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2 years ago

Why does the moon Appear in the daytime????/?//

2 answers:

8 0

Answer: We can see the moon during the day for the same reason we see the moon at night. The surface of the moon is reflecting the sun's light into our eyes

Explanation:

VladimirAG [237]2 years ago

3 0

Answer:

The moon is visible in daytime nearly everyday

You might be interested in

Who is john Dalton ?

nata0808 [166]

Answer:

He is a English chemist physicist and meteorologist. He is best known for the atomic theory

3 0

3 years ago

Use Q = mcAT

mr Goodwill [35]

Answer:

1).....for the specific heat capacity(c) of water is 4200kg/J°C..

....guven mass(m)=320g(0.32kg)

...change in temperature(ΔT) =35°C

from the formula

Q=mcΔT

Q=0.32Kg x 4200kg/J°C x 35°C

Q=47,040Joules

5 0

3 years ago

Consider an electron with a mass of 9.11 x 1051 kg and a 100.0 g tennis ball that are both moving with a velocity of 70.0 m s1.

MrRissso [65]

Answer:

(a) 6.38 × 10⁻²⁹ kg·m·s⁻¹; (b) 7.00 kg·m·s⁻¹; (c) 82.7 µm; (d) 7.53 × 10⁻³⁴ m;

(e) Δx ∝ 1/m

Explanation:

(a) Momentum of electron

p = mv = 9.11 × 10⁻³¹ kg × 70.0 m·s⁻¹ = 6.38 × 10⁻²⁹ kg·m·s⁻¹

(b) Momentum of tennis ball

p = mv = 0.1000 kg × 70.0 m·s⁻¹ = 7.00 kg·m·s⁻¹

(c) Δx for electron

Δp = 0.010p = 0.010 × 6.38 × 10⁻²⁹ kg·m·s⁻¹ = 6.38 × 10⁻³¹ kg·m·s⁻¹

$\begin{array}{rcl}\Delta x \Delta p & \geq & \dfrac{h}{4 \pi}\\\\\Delta x \times 6.38 \times 10^{-31} \text{ kg$\cdot$m$\cdot$s$^{-1}$} & \geq & \dfrac{6.626 \times 10^{-34} \text{ kg$\cdot$m$^{2}$s}^{-1}}{4 \pi}\\\\\Delta x \times 6.38 \times 10^{-31} & \geq & 5.273 \times 10^{-35} \text{ m}\\\Delta x & \geq & \dfrac{5.273 \times 10^{-35} \text{ m}}{6.38 \times 10^{-31}}\\\\ & \geq&8.27 \times10^{-5} \text{ m}\\ &\geq&\textbf{82.7 $\mu$m}\\\end{array}$

(d) Δx for tennis ball

Δp = 0.010p = 0.010 × 7.00 kg·m·s⁻¹ = 0.0700 kg·m·s⁻¹

$\begin{array}{rcl}\Delta x \times 0.0700 & \geq & 5.273 \times 10^{-35} \text{ m}\\\Delta x & \geq & \dfrac{5.273 \times 10^{-35} \text{ m}}{0.0700}\\\\ &\geq& \textbf{7.53 $\mathbf{\times 10^{-34}}$ m}\\\end{array}$

(e) Relative uncertainty

Both particles are travelling at the same speed, so,

ΔxΔp = Δx × mv = mvΔx = constant

v is constant, so

Δx ∝ 1/m

Thus, the larger the mass of an object, the smaller the uncertainty in its velocity.

5 0

3 years ago

What is another name for this equipment and what does it do to reduce the impact of coal use?

slega [8]

Coal-fire power plants.

I hope this helps ya!

5 0

3 years ago

How many molecules are in 3 moles of potassium bromide (KBr)

sattari [20]

Answer:

Your strategy here will be to use the molar mass of potassium bromide,

KBr

, as a conversion factor to help you find the mass of three moles of this compound.

So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.

Potassium bromide is an ionic compound that is made up of potassium cations,

, and bromide anions,

−

. Essentially, one formula unit of potassium bromide contains a potassium atom and a bromine atom.

Use the periodic table to find the molar masses of these two elements. You will find

For K:

39.0963 g mol

−

For Br:

79.904 g mol

−

To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements

M KBr

39.0963 g mol

−

79.904 g mol

−

≈

119 g mol

−

So, if one mole of potassium bromide has a mas of

119 g

m it follows that three moles will have a mass of

moles KBr

⋅

molar mass of KBr



119 g

mole KBr

357 g

You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs

mass of 3 moles of KBr

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

360 g

∣

−−−−−−−−−

Explanation:

answer: 360 g 

6 0

3 years ago