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3 years ago

infer whether a planet with active volcanoes would have more craters than a planet without active volcanoes

1 answer:

7 0

I predict that the planet with active volcanoes will have more craters, due to the volcanic explosions, as opposed to a planet without active volcanoes.

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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

(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

How to solve 30(a)<br><br> Give solution asap.

expeople1 [14]

Answer:

They will not meet

h-hX=1.2*g*t²

hX=v0*t-(1/2*g*t²)

Explanation:

fall h=1/2*g*t²

elevation time if v0=20 m/s te=v0/g=20 m/s /9.81 m/s²=2.0387s

hmax=v0²/(2*g)=(400 m²/s²)/19.62 m/s²2=20.387 m

free fall

t=2.0387s yields hX=1/2*g*t²=20.387 m

h-hX=200m - 20.387 m=179,613 m.

so, the second body has not enough initianoal speed to reach a meeting point

5 0

3 years ago

Question :

irakobra [83]

A rubber ball and a stone of the same size are examples which will have more inertia and is therefore denoted as option A.

<h3>What is Inertia?</h3>

This is referred to as the property exhibited by a body in which it has the tendency to remain at rest or in uniform motion.This property is dependent on the mass of the substance as we can deduce that the greater the mass, the greater the inertia and vice versa.

The size of a rubber ball and stone will have different masses in which that of the stone will be greater. This is as a result of the difference in the nature of the substances which are used to make both items mentioned above.

This is therefore the reason why a rubber ball and a stone of the same size as having more inertia(mass) where chosen as the most appropriate choice in this scenario.

Read more about Inertia here brainly.com/question/1140505

#SPJ1

5 0

1 year ago

1. What are the two factors affecting friction

bija089 [108]

Answer: a.) Roughness of the surfaces in contact with each other .
Higher the roughness of surfaces in contact with each other, greater is the friction between bodies. Force of friction will be less between smooth surfaces.

b.) Weight of the sliding/rolling body: greater the weight of the moving body on the surface, more is the force of friction on the body by the surface.

I hope this helps

5 0

1 year ago

Can anyone please help me with the steps

KIM [24]

Answer:

A) v_average = - 10 km / h, B) v = 1.6 m / s, v = 17.6 m / s

Explanation:

A) the average speed is the average speed of a body, if we assume that the direction of going up the hill is positive

v₁ = 40 km / h

v₂ = - 60 km / h

the average speed is

v_average = $\frac{v_1 + v_2}{2}$

v_average = ( 40 - 60)/2

v_average = - 10 km / h

B) in this case they indicate the acceleration a = 3.2 m / s² and the velocity vo = 9.6 m / s

i) the speed for 2.5 s above

v = v₀ + a t

as the time is earlier

t = - 2.5 s

we substitute

v = 9.6 - 3.2 2.5

v = 1.6 m / s

ii) the velocity for a subsequent time of 2.5 s

t = 2.5 s

we substitute

v = 9.6 + 3.2 2.5

v = 17.6 m / s

3 0

2 years ago