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3 years ago

List three ways the earth is different today from when it was first formed

2 answers:

8 0

<span>Huge mountains have formed, been destroyed, and replaced with new mountains. The oceans have opened up and moved around the globe. The continents have moved around, split apart from each other, and collided with each other, until finally reaching their present locations.

Hope I hlped :)

-Hind</span>

Law Incorporation [45]3 years ago

6 0

It is polluted, its bigger, and it has way more land.

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A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsul

jeka94

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

<h3>Given:</h3>

Mass of the astronaut, $m_a$ = 126 kg

Speed he acquires, $v_{a}$ = 2.70 m/s

Mass of the space capsule, $m_{c}$ = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

$P_f = m_av_a + m_cv_c$

$P_I$ = 0

$m_av_a + m_cv_c = 0$

$v_c =\frac{- m_a v_a}{m_c}}\\\\$

$= \frac{126* 2.70}{1800}$

$= - 0.189$ m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

= 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

= 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = $\frac{1}{2} m v^2$

$= \frac{1}{2}$ × 126 × $(2.70) ^2$

= 459.27 J

The Kinetic Energy of the capsule;

K.E = $\frac{1}{2} m v^2$

= $\frac{1}{2}$ ×1800× $(0.189) ^2$

= 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Learn more about kinetic energy here:

brainly.com/question/26520543

#SPJ1

3 0

2 years ago

A bullet of mass 0.1 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 3 kg that is sitting at re

Xelga [282]

Answer:

(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s

(b) the kinetic energy of the bullet plus the block before the collision is 500J

Explanation:

Given;

mass of bullet, m₁ = 0.1 kg

initial speed of bullet, u₁ = 100 m/s

mass of block, m₂ = 3 kg

initial speed of block, u₂ = 0

Part (A)

Applying the principle of conservation linear momentum, for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the block after the bullet embeds itself in the block

(0.1 x 100) + (3 x 0) = v (0.1 + 3)

10 = 3.1v

v = 10/3.1

v = 3.226 m/s

Part (B)

Initial Kinetic energy

Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²

Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)

Ki = 500 + 0

Ki = 500 J

Part (C)

Final kinetic energy

Kf = ¹/₂m₁v² + ¹/₂m₂v²

Kf = ¹/₂v²(m₁ + m₂)

Kf = ¹/₂ x 3.226²(0.1 + 3)

Kf = ¹/₂ x 3.226²(3.1)

Kf = 16.13 J

6 0

3 years ago

Can you explain hooks law in simple words?

dedylja [7]

Answer:

The extension of a material or a spring is its increase in length when pulled. Hooke’s Law says that the extension of an elastic object is directly proportional to the force applied to it. In other words:

Explanation:

7 0

2 years ago

Read 2 more answers

what is the total distance traveled by a bike rider who rides for three hours at 40 km/hr and then two more hours at 50 km/hr?

marta [7]

Answer:

220 km

Explanation:

3 hours at 40 km/hr

40 km = 1 hour

40 km times 3 = 120 km

2 hours at 50 km/hr

50 km = 1 hour

50 times 2 = 100 km

Total distance

120 km + 100 km = 220 km

4 0

3 years ago

A particle of mass m moves under a conservative force with potential energy V ( x )= cx/(x2+a2),where c and a are positive const

Anvisha [2.4K]

Answer:

The position of stable equilibrium is -a

And the period of small oscillations must be: c/(ma^3)

Explanation:

Since the potential is:

$V(x) = \frac{c x}{a^2+x^2}$

We first look for a position of stable equilibrium. This posiiton must satisfy two considtions, that the first derivative of the potential must vanish at this point and the second derivative must be positive.

$V'(x) = \frac{c}{a^2+x^2}-\frac{2 c x^2}{\left(a^2+x^2\right)^2}$

Which vanish for

x = a ; x =-a

The second derivative of V(x) is:

$V''(x) = \frac{8 c x^3}{\left(a^2+x^2\right)^3}-\frac{6 c x}{\left(a^2+x^2\right)^2}$

And:

$V''(a) = -\frac{c}{2 a^3}\\V''(-a) = \frac{c}{2 a^3}\\$

Therefore:

The position of stable equilibrium is -a

And the period of small oscillations must be:

$\omega = \sqrt{2 V''(-a)/m} = \sqrt{\frac{c}{a^3 m}}$

(c/(ma^3))^1/2

Let's find the maximum amplitude if the particle starts at this point with velocity v

If this is the case, the total energy will be:

(mv^2)/2

And the maximum amplitude will be

x = a^3/c mv^2 = (m v^2 a^3)/ c

7 0

3 years ago