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muminat
3 years ago
11

2 ways to change frictional force between 2 objects

Physics
2 answers:
mestny [16]3 years ago
8 0
It can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can.
Zanzabum3 years ago
5 0
In order to change the frictional force between two solid surfaces, it can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can. 
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4. A plane accelerates down a runway due to the constant resultant force of
Doss [256]

Answer:

See below

Explanation:

Net force acting on the plane after overcoming frictional forces is

2.7 x 10^6   -  6.8 x 10^4   -  8.5 x 10^5 = 1 782 000 N

Work = F x d

        = 1 782 000 x 1400 m = 2.5 x 10^9   J       or  2.5 x 10^6  kJ

4 0
1 year ago
A balloon is a sphere with a radius of 5.0 m. the force of air against the walls of the balloon is 45 n. what is the air pressur
Mazyrski [523]

The air pressure inside the balloon is: 0.1432 Pa

The formulas and procedures that we will use to solve this problem are:

  • a = 4 * π * r²
  • P = F/a

Where:

  • a = area of the sphere
  • r = radius
  • π = mathematical constant
  • P = Pressure
  • F = Force
  • a = surface area

Information about the problem:

  • r = 5.0 m
  • F = 45 N
  • 1 Pa = N/m²
  • 1 N = kg * m/s²
  • a=?
  • P=?

Using the formula of the sphere area we get:

a = 4 * π * r²

a = 4 * 3.1416 * (5.0 m)²

a = 314.16 m²

Applying the pressure formula we get:

P = F/a

P = 45 N/314.16 m²

P = 0.1432 Pa

<h3>What is pressure?</h3>

It is a physical quantity that expresses the force applied on the area of a surface.

Learn more about pressure at: brainly.com/question/26269477

#SPJ4

7 0
1 year ago
How many kilocalories are generated when the brakes are used to bring a 1200-kg car to rest from a speed of 95 km/h ? 1 kcal = 4
timofeeve [1]
<span>Answer: 1st, identify the givens and the unknown - this will give you parameter of what concept and formula are you going to use. Given: m= 1200kg v initial = 95km/hr v final = 0 2nd, focus on the units - in most cases units speak for the concept the unit of the unknown is kcal, thus its the unit of energy or work so, W = ? 3rd, provide the appropriate formula - give formula or equation that the given and the unknown are present since W = delta K.E =delta P.E W= 0.5m( vf^2 - vi^2) ---> best formula 4th, Substitute the given to the formula since 1 Joule = 1Nm 1N = 1kgms^-2 1cal = 4.19 J we express first 95 km/hr to m/s 95km/hr x 1000m/1km x 1hr/3600sec = 26.39 m/sec W= 0.5(1200kg)[(0^2- (26.39m/sec)^2] W=600 kg(0 - 696.43m^2/s^2) W=600kg(-696.43m^2/s^2) W=417859.3Nm or 417859.3 J W = 417859.3 J x 1 cal /4.19 J W = 99,727.7 cal or 99.728 kcal</span>
4 0
3 years ago
Discuss how a photon (aka the light particle) can be affected by gravity despite being massless.
Over [174]

Explanation:

Light is clearly affected by gravity, just think about a black hole, but light supposedly has no mass and gravity only affects objects with mass. On the other hand, if light does have mass then doesn't mass become infinitely larger the closer to the speed of light an object travels.

4 0
3 years ago
1.
ale4655 [162]

Answer:

5.0 atm

Explanation:

P₁V₁=P₂V₂

P₁V₁/V₂=P₂

(1)(2.5)/(0•50)=P₂

P₂=5

Pressure is now 5.0 atm

8 0
3 years ago
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