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Alexxx [7]
3 years ago
5

PLSSSSSSS HELP MEHHHHH

Physics
1 answer:
Akimi4 [234]3 years ago
6 0

Answer:

I think no 3rd one is correct

Explanation:

It's my opinion maybe it's wrong

<em>*</em><em>*</em><em>*</em><em>*</em><em>*</em><em>*</em><em>Sorry</em><em> </em><em>if</em><em> </em><em>it</em><em> </em><em>is</em><em> </em><em>wrong</em><em> </em><em>*</em><em>*</em><em>*</em><em>*</em><em>*</em><em>*</em>

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What happens to ar object in free fall?​
Inessa05 [86]
Energy from the gravitational potential store in converted to kinetic energy. Air friction acts against the object, dissipating some energy as heat or sound. The object will continuously accelerate until the acceleration is equal to the air friction acting against it. This is when it reaches terminal velocity
5 0
3 years ago
Which of the following is not a reason why plants need water? *
solong [7]

Answer:

C

Explanation:

the plant cools itself down by allowing water to evaporate from their leaves so it doesn't need water to cool down

5 0
3 years ago
In calisthenics, the resistance is ultimately provided by __________. A. free weights B. springs and elastic bands C. gravity D.
tangare [24]

Answer:

I think the answer is b am sorry if it is wrong

Explanation:

5 0
3 years ago
Read 2 more answers
1. A rocket is fired vertically from the launch pad with a
zavuch27 [327]

Answer:

Max height= 36000 meters

Total Time = 120 seconds

Explanation:

0 = U - at

U = at

U= 20*60

U= 1200 m/s

MAX altitude would be

(U²Sin²tita)/2g

Max height= 1200² *( SIN90)²/(2*20)

Time of FLIGHT

2 * 1200/20

2400/20

120 sec onds

7 0
3 years ago
Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°
lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía (S), medida en joules por Kelvin, tenemos la siguiente expresión:

dS = \frac{\delta Q}{T} (1)

Donde:

Q - Ganancia de calor, en joules.

T - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

dS = \frac{L_{v}}{T}\cdot dm (1b)

Donde:

m - Masa del sistema, en kilogramos.

L_{v} - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

\Delta S = \frac{\Delta m \cdot L_{v}}{T}

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que m = 0.50\,kg,L_{v} = 2.7\times 10^{5}\,\frac{J}{kg} and T = 630.15\,K, entonces el cambio de entropía es:

\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}

\Delta S = 214.235 \,\frac{J}{K}

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0
3 years ago
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