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mote1985 [20]
3 years ago
5

Which of the following scenerios fits all of the criteria for the two-source interference equations to be valid?

Physics
1 answer:
vekshin13 years ago
8 0

Answer:

answer the correct  is B

Explanation:

For the interference phenomenon to occur, some conditions must be met.

* You must have a light in phase and coherent, for this you can for a light from an incandescent source through a single slit, the light that comes out is coherent

* This light must strike two slits

* the light that passes through the slits must hit a distant screen and be able to see the phenomenon of constructive interference

when examining the different answer the correct one is B

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As an airplane leaves the runway and goes into the air, has acceleration occurred?
malfutka [58]
A, yes because the plane is using air resistance and acceleration is increasing while it goes up. Although you don’t know speed, still yes.
6 0
2 years ago
In a collision, a car of mass 1000kg travelling at 24m/s comes to rest in 1.2s calculate
kompoz [17]

Answer:

a. 24,000

b. -20,000 N

Explanation:

a. p = m•v

1,000•24 = 24,000

1,000•0 = 0

∆p = 24,000

b. F = m•a

a = ∆v/∆t

a = -24/1.2

a = -20 m/s²

F = 1,000•(-20)

F = -20,000

8 0
2 years ago
Which best describes friction?
tangare [24]

Answer:

a constabt force that acts on object that rub together

Explanation:

It is because Friction is the rubbing of one body against another body

5 0
2 years ago
While unrealistic, we will examine the forces on a leg when one falls from a height by approximating the leg as a uniform cylind
Leno4ka [110]

Answer:

Part A: 7.75 m/s

Part B: 2330.8 kN

Part C: 24.03 kN

Part D: 4.8 kN

Part E: 1.7\times 10^{9} Dyn/cm^{2}

Part F: Option D

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground.

Explanation:

<u>Part A </u>

From the fundamental kinematic equation

v^{2}=u^{2}+2gh where v is the velocity of the man just before hitting the ground, g is acceleration due to gravity, u is initial velocity, h is the height.

Since the initial velocity is zero hence

v^{2}=2gh

v=\sqrt 2gh

Substituting 10 m/s2 for g and 3 m for h we obtain

v=\sqrt 2\times 10\times 3 =\sqrt 60= 7.745967\approx 7.75 m/s

<u>Part B </u>

Force exerted by the leg is given by

F=PA where P is pressure, F is force, A is the cross-section of the bone

A=\frac {\pi d^{2}}{4}

Substituting 2.3 cm which is equivalent to 0.023m for d and 1.7\times10^{8} N/m2 for P we obtain the force as

F=PA=1.7\times10^{8}*\frac {\pi (0.023)^{2}}{4}= 2330818.276\approx 2330.8 kN

<u>Part C </u>

The fundamental kinematic equation is part (a) can also be written as

v^{2}=u^{2}+2a\triangle x and making a the subject then

a=\frac {v^{2}-u^{2}}{2\triangle x} where a is acceleration and \triangle x is the change in length

Substituting the value obtained in part a, 7.75 m/s for v, u is zero and 1cm which is equivalent to 0.01 m for \triangle x then  

a=\frac {7.75^{2}-0^{2}}{2\times 0.01}= 3003.125 m/s^{2}

Force exerted on the man is given by

F=ma=80\times 3003.125= 240250 N\approx 24.03 kN

<u>Part D </u>

The fundamental kinematic equation is part (a) can also be written as

v^{2}=u^{2}+2a\triangle h and making a the subject then

a=\frac {v^{2}-u^{2}}{2\triangle h} where a is acceleration and \triangle h is the change in height

Also, force exerted on the man is given by F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}

Substituting 80 Kg for m, 50 cm which is equivalent to 0.5m for \triangle h and other values as used in part c

F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}=80\times \frac {7.75^{2}-0^{2}}{2\times 0.5}= 4805 N\approx 4.8 kN

<u>Part E </u>

P=1.7\times 10^{8}=1.7\times 10^{8}\times (\frac {10^{5} Dyn}{10^{4} cm^{2}}=1.7\times 10^{9} Dyn/cm^{2}

Part F

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground

7 0
2 years ago
A rescue plane spots a survivor 132 m directly below and releases an emergency kit with a parachute. If the package descends at
borishaifa [10]

Answer: 19.15 meters on the waves away from the survivor.

Explanation:

7 0
3 years ago
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