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Sedbober [7]
2 years ago
10

Myron was trying to light a match. The first time he struck the match, the match did not light. The second time he tried, the ma

tch lit successfully.
Which most likely occurred to make the match light?

Myron increased the friction by decreasing the surface area.
Myron increased the friction by increasing speed.
Myron decreased the friction by making the match aerodynamic.
Myron decreased the friction by lubricating the match.
Physics
2 answers:
Andru [333]2 years ago
8 0

Answer:

B. Myron increased the friction by increasing speed.

Explanation:

i got it correct on edge

Volgvan2 years ago
4 0

Answer:

The answer is B i think

Explanation:

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A 1 kg flashlight is dropped from rest and falls to the floor without air resistance . At the point during its fall, when it is
alexgriva [62]

Answer:

The speed of the flashlight at that point is 3.7 m/s

Explanation:

When an object of mass M is at a height H above the ground, the potential energy of the object is:

U = M*H*g

Where g is the gravitational acceleration, g = 9.8 m/s^2

And for an object with velocity v, the kinetic energy is:

K = (M/2)*v^2

We know that when the flashlight of mass  1kg is 0.7 meters above the ground, the potential energy is equal to the kinetic energy, then:

M = 1kg

H = 0.7m

g = 9.8 m/s^2

Replacing these in the equations, we get:

U = K

(1kg)*(0.7m)*(9.8 m/s^2) = ((1kg)/2)*v^2

As the mass factor appears in both sides, we can remove it:

(0.7 m)*(9.8 m/s^2) = (v^2)/2

Now we can multiply both sides by 2:

2*(0.7 m)*(9.8 m/s^2) = v^2

Now let's apply the square root to both sides:

√(2*(0.7 m)*(9.8 m/s^2)) = v = 3.7 m/s

8 0
3 years ago
A car traveling 34 mi/h accelerates uniformly for 4 s, covering 615 ft in this time. What was its acceleration? Round your answe
Contact [7]

Answer:

51.94 ft/s²

257.63 ft/s

Explanation:

t = Time taken = 4 s

u = Initial velocity = 34 mi/h

v = Final velocity

s = Displacement = 615 ft

a = Acceleration

Converting velocity to ft/s

34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2

Acceleration is 51.94 ft/s²

v=u+at\\\Rightarrow v=49.87+51.94\times 4\\\Rightarrow v=257.63\ ft/s

Final velocity at this time is 257.63 ft/s

5 0
3 years ago
Plate tectonics is the scientific theory explaining the movement of the earths crust. The movement of these tectonic plates is l
zzz [600]

Answer:

Continental drift theory describes the long term effect of plate tectonics.

Explanation:

The long term result of plate tectonic movement is the continental drift. The continents of Earth lay on tectonic plates, that are in motion and interaction via plate tectonics. The drift of the Earths continent is an ongoing process evident in the rift valleys and seafloor spreading zones.

The theory that  the Earth's continents are dynamic and have drifted relative to each other is known as continental drift which correlates with the theory of plate tectonics.

Every year, the Earth's outer shell plates are displaced by a small amount due to the heat coming from the Earths interior via convection currents.

7 0
3 years ago
In which of these examples is the greatest movement occurring?
vlada-n [284]
You need to provide a picture or tell us the examples... we can’t see what you see
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3 years ago
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A depiction of a famous scientific experiment is given. Consider how the beam changes when the magnet is off compared to when th
iogann1982 [59]

Answer:

The beam used is a negatively charged electron beam with a velocity of

v = E / B

Explanation:

After reading this long statement we can extract the data to work on the problem.

* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.

* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced

           F_{e} =  F_{m}

           q E = qv B

           v = E / B

this configuration is called speed selector

They ask us what type of beam was used.

The beam used is a negatively charged electron beam with a velocity of v = E / B

6 0
3 years ago
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