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Firdavs [7]
1 year ago
15

2) A tomato is launched upwards with a velocity of 27 m/s. What is its height after 2.1 seconds in the air?

Physics
1 answer:
olga_2 [115]1 year ago
8 0

The height of the tomato in air after 2.1 seconds is 78.31 m.

<h3 /><h3>What is height?</h3>

Height can be defined as vertical distance either between the top and bottom of something or between a base and something above it.

To calculate the height of the tomato after 2.1 seconds, we use the formula below.

Formula:

  • H = ut+gt²/2............. Equation 1

Where:

  • H = Height of the tomato
  • u = Initial velocity = 27 m/s
  • g = Acceleration due to gravity = Constant = 9.8 m/s²
  • t = 2.1 seconds

Substitute these values into equation 1

  • H = (27×2.1)+(9.8×2.1²)/2
  • H = 56.7+21.609
  • H = 78.31 m

Hence, the height of the tomato after 2.1 seconds is 78.31 m.

Learn more about height here: brainly.com/question/28122539

#SPJ1

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When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas
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<h2>Answer: its temperature must increase.</h2>

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure \Delta P is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

\Delta U=\Delta Q-\Delta W   (1)

Where:

\Delta U is the internal energy

\Delta Q is the heat transferred

\Delta W is the work

Now, for an isobaric process:

\Delta W=P\Delta V    (2)

Where:

P is the pressure (<u>always positive</u>)

\Delta V is the volume variation of the gas

<u />

<u>Here we have two possible results:</u>

-If the gas expands (positive \Delta V), the work is positive.

-If the gas compresses (negative \Delta V), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

\Delta U=\Delta Q - P\Delta V   (3)

Clearing \Delta Q:

\Delta Q=\Delta U+P \Delta V    (4)

Then, for an ideal gas in an isobaric process, part of the heat (Q) added to the system will be used to do work (positive in this case) and the other part <u>will increase the internal energy</u>, hence <u>the temperature will increase as well.</u>

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3 years ago
A thin walled spherical shell is rolling on a surface. What
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=\frac{1/3}{5/6} = 0.4

Explanation:

Moment of inertia of given shell= \frac{2}{3} MR^2

where

M represent sphere mass

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we know linear speed is given as v = r\omega

translational K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2

rotational K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{2}{3} MR^2 \omega^2

total kinetic energy will be

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K.E =\frac{5}{6} MR^2 \omega^2

fraction of rotaional to total K.E

=\frac{1/3}{5/6} = 0.4

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Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s
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Answer:

42m/s

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Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}

with this value of vo you calculate the max time:

t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

- - - - - - - - - - - - -- - - - - - - - - - - - - -

TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

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