The law of conservation of mass applies to all chemical reactions.

A Cat is on a balcony floor (90cm below the railing), keenly eyeing a butterfly hovering 60 cm above the railing. With what spee

GenaCL600 [577]

We will have the following:

First, the equation to use is the following:

$d=v_ot+\frac{1}{2}at^2$

Now, we transform the total distance the cat would need to travel:

$90\operatorname{cm}+60\operatorname{cm}=150\operatorname{cm}\cdot\frac{1m}{100\operatorname{cm}}=1.5m$

So, the cat would need to travel 1.5 meters. ("d" in the equation).

Now, using the speed given we determine the time it would take the cat to traverse the 1.5 meters:

$t=\frac{1.5m\cdot1s}{0.45m}\Rightarrow t=\frac{10}{3}\Rightarrow t=3.333\ldots$

So, the time it would take the cat to traverse the distance will be approximately 3.33 seconds.

Now, we know that the acceleration will be given by Earth's gravity, so:

$1.5m=v_0(\frac{10}{3}s)+\frac{1}{2}(-\frac{9.8m}{s^2})(\frac{10}{3}s)^2\Rightarrow1.5m=v_0(\frac{10}{3}s)+(-\frac{490}{9}m)$ $\Rightarrow\frac{1007}{18}m=v_0(\frac{10}{3}s)\Rightarrow v_0=\frac{1007}{60}\frac{m}{s}\Rightarrow v_0=16.78333\ldots$

So, the initial vvelocity the cat must leave the floor in order to arrive at the butterfly with the optimum pouncing speed of 0.45 m/s is approximately 16.78 m/s or exactly 1007/60 m/s.

5 0

2 years ago

A 2.0 kilogram ball above the floor has 39 joules of potential energy the balls height is about meters above the floor

Akimi4 [234]

Answer:

the height h is 1.95 m above the floor

Explanation:

m = 2.0 kilogram

Ep = 39 J

g = 10 m/s²

h = ?

Ep = mgh

h = Ep ÷ mg

h = 39 ÷ (2×10)

h = 39÷20

h = 1.95 m

8 0

3 years ago

Read 2 more answers

The trajectory of a rock ejected from the Kilauea volcano, with a velocity of magnitude 6.4 m/s and at angle 2.2 degrees above t

Vika [28.1K]

Answer:

v = 8.03 m / s

Explanation:

This is a missile throwing exercise.

y = y₀ + $v_{oy}$ t - ½ g t²

indicates that y = -1.2 m and the initial velocity is

v_{oy} = v₀ sin θ

v_{oy} = 6.4 sin 2.2

v_{oy} = 0.2457 m / s

we substitute in the equation

-1.2 = 0.2457 t - ½ 9.8 t²

4.9 t² - 0.2457 t - 1.2 = 0

t² - 0.05014 t -0.2449 = 0

we solve the quadratic equation

t = [0.05014 ±√ (0.05014² + 4 0.2449)] / 2

t = [0.05014 ± 0.9910] / 2

t₁ = 0.5206 s

t₂ = -0.47 s

time must be a positive magnitude therefore the correct answer is

t = 0.5206 s

with this time we can calculate the vertical speed when the rock hits the ground

v_{y} = v_{oy} - gt

v_{y} = 0.2457 - 9.8 0.5206

v_{y} = -4.856 m / s

the negative sign indicates that the speed is down

horizontal velocity is constant, due to no acceleration

vₓ = v₀ₓ = v₀ cos 2,2

v₀ₓ = 6.4 cos 2.2

v₀ₓ = 6.395 m / s

therefore let's use Pythagoras' theorem to find the velocity

v = √ (vₓ² + $v_{y}^{2}$ )

v = √ (6,395² + 4,856²)

v = 8.03 m / s

the direction can be found with trigonometry

tan θ = v_{y} / vₓ

θ = tan⁻¹ (-4,856 / 6,395)

θ = - 37

the negative sign indicates that it is half clockwise from the x axis

4 0

3 years ago

Chapter 05, Problem 15 Multiple-Concept Example 7 and Concept Simulation 5.2 review the concepts that play a role in this proble

Llana [10]

Answer:

Explanation:

The question relates to motion on a circular path .

Let the radius of the circular path be R .

The centripetal force for circular motion is provided by frictional force

frictional force is equal to μmg , where μ is coefficient of friction and mg is weight

Equating cenrtipetal force and frictionl force in the case of car A

mv² / R = μmg

R = v² /μg

= 26.8 x 26.8 / .335 x 9.8

= 218.77 m

In case of moton of car B

mv² / R = μmg

v² = μRg

= .683 x 218.77x 9.8

= 1464.35

v = 38.26 m /s .

3 0

3 years ago

A graduated cylinder contains 62 ml of water. When a small metal block is added to the water, the volume of the water increases

lawyer [7]

Answer:

The density of the block is 7.4g/ml.

Explanation:

We can determine the volume of the metal block by taking the difference between the volumes measured in the graduated cylinder:

$V_{block}=65.5ml-62ml\\\\V_{block}=3.5ml$

Now, as we know that the average density of an object is calculated dividing its mass by its volume, we can calculate the density ρ of the metal block using the expression:

$\rho_{block}=\frac{m_{block}}{V_{block}}\\\\\rho_{block}=\frac{26g}{3.5ml}\\\\\rho_{block}=7.4\frac{g}{ml}$

Finally, it means that the density of the metal block is 7.4g/ml.

3 0

3 years ago

Read 2 more answers