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3 years ago

Rita throws a ball straight up into the air and catches it at the

Physics

2 answers:

Kaylis [27]3 years ago

6 0

(1) The potential energy at the top of the ball’s motion is 18 J.

(3) The kinetic energy increases as the potential energy decreases.

(4) The kinetic energy decreases as the potential energy increases.

(5) The total mechanical energy of the ball stays constant.

Explanation:

The total mechanical energy of the ball is equal to the sum of its kinetic energy (K, energy due to the motion) and its potential energy (U, energy due to the height of the ball). Mathematically:

$E=K+U$

In absence of friction, the mechanical energy of the ball is conserved, so in this case, it is always equal to 18 J. Let's now use this information to analyze each of the given statements:

(1) The potential energy at the top of the ball’s motion is 18 J. --> TRUE. In fact, at the top, the ball's speed becomes zero, so its kinetic energy is zero: K = 0. This means that all the mechanical energy of the ball is potential energy, therefore

E = U = 18 J

(2) The kinetic energy is less when the ball is thrown than when it is caught. --> FALSE. As we said, in absence of friction, the mechanical energy is conserved, therefore it always remains equal to 18 J.

(3) The kinetic energy increases as the potential energy decreases. --> TRUE. As we said, the sum of potential+kinetic energy remains constant:

E = K + U = 18 J

therefore, when the potential energy decreases, the kinetic energy increases.

(4)The kinetic energy decreases as the potential energy increases. --> TRUE. For the same reason described in (3).

(5)The total mechanical energy of the ball stays constant. --> TRUE. As we said at the beginning, the total mechanical energy is constant.

(6) The mechanical energy decreases as the ball moves up and increases as the ball comes down. --> FALSE. As we said, the mechanical energy remains constant, so it cannot change.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

vfiekz [6]3 years ago

4 0

Answer:

1, 3, 4, 5

Explanation:

if this helps just leave a thanks an friend me if it did.

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3 years ago

An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in

Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

$T {V}^{\gamma - 1} = \text{constant}$

$\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}$

Given:-

${T}_{1} = 275 \; K$

${T}_{2} = T \left( \text{say} \right)$

${V}_{1} = V$

${V}_{2} = 6V$

$\gamma = \cfrac{5}{3} \;$ (the gas is monoatomic)

$\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}$

$\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}$

T = 275 $\times$ 0.30

T = 83 K.

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3 years ago

A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS

Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;

transferred energy + generated energy = 0

(radiation + convection) + generated energy = 0

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0$

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s$

(a) the electrical power generated for still summer day

$T_s = T_{\infty} = 35 ^oC = 308 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W$

(b)the electrical power generated for a breezy winter day

$T_s = T_{\infty} = -15 ^oC = 258 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k$