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3 years ago

Which would sound travel faster through: the ocean, the air, or a rock? Why?

Physics

1 answer:

8090 [49]3 years ago

7 0

Answer:

the air

Explanation:

this is because sound is carried away by particles which are vibrating.

the lighter the particles, the faster the vibration and also the faster the sound wave.

air is the lightest among the options been given.

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A spring scale hung from the ceiling stretches by 6.4 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl

dolphi86 [110]

In the first case, the force acting on the spring is the weight of the mass:
$F=mg=(2.0 kg)(9.81 m/s^2)=19.6N$
This force causes a stretching of $x=6.4 cm=0.064 m$ on the spring, so we can use these data to find the spring constant:
$k= \frac{F}{x}= \frac{19.6 N}{0.064 m}=306.3 N/m$

In the second case, the first mass is replaced with a second mass, whose weight is
$F=mg=(2.5 kg)(9.81 m/s^2)=24.5 N$
And since we know the spring constant, we can calculate the new elongation of the spring:
$x= \frac{F}{k}= \frac{24.5 N}{306.3 N/m}=0.080 m=8.0 cm$

5 0

3 years ago

For the following elementary reaction 2br• -> br2-. The rate of consumption of the reaction and the rate of formation of prod

Scorpion4ik [409]

Answer: $-\frac{1}{2}\times \frac{d[Br^.]}{dt}=+\frac{d[Br_2]}{dt}$

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

$2Br^.\rightarrow Br_2$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{d[Br^.]}{2dt}$

or $Rate=+\frac{d[Br_2]}{dt}$

Thus $-\frac{d[Br^.]}{2dt}=+\frac{d[Br_2]}{dt}$

4 0

3 years ago

When two plates of differing densities collide, how might the density of each plate affect which plate is pulled beneath the oth

OlgaM077 [116]

Pull the plates apart and you will knwo what it is lmaoo

5 0

3 years ago

Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only

Ne4ueva [31]

Answer:

(a)Therefore the highest altitude attained by the object is =576 ft .

(b)Therefore the object takes 6 sec to fall to the ground.

Explanation:

Initial velocity: Initial velocity is a velocity from which an object starts to move.

u is usually used for notation of initial notation.

Final velocity: Final velocity is a velocity of an object after certain second from starting.

The final velocity is denoted by v.

Acceleration: The difference of final velocity and initial velocity per unit time

The S.I unit of acceleration is m/s².

(a)

Given that u= 128 ft\sec and g = 32 ft/sec².

At highest point the velocity of the object is 0 i.e v=0

Since the displacement is opposite to the gravity.

Therefore acceleration( a)= -g = -32 ft/sec².

To find the time this to happen we use the following formula

$v=u+at$

Here v=0

⇒0=128+(-32) t

⇒32t=128

⇒t = 4 sec

To determine the height we use the following formula

$s=ut+\frac{1}{2} at^2$

$\Rightarrow s= (128\times4)+\frac{1}{2}\times (-32) \times4^2$

⇒s= 256 ft

Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .

(b)

At the highest point the velocity of the object is 0.

so u=0. a=g= 32 ft/sec² [ since the direction of gravity and the displacement are same] s= 576 ft

To determine the time to fall we use the following formula

$s=ut+\frac{1}{2} at^2$

$\Rightarrow 576 = (0\times t)+\frac{1}{2} \times 32 \times t^2$

$\Rightarrow 16\times t^2=576$

$\Rightarrow t^2=\frac{576}{16}$

$\Rightarrow t^2=36$

⇒t=6 sec

Therefore the object takes 6 sec to fall to the ground.

8 0

3 years ago

Three equal 1.55-μC point charges are placed at the corners of an equilateral triangle whose sides are 0.500 m long. What is the

kati45 [8]

Answer:

0.12959085 J

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = 1.55 μC

d = Distance between charge = 0.5 m

Electric potential energy is given by

$U=k\dfrac{q^2}{d}$

In this system with three charges which are equidistant from each other

$U=k\dfrac{q^2}{d}+k\dfrac{q^2}{d}+k\dfrac{q^2}{d}$

$\\\Rightarrow U=k\dfrac{3q^2}{d}\\\Rightarrow U=8.99\times 10^9\times \dfrac{3\times (1.55\times 10^{-6})^2}{0.5}\\\Rightarrow U=0.12959085\ J$

The potential energy of the system is 0.12959085 J

6 0

3 years ago