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3 years ago

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Which statement is true about vectors? a. All quantities in physics are not vectors. b. A vector may have either magnitude or di

rection. c. The vector's length should be proportional to its magnitude. d. Two vectors can be added only if they have the same direction.

1 answer:

dlinn [17]3 years ago

8 0

Yeah yeah ok yeah I saw

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Which has more kinetic energy a bowling ball or a soccer ball

lawyer [7]

A bowling ball because it is heavier and it has more air force going against it<span />

6 0

3 years ago

Read 2 more answers

which of the following statements describes the formation of acid rain?...A. acid rain forms from human activities and lowers th

olga_2 [115]

The correct answer is A. Acid rain forms from human activities and lowers the pH of rainwater below 7

Explanation:

The term "acid rain" is used to refer to rainwater with a low pH or a high acidity, this includes rainwater with a pH below 7 (solutions with a pH under 7 are acidic). In terms of formation, acid rain is mainly the result of human activities such as factories or cars that release pollutants into the atmosphere including nitric and sulfuric acids and these alter the natural pH of rain and makes it more acidic. Additionally, acid rain has negative consequences such as the death of fishes and other organisms in lakes, rivers, etc. because the acidity is toxic to many organisms. Thus, acid rain is the result of human activities and it lowers the pH of rain (Option A).

7 0

3 years ago

Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The ga

Radda [10]

Answer:

a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm

Explanation:

For isothermal expansion PV = constant

So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,

So, P₁V₁ = P₂V₂

P₂ = P₁V₁/V₂

Since V₂/V₁ = 0.19,

P₂ = P₁V₁/V₂

P₂ = 1 atm (1/0.19)

P₂ = 5.26 atm

For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas

So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,

So, P₂V₂ⁿ = P₃V₃ⁿ

P₃ = P₂V₂ⁿ/V₃ⁿ

P₃ = P₂(V₂/V₃)ⁿ

Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19

1/0.19,

P₃ = P₂(V₂/V₃)ⁿ

P₃ = 5.26 atm (0.19)⁽⁵/³⁾

P₃ = 5.26 atm × 0.0628

P₃ = 0.33 atm

Using the ideal gas equation

P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)

P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

T₃ = (0.33 atm/1 atm)(1)273 K

T₃ = 90.1 K

So,

a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

d. The lowest pressure attained by the gas is P₃ = 0.33 atm

6 0

3 years ago

F = 2.0*10^20 N,

natka813 [3]

$F=\dfrac{Gm_1m_2}{r^2}$

With the given values of $F,G,m_1,r$ , we have

$2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}$

Try dealing with the powers of 10 first: On the right, we have

$\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}$

Meanwhile, the other values on the right reduce to

$\dfrac{6.67\times5.98}{3.8^2}\approx2.76$

Then taking units into account, we end up with the equation

$2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2$

Now we solve for $m_2$ :

$m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}$

$m_2=7.25\times10^{22}\,\mathrm{kg}$

or, if taking significant digits into account,

$m_2=7.3\times10^{22}\,\mathrm{kg}$

5 0

3 years ago

The rectangular boat shown below has base dimensions 10.0 cm × 8.0 cm. Each cube has a mass of 40 g, and the liquid in the tank

Paladinen [302]

When boat is sunk into the liquid the net buoyancy on the boat is counterbalanced by weight of the boat

So here weight of the boat = Buoyancy force

let say boat is sunk by distance "h"

now we can say

$F_b = \rho * V * g$

$F_b = 1000*0.10 * 0.08 * h * 9.8$

now by above force balance equation we can write

$m*g = F_b$

$0.040 * 9.8 = 1000 * 0.10 * 0.08 * h * 9.8$

$0.040 = 8h$

$h = 5 * 10^{-3} m$

so boat will sunk by total 5 mm distance

8 0

3 years ago