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3 years ago

What is the reason behind superposition?why waves behave this way?

Physics

1 answer:

marshall27 [118]3 years ago

4 0

Explanation:

Most waves appear complex because they result from two or more simple waves that combine as they come together at the same place at the same time—a phenomenon called superposition. Waves superimpose by adding their disturbances; each disturbance corresponds to a force, and all the forces add

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A 3 kg bowling ball is thrown onto a mattress. If it takes 0.3 seconds to stop the ball using a force of 24 N, what was the init

enyata [817]

Answer:

-24 m/s

Explanation:

mass of the bowling ball = 3 kg

time (t) = 0.3 seconds

Force = 24 N

initial velocity u = ???

We know that;

Force = mass × acceleration (a)

So;

24 = 3 × a

a = 24/3

a = 8 m/s²

Also;

From equation of motion; acceleration is given by the relation;

$a =\frac{v-u}{t}$

if v = 0

then ;

$8 = \frac{0-u}{0.3}$

24 = 0- u

u = -24 m/s

Thus; the initial velocity of the bowling ball when it first touched the mattress = -24 m/s

7 0

3 years ago

9) If a wave has a speed of 362 m/s and a period of 4.17 s, what is its wavelength?

kakasveta [241]

Answer:

Option B (1.51 m)

Explanation:

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5 0

2 years ago

Read 2 more answers

A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height

AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

$E_i=\frac{1}{2}m_1v_1^2$

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

$E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh$

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

$\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s$

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

$p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3$

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

$v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s$

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

$E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J$