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Rashid [163]
3 years ago
13

N

Physics
1 answer:
9966 [12]3 years ago
7 0

Answer: de?

Explanation:

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Two waves are shown below. When the crest of wave A meets the crest of wave B, ____________ interference will occur.
Lana71 [14]

Answer: The Answer: Destructive

Explanation: hope This helped

5 0
3 years ago
answers Collision derivation problem. If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the in
Natasha2012 [34]

Answer:

4.8967m

Explanation:

Given the following data;

M = 0.2kg

∆p = 0.58kgm/s

S(i) = 2.25m

Ratio h/w = 12/75

Firstly, we use conservation of momentum to find the velocity

Therefore, ∆p = MV

0.58kgm/s = 0.2V

V = 0.58/2

V = 2.9m/s

Then, we can use the conservation of energy to solve for maximum height the car can go

E(i) = E(f)

1/2mV² = mgh

Mass cancels out

1/2V² = gh

h = 1/2V²/g = V²/2g

h = (2.9)²/2(9.8)

h = 8.41/19.6 = 0.429m

Since we have gotten the heigh, the next thing is to solve for actual slant of the ramp and initial displacement using similar triangles.

h/w = 0.429/x

X = 0.429×75/12

X = 2.6815

Therefore, by Pythagoreans rule

S(ramp) = √2.68125²+0.429²

S(ramp) = 2.64671

Finally, S(t) = S(ramp) + S(i)

= 2.64671+2.25

= 4.8967m

3 0
3 years ago
Does water have mass? How do you know?
Bess [88]

every thing on this earth has it,s own mass

8 0
3 years ago
Read 2 more answers
A box of mass m1 = 20.0 kg is released from rest at a warehouse loading dock and slides down a 3.0 m-high frictionless chute to
Harrizon [31]

´To develop this problem we will use the concepts related to the conservation of momentum and the application of energy conservation equations to find the velocity of the mass after the collision, like this:

Velocity of the mass m_1 just before the collision

v_1 = \sqrt{2gh}

v_1 = \sqrt{2(9.8)(3)}

v_1 = 7.67m/s

Therefore the momentum just before collision would be

p_2 = m_1v_1+40(0)\\p_1 = 20*7.67+40(0)\\p_1 = 153.36kg \cdot m/s

Momentum after the collision

p_1 = 20*u_1+40u_1\\p_1 = 60u_1

Since the momentum is conserved we have that

153.36= 60u_1

u_1 = \frac{153.36}{60}

u_1 = 2.56m/s

The velocity of mass m_2 after the collision is given by

v_2 = \frac{2m_1}{m_1+m_2} u_1

v_2 = \frac{2(20)}{20+40}(2.56)

v_2 = 1.71m/s

Therefore the change in momentum of mass 2 is

p_2 = m_2v_2

p_2 = 40*1.71

p_2 = 68.4kg\cdot m/s

Therefore the impulse acting on m2 during the collision between the two boxes is p = 68.4kg\cdot m/s

8 0
4 years ago
What is the correct formula for finding the frequency of an electromagnetic wave
Licemer1 [7]
Answer:
f = c / λ

Explanation:
Electromagnetic waves are types of periodic waves. They propagate with the same speed as light (3 * 10⁸ m/sec).
Therefore:
velocity of wave = c

Now, the equation that relates speed of wave and its frequency is as follows:
c = λf
where:
c is the speed of wave
f is the frequency of the wave
λ is the wavelength of the wave

Solve the above equation for frequency, we will end up with:
f = c / λ


3 0
3 years ago
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