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3 years ago

The volume of a gas depends upon its mass true or false

1 answer:

4 0

I would say false because the volume of gas depends on several things the amount of matter in the volume , the pressure and the temperature. So the answer would be false in my opinion. Because mass is not the only factor

You might be interested in

What is the weight of a 2.06 kg text book. show work please

salantis [7]

The weight of anything is (mass) x (gravity in the place where the thing is).

   Gravity on Earth is about 9.8 newtons per kilogram, so on Earth,
   the weight of the book is

(2.06 kg) x (9.8 N/kg) = <em>20.19 newtons</em> .

Another way to get around it is:

   On Earth, 1 kg of mass weighs about 2.205 pounds, so on earth,
   the weight of the book is

   (2.06 kg) x (2.205 pounds/kg) = <em>4.54 pounds </em>.

7 0

4 years ago

A loudspeaker, mounted on a tall pole, is engineered to emit 75% of its sound energy into the forward hemisphere, 25% toward the

Naily [24]

Answer:

"0.049 W" is the correct answer.

Explanation:

According to the given question,

$r = \sqrt{(3.5)^2+(2.5)^2}$

$=\sqrt{8.5}$

$SL=85$

As we know,

⇒ $SL=10 \ log(\frac{I}{I_o} )$

$85=10 \ log(\frac{I}{10^{-12}} )$

$I=3.162\times 1^{-4} \ W/m^2$

Now,

⇒ $P_{front} = I(2\pi r^2)$

$=(3.162\times 10^{-4})(2\pi\times 18.5)$

$=0.0368 \ W$

$=0.75 \ P$

or,

$=0.049 \ W$

7 0

3 years ago

How long does it take light from a camera to travel to the face of someone standing 7 meters away?

mestny [16]

The time taken for the light to travel from the camera to someone standing 7 m away is 2.33×10¯⁸ s

Speed is simply defined as the distance travelled per unit time. Mathematically, it is expressed as:

<h3>Speed = distance / time </h3>

With the above formula, we can obtain the time taken for the light to travel from the camera to someone standing 7 m away. This can be obtained as follow:

Distance = 7 m

Speed of light = 3×10⁸ m/s

Time = Distance / speed

Time = 7 / 3×10⁸

Therefore, the time taken for the light to travel from the camera to someone standing 7 m away is 2.33×10¯⁸ s

Learn more: brainly.com/question/14988345

8 0

2 years ago

Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0

BaLLatris [955]

Answer:

Part a)

$r = 6.96 \times 10^6 m$

Part b)

$F_g = 3.004 \times 10^3 N$

Part c)

$a = 0.88 m/s^2$

Part d)

$v = \frac{2480}{2} = 1240 m/s$

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

$v = 2480 m/s$

now we will have

$v = \sqrt{\frac{GM_{mars}}{r}}$

now we have

$M_{mars} = 6.4191 \times 10^{23} kg$

$R_{mars} = 3.397 \times 10^6 m$

now we have

$2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}$

$r = 6.96 \times 10^6 m$

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

$F_g = \frac{GM_{mars} m}{r^2}$

$F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}$

$F_g = 3.004 \times 10^3 N$

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

$a = \frac{F_g}{m}$

$a = \frac{3004}{3400}$

$a = 0.88 m/s^2$

Part d)

As we know by III law of kepler

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

here we know that T2 = 8 T1

$(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}$

$\frac{r_1}{r_2} = (\frac{1}{2})^2$

so we have

$r_2 = 4r_1$

as we know that speed is given as

$v = \sqrt{\frac{GM}{r}}$

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

$v = \frac{2480}{2} = 1240 m/s$

7 0

4 years ago

A student of mass 63.4 kg, starting at rest, slides down a slide 21.2 m long, tilted at an angle of 26.1° with respect to thehor

Vera_Pavlovna [14]

Answer:

fr = 65.46 N , a = 8.74 m / s² and vf = 19.25 m / s

Explanation:

We write a reference system with an axis parallel to the slide and gold perpendicular axis, in this system we decompose the weight

sin 21.2 = Wx / W

cos21.2 = Wy / W

Wx = W sin21.2

Wy = W cos 21.2

We form Newton's equations

X axis

Wx -fr = m a

Y Axis

N- Wy = 0

N = Wy

fr = μ N

fr = μ (W cos 21.2)

fr = 0.113 63.4 9.8 cos 21.2

fr = 65.46 N

We replace and calculate the acceleration

W sin 21.2 - μ W cos 21.2 = m a

a = g (sin21.2 - μ cos 21.2)

a = 9.8 (without 21.2 - 0.113 cos 21.2)

a = 8.74 m / s²

This acceleration is along the slope of the slide, so we can calculate the distance

d = 21.2 m

vf² = v₀² + 2 a d

vf² = 0 + 2 a d

vf = √(2 8.74 21.2)

vf = √ (370,576)

vf = 19.25 m / s

3 0

3 years ago