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3 years ago

12

Q1. Fill in the blanks .(a) Monsoon winds carry .................... (dust/rain)(b) Moving air can provide a ...................

. (gravity/lift).(c) An image formed by a concave __________ cannot be obtained on a screen. .(d) An image formed by a __________ mirror is always of the same size as that of theobject

1 answer:

Musya8 [376]3 years ago

3 0

Monsoon is a blah blah hahahahah adding for someone

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Which is the quickest way to accurately take your heart rate

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I would say by putting two fingers under your chin or putting two fingers on the back of your wrist, hope i helped ! :)

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3 years ago

gas has a volume of 185 ml and pressure of 310 mm hg. The desiered volume is 74.0 ml. What is the required new pressure

Mamont248 [21]

Answer:

The required new pressure is 775 mm hg.

Explanation:

We are given that gas has a volume of 185 ml and a pressure of 310 mm hg. The desired volume is 74.0 ml.

We have to find the required new pressure.

Let the required new pressure be ' $\text{P}_2$ '.

As we know that Boyle's law formula states that;

$P_1 \times V_1 = P_2 \times V_2$

where, $P_1$ = original pressure of gas in the container = 310 mm hg

$P_2$ = required new pressure

$V_1$ = volume of gas in the container = 185 ml

$V_2$ = desired new volume of the gas = 74 ml

So, $P_2 = \frac{P_1 \times V_1}{V_2}$

$P_2 = \frac{310 \times 185}{74}$

= 775 mm hg

Hence, the required new pressure is 775 mm hg.

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3 years ago

The earplug can reduce the sound level to about 18 decibels (dB). What percentage reduction is this intensity?

zlopas [31]

Answer:

1 x 10 -10 whisper at 1m distance.

Explanation:

Properly fitted ear plugs an reduce noise form 15-30db. Although they are better for low frequency

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An airplane is moving at 350 km/hr. If a bomb is

bogdanovich [222]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final height

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb's initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity </h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

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The answer should be B, Condense.

Freezing is when a liquid transforms into a solid.
Sublimation is when a solid goes straight to a gas.
Evaporation is when a liquid transforms into a gas.

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