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MrRa [10]
2 years ago
15

Romeo lanza suavemente guijarros a la ventana de julieta y quiere que los guijarros golpeen la ventana solo con con un component

e horizontal de velocidad el esta parado en el extremo de un jardin de rosas 4.5m por abajo de la ventana y a 5.0m de la base de la pared cual es la rapidez de los guijarros cuando golpean la ventana
Physics
1 answer:
Yuki888 [10]2 years ago
6 0

Answer:

5.219\,\frac{m}{s}

Explanation:

Las condiciones del problema requieren el cálculo de la rapidez inicial de los guijarros. Se sabe que el componente vertical de la rapidez final es cero. Por tanto, el tiempo se determina a continuación: (The conditions of this problems require the calculation of the initial speed of the peebles. It is known that vertical component of the final speed is zero. Therefore, the time is determined herein:).

(0\,\frac{m}{s})^{2} = v_{o,y}^{2} - 2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.5\,m)

v_{o,y} = 9.395\,\frac{m}{s}

0\,\frac{m}{s} = 9.395\,\frac{m}{s} - \left(9.807\,\frac{m}{s^{2}} \right)\cdot \Delta t

\Delta t = 0.958\,s

Además, se determina el componente horizontal de la rapidez inicial (Likewise, the horizontal component of the initial speed is determined):

v_{o,x} = \frac{5\,m}{0.958\,s}

v_{o,x} = 5.219\,\frac{m}{s}

El guijarro tiene una rapidez de 5.219\,\frac{m}{s} cuando golpea la ventana (The peeble has a speed of  5.219\,\frac{m}{s} when it hits the window).

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A projectile follows a straight-line path instead of a parabolic trajectory. Which could be the launch angle? would it be 90 or
Gnesinka [82]
<em>projectile can only follow the straight line path when it is launched upward straightly so the correct option is <u>90 degree with respect to horizontal x -axis ..:)</u></em>
7 0
2 years ago
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There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame
slavikrds [6]

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Smaller diameter paintball = 5 cm

d₂ = Larger diameter paintball = 9 cm

V₂ = Volume of larger diameter paintball

Volume of smaller diameter paintball

V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3

Similarly

V_2=\frac{4}{24}\pi d_2^3

Dividing the above two equations, we get

\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3

∴ The larger one hold 163.296 cm³ of paint

5 0
2 years ago
Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. Wh
Alborosie

Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

                v = u + at

                   = 6.8 + 4.5 x 3.2

                   = 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

The initial velocity of Lisa, u = 12 m/s

The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

                          v² = u² +2aS

                         S = (v² - u²) / 2a

                            = (26² -12²) / 2 x 4.2

                            = 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

                        s = ut + 1/2 at²

                           =38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

                           = 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

4) Given,

The initial velocity of the car, u = 24.2 m/s

The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

                         v = u + at

∴                        a = (v - u) / t

                            = (11.9 - 24.2) / 2.85

                            = -4.32 m/s²

Hence, the deceleration of the car, a = = -4.32 m/s²

5 0
2 years ago
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a partially inflated weather balloon has a volume of 1.56 * 10^3 L and a pressure of 98.9 lPa. What is the volume of the balloon
e-lub [12.9K]
To solve the problem, we enumerate all the given first. Then the required and lastly the solution.

Given:

V1= 1.56x10^3 L = 1560 L                P2 = 44.1 kPa
P1 = 98.9 kPa

Required:  V2

Solution:

Assuming the gas is ideal. Ideal gas follows Boyle's Law which states that at a given temperature the product of pressure and volume of a gas is constant. In equation,

PV = k

Applying to the problem, we have
 
P1*V1 = P2*V2
(98.9 kPa)*(1560 L) = (44.1 kPa)*V2
V2 = 3498.5 L

<em>ANSWER: V2 = 3498.5 L</em>
7 0
2 years ago
A 4kg block is sliding on a horizontal frictionless floor at a speed of 2.5ms and runs into a horizontal spring. The spring has
alexgriva [62]

Answer:

Explanation:

Given

mas of block m=4\ kg

speed of block v=2.5\ m/s

spring constant k=30\ N-m

As the mass collides with the spring its kinetic energy is converted to the Elastic Potential energy of the spring

\frac{1}{2}mv^2=\frac{1}{2}kx^2

x=v\sqrt{\frac{m}{k}}

x=2.5\times \sqrt{\frac{4}{30}}

x=0.912\ m

8 0
2 years ago
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