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Rzqust [24]
3 years ago
6

Which of these is a contact force?

Physics
2 answers:
egoroff_w [7]3 years ago
8 0
The correct answer is B.
Vera_Pavlovna [14]3 years ago
7 0
B is the answer because it causes impact and collides together.
You might be interested in
Why isn't Coulomb's law valid for dielectric objects, even if they are spherically symmetrical?
marshall27 [118]

Answer:

Explanation:

The "traditional" form of Coulomb's law, explicitly the force between two point charges. To establish a similar relationship, you can use the integral form for a continuous charge distribution and calculate the field strength at a given point.

In the case of moving charges, we are in presence of a current, which generates magnetic effects that in turn exert force on moving charges, therefore, no longer can consider only the electrostatic force.

4 0
3 years ago
A car initially traveling at 15.0 m/s accelerates at a constant rate of 4.50 m/s2 over a distance of 45.0 m. How long does it ta
anygoal [31]

Answer:

2.24 seconds

Explanation:

xf = xo + vo t + 1/2 at^2

45 = 0 + 15 t  + 1/2 (4.5) t^2

   2.25 t^2 + 15t - 45 = 0              Quadratic formula shows  t = 2.24 seconds

4 0
2 years ago
PLEASE HELP
Mashutka [201]

Answer:

It cannot be constant because if it does not change and each time it increases its strength and speed.

Explanation:

4 0
2 years ago
Two protons are released from rest when they are 0.720 nm apart. For related problem-solving tips and strategies, you may want t
Gnesinka [82]

Answer:

a) Speed of the electrons at maximum speed = (1.384 × 10⁴) m/s

The maximum speed occurs at the point where all of the initial potential energy is converted into kinetic energy.

b) Maximum acceleration of the protons = (2.660 × 10¹⁷) m/s²

The maximum acceleration occurs at the minimum distance apart for the two protons.

Explanation:

The maximum speed occurs when all the potential energy of the protons has been converted to kinetic energy.

The potential energy between the two protons at the instant of release is given by

U = (kq₁q₂/r)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

q₁ = q₂ = charge on a proton = q = (1.602 × 10⁻¹⁹) C

r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

U = (kq²/r) = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ (7.2 × 10⁻¹⁰) = (3.204 × 10⁻¹⁹) N/m or Joules

At the maximum speeds, the two protons will not possess any potential Energy, only kinetic energy.

The sum of kinetic and potential energies is always constant for the system

(Initial Kinetic Energy) + (Initial Potential Energy) = (Kinetic Energy at maximum speed) + (Potential Energy at maximum speed)

Initial Kinetic Energy of the system = 0 J (Since both protons were intially at rest)

Initial Potential Energy = (3.204 × 10⁻¹⁹) J

Kinetic Energy at maximum speed = Sum of the kinetic energies of the protons at this point = (½mv²) + (½mv²) = (mv²) J (Since theu are both protons, they have the same mass and the same speed at maximum speed)

Potential Energy at maximum speed = 0 J

0 + (3.204 × 10⁻¹⁹) = mv² + 0

mv² = (3.204 × 10⁻¹⁹)

m = mass of a proton = (1.673 × 10⁻²⁷) kg

v = speed of each of the protons at maximum speed = ?

v = √[(3.204 × 10⁻¹⁹) ÷ m]

v = √[(3.204 × 10⁻¹⁹) ÷ (1.673 × 10⁻²⁷)]

v = √(1.915 × 10⁸) = 13,838.8 m/s = (1.384 × 10⁴) m/s

b) Since the two protons repel each other and force of repulsion reduces as the dI stance between the protons increases, the maximum acceleration occurs at the minimum distance apart for the two protons.

Force of repulsion acting on each proton is given through Coulomb's law as

F = (kq₁q₂/r²)

And the force acting on each proton is obtainable using Newton's law that

F = ma

So, the acceleration of each proton at any time is obtainable through a relation of these 2 formulas.

ma = (kq₁q₂/r²)

a = (kq₁q₂/r²m)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

q₁ = q₂ = charge on a proton = q = (1.602 × 10⁻¹⁹) C

r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

m = mass of a proton = (1.673 × 10⁻²⁷) kg

a = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ [(7.2 × 10⁻¹⁰)² × (1.673 × 10⁻²⁷)]

a = (2.660 × 10¹⁷) m/s²

Hope this Helps!!!

5 0
2 years ago
It would be really helpful if u help me solving this question. PLEASE!!!
sweet [91]

Answer: The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

The room temperature will be the point at which the themometric substance remains constant when rising from ice point.

Explanation:

Apparatus available:

Unmarked thermometer

250 cm3 glass beaker

crushed ice 

water

heatproof mat 

clamp, boss and stand

meter rule

Added apparatus

Bunsen burner

Stirrer

Method

The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

Procedures

Set up the apparatus of illustrated in the attached figure.

Immerse the unmarked thermometer into the ice in the beaker.

When the level indicated by the thermometric substance remains steady after some time, a mark will be made at that point. This mark will corresponds to the ice point (lower fixed point) and is assigned the value of 0 °C.

You may add little water and continue to stir gently.

The themometric substance will start to rise and stop when it reaches room temperature. Mark the point but do not assign any value

Place the beaker on bunsen burner and boil the water. The themometric substance will continue to rise and remain constant at upper fixed point

This mark will corresponds to the steam point (upper fixed point) and is assigned the value of 100 °C.

Divide between the lower fixed point and upper fixed point into equal intervals. Then you can see the value of room temperature.

7 0
2 years ago
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