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3 years ago

Give one example of something you use or make at home that is an example of solubility.

2 answers:

6 0

Answer: I'm not 100% sure if this is correct, but I think an example of solubility would be putting sugar into hot coffee or boiling water.

Explanation:

Nana76 [90]3 years ago

5 0

Answer:

Salt can dissolve in water making it an excellent soluble for water

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An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

m is the mass of object

$m=\dfrac{W}{g}$

$m=\dfrac{2.45}{9.8}$

m = 0.25 kg

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}$

k = 24.09 N/m

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0

3 years ago

Radio waves travel at a speed of 300,000,000 m/s. WFNX broadcasts radio waves at a

yKpoI14uk [10]

Answer:

Wavelength, $\lambda=2.94\ m$

Explanation:

It is given that,

Speed of radio waves is $v=3\times 10^8\ m/s$

Frequency of radio waves is f = 101,700,000 Hz

We need to find the wavelength of WFNX’s radio waves. The relation between wavelength, frequency and speed of a wave is given by :

$v=f\lambda$

$\lambda$ is wavelength

$\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{3\times 10^8}{101,700,000}\\\\\lambda=2.94\ m$

So, the wavelength of WFNX’s radio waves is 2.94 m.

3 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

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

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

$\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63$ seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

$D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m$ .

Now, for the maximum height, H, applying the equation of motion as

$v^2=u^2+2as$

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, $0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H$

$\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}$

$\Rightarrow H = 3.25 m$ .

Hence, the maximum height covered is 3.25 m.

8 0

3 years ago

What type of glass absorbs light and does not allow light to pass through it?

sesenic [268]

opaque glass does not allow light to pass through it.

8 0

3 years ago

Suppose the glass paperweight has index of refraction n=1.38. a) find the value of θ for which the reflection on the vertical su

zimovet [89]

Answer:

a)θ=71.89°

b)NO

Explanation:

Given that

For glass n= 1.38

We know that for air n'=1

The angle for total internal reflection θc given as

sin θc=n'/n

By putting the values

sin θc=n'/n

sin θc=1/1.38

θc=46.43°

n'sinθ = n sinθref

sinθref = cosθc

n'sinθ = n cosθc

1 x sinθ =1.38 x cos 46.43°

θ=71.89°

8 0

3 years ago