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melomori [17]
3 years ago
5

How is this please help me​

Physics
1 answer:
navik [9.2K]3 years ago
8 0

Answer:

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During which stage does the birth rate begin to decline?
sveticcg [70]

Answer:

During stage 3 - late expanding (of demogrpahic transition model)

Explanation:

During stage 3, birth rate begins to decline as infant mortality is lower and women have more access to education, family planning, and contraceptives.  Children are not needed as "free labor" as they might have been in earlier stages.

3 0
3 years ago
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Which of these substances has the greatest index of refraction?
Inessa05 [86]
D. diamond, is the best option..
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Consider three scenarios in which a particular box moves downward under the pull of gravity :
luda_lava [24]

Answer:

1. True  WA > WB > WC

Explanation:

In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.

A) Work is the product of force by distance and the cosine of the angle between them

    WA = W h cos 0

   WA = mg h

B) On a ramp without rubbing

     Sin30 = h / L

     L = h / sin 30

     WB = F d cos θ  

     WB = F L cos 30

     WB = mf (h / sin30) cos 30

     WB = mg h ctan 30

C) Ramp with rubbing

    W sin 30 - fr = ma

   N- Wcos30 = 0

   W sin 30 - μ W cos 30 = ma

    F = W (sin30 - μ cos30)

   WC = mg (sin30 - μ cos30) h / sin30

   Wc = mg (1 - μ ctan30) h

When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum

Let's review the claims

1. True The work of gravity is the greatest and the work where there is friction is the least

2 False. The job where there is friction is the least

3 False work with rubbing is the least

4 False work with rubbing is the least

5 0
3 years ago
A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?
garri49 [273]
<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>
6 0
3 years ago
g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t
jok3333 [9.3K]

Answer:

(a)    f = 0.58Hz

(b)    vmax = 0.364m/s

(c)    amax = 1.32m/s^2

(d)    E = 0.1J

(e)    x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}            (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

v_{max}=\omega A=2\pi f A      (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

a_{max}=\omega^2A=(2\pi f)^2 A

a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J

The total energy is 0.1J

(e) The displacement as a function of time is:

x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)

6 0
2 years ago
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