Physics, 18.10.2020 16:01 tryintopassenioryear

A ball is thrown upwards into the air with an initial velocity of 17.5 m/s. What is the maximum height that the ball traveled
during that time (assuming there is no air resistance). Round your
answer to two decimals places.

Answers: 2

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Answer from: Aced

y = 15.61 [m]

Explanation:

To solve this problem we must use the following kinematics equation. We must bear in mind that the acceleration of gravity is negative since the direction of gravity is opposite to the direction of motion.

$v_{f} ^{2} = v_{i} ^{2} -(2*g*y)$

where:

Vf = final velocity = 0

Vi = initial velocity = 17.5 [m/s]

g = gravity acceleration = 9.81 [m/s^2]

y = height [m]

The final speed is zero, since the ball reaches the maximum height its speed will be zero since it cannot be lifted up more.

0 = (17.5)^2 - (2*9.81*y)

306.25 = 2*9.81*y

y = 15.61 [m]

Answer from: Quest

it is given that trillions of microbes make up to 1% to 3% of the body mass.

let's take average of this percentage: 2%

consider the average mass of the human body as 70 kg.

then, trillions of microbes would have mass $= 2\%\hspace{1mm}of \hspace{1mm} 70\hspace{1mm} kg=\frac {2}{100}\times 70 =1.4\hspace{1mm} kg$

therefore, 100 trillion micro-organisms would weigh:

$\rightarrow 100\times1.4 kg = 140 kg$

this is an approximate value and could change depending on the mass of the human body and percentage of the microbes present in it.

average mass per microorganism = $\frac{1.4 kg}{10^12}=1.4\times10^{-12} kg$

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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A ball is thrown upwards into the air with an initial velocity of 17.5 m/s. What is the maximum hei...

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