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1 year ago

9

URGENT Give three examples of models that you see every day

1 answer:

umka2103 [35]1 year ago

7 0

Models are very common. The ingredients list on a bottle of ketchup is a model of its contents, and margarine is a model of butter. A box score from a baseball game is a model of the actual event. A trial over an automobile accident is a model of the actual accident.

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Describe one way in which light acts like a wave and one way in which light acts like a particle.

monitta

Answer:

Quantum mechanics tells us that light can behave simultaneously as a particle or a wave. ... When UV light hits a metal surface, it causes an emission of electrons. Albert Einstein explained this "photoelectric" effect by proposing that light – thought to only be a wave – is also a stream of particles.

Explanation:

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2 years ago

A little help with this one

Aloiza [94]

The answer would be A because as you get further away from sound the intensity decreases.

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2 years ago

A train moves uniform velocity of 36km/h for 15 sec.find distance

vesna_86 [32]

Answer:

S = V × t

Explanation:

first convert 36km/h to m/s

36km/h=36*5/18=180/18=10m/s

d =s*t

=10*10=100m

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2 years ago

An elevator filled with passengers has a mass of 1603 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 f

Galina-37 [17]

Answer:

T = 17649.03 N = 17.65 KN

Explanation:

The tension in the cable must be equal to the apparent weight of the passenger. For upward acceleration:

$T = W_A = m(g+a)\\$

where,

T = Tension in cable = ?

$W_A$ = Apparent weight

m = mass = 1603 kg

g = acceleration due to gravity = 9.81 m/s²

a = acceleration of elevator = 1.2 m/s²

Therefore,

$T = (1603\ kg)(9.81\ m/s^2+1.2\ m/s^2)\\\\$

<u>T = 17649.03 N = 17.65 KN</u>

8 0

2 years ago

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

<u>Using the equation of motion:</u>

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

<u>Now we find the force along the slide due to the body weight:</u>

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

<em><u>Hence the net force along the slide:</u></em>

$F_R=71.4\ N$

<em>Now the acceleration of John:</em>

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

<u>Now the new velocity:</u>

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

2 years ago