What are you calculating when forces are added together

Which scientist called what he observed cells

never [62]

Answer:

Robert Hooke

Explanation:

The cell was first discovered by Robert Hooke in 1665 using a microscope. The first cell theory is credited to the work of Theodor Schwann and Matthias Jakob Schleiden in the 1830s.

7 0

4 years ago

There is evidence that elephants communicate via infrasound, generating rumbling vocalizations as low as 14 Hz that can travel u

amid [387]

Answer:

$L = 96.2 dB$

Explanation:

As we know that the level of intensity is given as

$\beta = 10Log(\frac{I}{I_o})$

here we know

$\beta = 104 dB$

$104 = 10Log(\frac{I}{10^{-12}})$

$I = 0.025 W/m^2$

now the sound is travelling in all possible directions so we have

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$\frac{0.025}{I_2} = \frac{10^2}{4.1^2}$

$I_2 = 4.2 \times 10^{-3} W/m^2$

now for level of sound we have

$L = 10Log(\frac{4.2 \times 10^{-3}}{10^{-12}})$

$L = 96.2 dB$

5 0

3 years ago

Uranium-238 (U238) has three more neutrons than uranium-235 (U235). Compared to the speed of sound in a bar of U235, is the spee

natta225 [31]

Answer:

The correct answers to then question are

A and B

Explanation:

As highlighted, the atoms of the two isotopes are chemically identical and they have similar characteristics due to the presence of equal number of protons and electrons (92 each). Therefore factors influenced by the chemistry of the atoms of the isotopes such as the distance between atoms and the stiffness of the inter-atomic spring is identical for both $U_{235}$ and $U_{238}$

An analogy of the spring constant and the inter-atomic distance is to consider the atoms of the metal to be interconnected by elastic material and as such the stiffness of the elastic material and the inter-atomic distance are determined by surface factors occurring on the surface

7 0

3 years ago

Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (

alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b) $E_1=\dfrac{\sigma}{\epsilon_0}$

c) $E_2=\dfrac{\sigma}{\epsilon_0}$

Explanation:

Let $\sigma$ be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

$E\times 2A=\dfrac{\sigma A}{\epsilon_0}\\E=\dfrac{\sigma}{2\epsilon_0}$

The Field will be away from the sheet and perpendicular to it.

a) The Electric Field between them

$E_1=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}\\=0$

b)The Electric Field to the right of the sheets

$E_1=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

c)The Electric Field to the left of the sheets

$E_2=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

3 0

3 years ago

If the car passes point A with a speed of 20 m/s and begins to increase its speed at a constant rate of at = 0.5 m/s2 , determin

IceJOKER [234]

Answer:

$1.68 \frac{m}{s^2}$

Explanation:

Please find the image for the question as attached file.

Solution -

Given -

First of all we will calculate the velocity at point C,

As per newton's third law of motion-

$V_C^2 = V_A^2 + 2 a_t (S_C - S_A)\\$

Substituting the given values in above equation, we get -

$V_C^2 = 20^2 + 2*0.5*(100-0)\\V_C = 22.361 \frac{m}{s}$

Now we will determine the radius of curvature for the curve shown in the attached image

$Y = 16 - \frac{1}{625} X^2\\$

Differentiating on both the sides, we get -

$\frac{dy}{dx} = -3.2 (10^-3) X\\\frac{d^2y}{d^2x} = -3.2 (10^-3)\\Curve = \frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}} }{\frac{d^2y}{d^2x}} \\Curve = 312.5$ meter

Acceleration on curved path

$a = \frac{V_C^2}{Curve} \\a = \frac{22.361^2}{312.5} \\a= 1.60 \frac{m}{s^2}$

Final acceleration

$a_f = \sqrt{0.5^2 + 1.6^2} \\a_f = 1.68\frac{m}{s^2}$

5 0

3 years ago