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3 years ago

Identify three pieces of evidence that indicate the continents were once closer than they are today

Physics

1 answer:

nadya68 [22]3 years ago

3 0

Answer:

fit of the continents, paleoclimate indicators, truncated geologic features, and fossils.

Explanation:

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If a refrigerator is a heat pump that follows the first law of thermodynamics, how much heat was removed from food inside of the

Anna007 [38]

Idk I haven’t learned about this yet

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4 years ago

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A sound wave travels twice as far in neon (Ne) as it does it krypton (Kr) in the same time interval. Both neon and krypton can b

avanturin [10]

Answer:

281.6 K

Explanation:

The speed of sound in an ideal gas is given by c = √(γKT/m).

From the question speed of sound in Ne, c₁ = 2c₂ speed of sound in Kr

c₁ = √(γKT₁/m₁) and c₂ = √(γKT₂/m₂)

So √(γKT₁/m₁) = 2√(γKT₂/m₂) where T₁, m₁ and T₂, m₂ are the temperatures and atomic masses of Neon and Krypton respectively.

So, √(T₁/m₁) = 2√(T₂/m₂)

(T₁/m₁) = 4(T₂/m₂) (squaring both sides)

T₁ = 4(T₂m₁/m₂)

Given that m₁ = 20.2 u , m₂ = 83.8 u, T₂ = 292 K

T₁ = 4(292 × 20.2/83.8) K = 23593.6/83.8 = 281.55 K ≅ 281.6 K

6 0

3 years ago

Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Sim

Nonamiya [84]

Answer:

The distance covered by puck A before collision is $z = 8.56 \ m$

Explanation:

From the question we are told that

The label on the two hockey pucks is A and B

The distance between the two hockey pucks is D 18.0 m

The speed of puck A is $v_A = 3.90 \ m/s$

The speed of puck B is $v_B = 4.30 \ m/s$

The distance covered by puck A is mathematically represented as

$z = v_A * t$

=> $t = \frac{z}{v_A}$

The distance covered by puck B is mathematically represented as

$18 - z = v_B * t$

=> $t = \frac{18 - z}{v_B}$

Since the time take before collision is the same

$\frac{18 - z}{V_B} = \frac{z}{v_A}$

substituting values

$\frac{18 -z }{4.3} = \frac{z}{3.90}$

=> $70.2 - 3.90 z = 4.3 z$

=> $z = 8.56 \ m$

8 0

3 years ago

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between

Klio2033 [76]

Answer:

a) $u=0.05602\ m.s^{-1}$

b) $u'=0.0169\ m.s^{-1}$

Explanation:

Given:

mass of the ball, $m=0.4\ kg$

horizontal velocity of the ball, $u=10\ m.s^{-1}$

mass of the person, $M=71\ kg$

Using the law of conservation of momentum:

$m.v=(M+m).u$

$0.4\times 10=(71+0.4)\times u$

$u=0.05602\ m.s^{-1}$

Given:

rebound velocity of the ball, $v'=7\ m.s^{-1}$

Using conservation of momentum,

$m.v+M.u=m.v'+M.u'$

$0.4\times10+71\times 0=0.4\times 7+71\times u'$

$u'=0.0169\ m.s^{-1}$

4 0

3 years ago

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A 102 kg box is pushed on a linoleum floor with an acceleration of 2.5 ms to the second power. The force of the kinetic movie fr

maria [59]

A force is the push or pull acting on an object as a result of its interaction with another object. Here the net force acting on the box is 297 N

From the question

Given that,

Mass of the box ,m = 102 kg

Acceleration of the box ,a = 2.5 m/s²

Frictional force acting between the box and the floor ,f = 42 N

Here we have to find the net force acting on the body, F = ?

By Newton's second law of motion,

∑F = ma,

where, F is the net force

m is the mass of the body

a is the acceleration of the moving body

If the body is considered moving in the x-direction then,

F-f = ma

F - 42 = 102 x 2.5

= 255

F = 255 + 42 = 297 N

To learn more about force : brainly.com/question/12800231

#SPJ1

8 0

2 years ago