor because

Q(charge) = CV  V = VOLTAGE , c = capacitance  

E = 1/2 eAV^2/ D  ( ENERGY STORED )

where D = distance between plates, e = dielectric, A = area of capacitor , V = potential difference

6 0
3 years ago
What is one factor that helps keep the planets and other solar system objects in orbit around the sun?
pshichka [43]
The answer is D<span>.the gravitational force between the sun and each object in the solar system</span>
7 0
3 years ago
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What units must the constant G have in order for the u it’s F to be newtons (N)?
EleoNora [17]

The units for G must be [N][m^2][kg^{-2}]

Explanation:

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

F is the force

G is the gravitational constant

m_1, m_2 are the masses of the two objects

r is the separation between the objects

We know that:

  • The units of F are Newtons (N)
  • The units of m_1,m_2 are kilograms (kg)
  • The units of r are metres (m)

So, we can rewrite the equation in terms of G, to find its units:

G=\frac{Fr^2}{m_1 m_2}=\frac{[N][m]^2}{[kg][kg]}=[N][m^2][kg^{-2}]

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

5 0
2 years ago
An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
3 years ago
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