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3 years ago

A ball is thrown in the air with a vertical velocity of 12 m/s.

Physics

2 answers:

melamori03 [73]3 years ago

4 0

The answer to the question is gonna be A

Ipatiy [6.2K]3 years ago

4 0

Answer:

a) 1.22 s

b) 6 m/s

c) 7.35 m

Explanation:

Given:

v₀ = 12 m/s

a = -9.8 m/s²

a) Find t when v = 0 m/s.

v = at + v₀

(0 m/s) = (-9.8 m/s²) t + (12 m/s)

t = 1.22 s

b) v_avg = ½ (v₀ + v)

v_avg = ½ (12 m/s + 0 m/s)

v_avg = 6 m/s

c) Find Δx.

v² = v₀² + 2aΔx

(0 m/s)² = (12 m/s)² + 2 (-9.8 m/s²) Δx

Δx = 7.35 m

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To achieve a speed of 2 m/s, the bottle must be dropped at m. To achieve a speed of 3 m/s, the bottle must be dropped at m. To a

klio [65]

Answer:

$\begin{array}{l|l}\text{Speed}\; \mathrm{(m\cdot s^{-1})} & \text{Minimum Height\;(m)}\\\cline{1-2}\\[-1em] 2 & 0.204\\3&0.459\\4 & 0.815\\5 & 1.27 \\6 & 1.83\end{array}$ .

Assumptions:

The object is dropped in a free fall.
There's no air resistance.
The downward acceleration due to gravity is $\rm 9.81\;m\cdot s^{-2}$

Explanation:

Consider the "SUVAT" equation

$\displaystyle \frac{v^{2} - u^{2}}{2a} = x$ ,

where

$v$ is the final velocity,
$u$ is the initial velocity,
$a$ is the acceleration of the object, and
$x$ is the change in the object's position.

For example, if the bottle needs to achieve a speed of $v = \rm 2\; m\cdot s^{-1}$ by the time it reaches the ground,

$u = 0$ since the statement that the bottle is "dropped" implies a free fall.
$a = g = \rm 9.81\;m\cdot s^{-2}$ .

Apply the previous equation to find the minimum height, $x$ :

$\displaystyle x = \frac{v^{2} - u^{2}}{2a} = \rm \frac{\left(2\; m\cdot s^{-1}\right)^{2}}{2\times 9.81\; m\cdot s^{-2}} \approx 0.204\; m$ .

Replace the $v$ value and apply the formula to find the minimum height required to reach different final speeds.

8 0

3 years ago

Read 2 more answers

10 kg of R-134a fill a 1.115-m^3 rigid container at an initial temperature of -30∘C. The container is then heated until the pres

Sidana [21]

Answer : The final temperature an the initial pressure of gas is, 273.6 K and 177.6 kPa

Explanation :

First we have to calculate the initial pressure of gas.

As we know that, R-134a is 1,1,1,2-Tetrafluoroethane.

Using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT$

where,

P = pressure of gas = ?

V = volume of gas = $1.115m^3$

T = temperature of gas = $-30^oC=273+(-30)=243K$

R = gas constant = $8.314m^3Pa/mole.K$

w = mass of gas = 10 kg = 10000 g

M = molar mass of R-134a gas = 102.03 g/mole

Now put all the given values in the ideal gas equation, we get:

$P\times 1.115m^3=\frac{10000g}{102.03g/mole}\times (8.314m^3Pa/mole.K)\times (243K)$

$P=177587.9687Pa=177.6kPa$

Thus, the initial pressure of gas is, 177.6 kPa

Now we have to calculate the final temperature of gas.

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$