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umka2103 [35]
3 years ago
10

Where might you look on the internet to find good scientific information about illness

Physics
2 answers:
user100 [1]3 years ago
5 0
It depends on what illness and what country you are in but Mayo Clinic and Johns Hopkins are good sources.
Margarita [4]3 years ago
5 0
Just search "scientific information about illness"
Its simple thats what i do and i get A's
You might be interested in
A barbell is 1.5 m long. Three weights, each of mass 20 kg, are hung on the left and two weights of the same mass, on the right.
VMariaS [17]

Answer:

b. -11.6 cm

Explanation:

We have given parameters:

Length, l = 1.5 m = 150 cm

Mass of weight, m_1 = 20 kg

Width, x = 4 cm

Distance d = 4 cm

Mass of bar, m_{bar} = 5 kg

We are asked to find the center of mass from the mid-point, X_{CM} = ?

Since 3 weights are on the left and 2 weights are on the right, we know:

m_{left} = 3 * 20 = 60 kg

m_{right} = 2 * 20 = 40 kg

And also we know that, M = \frac{l}{2} = 150/2 = 75 cm

For the left side, center of mass is:

x_{left} = \frac{3 * 4}{2} = 6 cm

From the midpoint, the distance to the left is:

X_{left} = -(M - 4 - x_{left}) = -(75 - 4 -6) = -65 cm

For the right side, center of mass is:

x_{right} = \frac{2 * 4}{2} = 4 cm

From the midpoint, the distance to the right will be:

X_{right} = (M - 4 - x_{right}) = (75 - 4 - 4) = 67 cm

Hence,

X_{CM} = \frac{m_{right}*x_{right} + m_{left}*x_{left} }{m_{right} + m_{left} + m_{bar}} = \frac{40 * 67 - 60 * 65}{40 + 60 + 5} = -11.62 cm

4 0
3 years ago
Substance X is placed in a container with substance Y.
balandron [24]

Answer:

Substance X will rise due to convection.

Explanation:

4 0
3 years ago
A crane lifts a 425 kg steel beam vertically upward a distance of 66 m. How much work does the crane do on the beam if the beam
iogann1982 [59]

Answer:

W = 311074.5 [J]

Explanation:

In order to solve this problem we must analyze two parts, in the first part by means of Newton's second law we can determine the acceleration of the beam, remembering that the sum of the forces is equal to the product of mass by acceleration.

∑F = m*a

F = forces acting on the beam [N]

m = mass = 425 [kg]

a = acceleration = 1.8 [m/s²]

The forces acting on the beam are the force of the crane up (positive) and the weight of the beam down (negative)

F_{crane}-(425*9.81)= 425*1.8\\F_{crane}=4713.25 [N]

Now in the second part, we use the definition of work, which is equal to the product of the force applied in the direction of displacement, that is, the product of force by distance.

W=F*d

where:

W = work [J]

F = force = 4713.25 [N]

d = distance = 66 [m]

W=4713.25*66\\W=311074.5[J]

5 0
3 years ago
A sound source is located somewhere along the x-axis. Experiments show that the same wave front simultaneously reaches listeners
guapka [62]

Answer:

a)   x₀ = - 2 m  , b)     y = 4.47 m

Explanation:

A wave travels in the middle with constant speed, let's use the equation of uniform motion

    v = d / t

    t = d / v

The distance to the first listeners, see attached

    d₁ = x₀-x

     t = (x₀ +7) / v

The distance to the second listener

    d₂ = x - x₀

     t = (+ 3- x₀) / v

As the wave arrives at the same time, we can equal the two equations

     (x₀ +7) / v = (3 -x₀) / v

      x₀ + 7 = 3 - x₀

      2 x₀ = 3 - 7

      x₀ = -4/2

      x₀ = - 2 m

b) The time it takes for the wave to reach the listeners of the x-axis, where the speed of sound is 340 m / s

          t = 5/340

          t = 0.0147 s

Let's look for the distance the wave travels for the listener axis and

       v = d₃ / t

       d₃ = v.t

       d₃ = 340 * 0.0147

       d₃ = 5 m

For the distance component we use the Pythagorean triangle

      d₃² = x₀² + y²

      y² = d₃² - x₀²

     y = √ (d₃² -4)

      y = √ (5² -4)

     y = 4.47 m

6 0
3 years ago
A certain tuning fork vibrates at a frequency of 215 Hz while each tip of its two prongs has an amplitude of 0.832 mm. (a) What
Marysya12 [62]

Explanation:

It is given that,

Frequency of vibration, f = 215 Hz

Amplitude, A = 0.832 mm

(a) Let T is the period of this motion. It is given by the following relation as :

T=\dfrac{1}{f}

T=\dfrac{1}{215}

T=4.65\times 10^{-3}\ s

(b) Speed of sound in air, v = 343 m/s

It can be given by :

v=f\times \lambda

\lambda=\dfrac{v}{f}

\lambda=\dfrac{343\ m/s}{215\ Hz}

\lambda=1.59\ m

Hence, this is the required solution.

5 0
3 years ago
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