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3 years ago

TRUE OR FALSE? earth revolves around the sun tilted on its axis

Physics

2 answers:

Alex777 [14]3 years ago

6 0

The axis of the Earth's rotation is tilted relative to the plain of the Earth's revolution around the Sun.

The question is worded very poorly, but you'd have to say it's TRUE.

mina [271]3 years ago

4 0

True. The tilt of the axis is what leads to varying seasons around the year.

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HELP ME IM IN 3TH GRADE AND ONLINE SCHOOL IS POOP

snow_lady [41]

Answer:

1. evaporation

2. sublimation

3. melting

4. freezing

5. deposition

3. condensation

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3 years ago

7Which of the following terms describes how glaciers move?

tatuchka [14]

Answer:

D is the answer I think (0 w 0 )

Explanation:

7 0

4 years ago

Read 2 more answers

A 0.850 kg block is attached to a spring with spring constant 18 N/m . While the block is sitting at rest, a student hits it wit

Sveta_85 [38]

The block's speed at the point where x=0.25A is v = 31.95 cm/s.

<h3>What is Spring constant?</h3>

The spring stiffness is quantified by the spring constant, or k. For various springs and materials, it varies. The stiffer the spring is and the harder it is to stretch, the bigger the spring constant.

question is incomplete, this is the remaining statement

What is the amplitude of the subsequent oscillations? And What is the block's speed at the point where x=0.25A?

x = Asin(wt)

v = Aw coswt

at t = 0

w = sqrt(k/m)

v = Aw

A = v/w

A = 7.17 cm

part b )

E = 1/2mv^2 + 1/2kx^2 = 1/2kA^2

mv^2 + k(1/4A)^2 = 1/2kA^2

mv^2 + kA^2/16 = kA^2

mv^2 = kA^2 - kA^2/16

mv^2 = 15kA^2/16

v^2 = 15/16 * (k/m) * A^2

v^2 = 15/16 *w^2A^2

v = sqrt(15/16) * wA

v = 31.95 cm/s

to learn more about spring constant go to -

brainly.com/question/23885190

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2 years ago

A 4.0 Ohms resistor, an 8.0 Ohms resistor, and a 12.0 Ohms resistor are connected in series with a 24.0 V battery. Calculate the

Flura [38]

Answer:

RESISTANCE=(4+8+12)Ohms

=24 ohms

V=IR

24= I ×24

I= 24/24

I= 1 A

CURRENT IS 1 A

8 0

3 years ago

Read 2 more answers

An object is moving along a straight line, and the uncertainty in its position is 1.90 m.

just olya [345]

Answer:

$2.78\times 10^{-35}\ \text{kg m/s}$

$6.178\times 10^{-34}\ \text{m/s}$

$0.31\times 10^{-4}\ \text{m/s}$

Explanation:

$\Delta x$ = Uncertainty in position = 1.9 m

$\Delta p$ = Uncertainty in momentum

h = Planck's constant = $6.626\times 10^{-34}\ \text{Js}$

m = Mass of object

From Heisenberg's uncertainty principle we know

$\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}$

The minimum uncertainty in the momentum of the object is $2.78\times 10^{-35}\ \text{kg m/s}$

Golf ball minimum uncertainty in the momentum of the object

$m=0.045\ \text{kg}$

Uncertainty in velocity is given by

$\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}$

The minimum uncertainty in the object's velocity is $6.178\times 10^{-34}\ \text{m/s}$

Electron

$m=9.11\times 10^{-31}\ \text{kg}$

$\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}$

The minimum uncertainty in the object's velocity is $0.31\times 10^{-4}\ \text{m/s}$ .

6 0

3 years ago