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4 years ago

The density of a substance is defined as its

Physics

1 answer:

Fantom [35]4 years ago

7 0

<h2>Answer</h2>

<u>Density is the mass per unit volume</u>

<h2>Explanation</h2>

Density is actually is the ratio of the mass of an object to its volume. It is used to compare the different objects on the basis of their weight. You can say that more the weight of an object more will be its density. For example, the density has the role in floating the ships on the water. When we insert the wooden piece inside water it pops up to the surface. The weight of the wood piece is more than the water and volume is greater which affects its density. When the object gets smaller the size on pressing, the density of the object increase.

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Which element is usually listed first in the formula of a binary

Elina [12.6K]

Answer: the element with the lowest electronegativity

Explanation:

3 0

3 years ago

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr

fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

$\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}$

The values are

$\displaystyle y_o=15\ m$

$\displaystyle v_o=14\ m/s$

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

$\displaystyle y=0$

$\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0$

Multiplying by 2

$\displaystyle 2y_o+2v_ot-gt^2=0$

Clearing the coefficient of $t^2$

$\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0$

Plugging in the given values, we reach to a second-degree equation

$\displaystyle t^2-2.857t-3.061=0$

The equation has two roots, but we only keep the positive root

$\displaystyle \boxed {t=3.69\ s}$

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

$\displaystyle Y_r=15+V_r.t$

$\displaystyle Y_r=15+14\times3.69$

$\displaystyle \boxed{Y_r=66.62\ m}$

3 0

3 years ago

Consider Compton Scattering with visible light.A photon with wavelength 500nm scatters backward(theta=180degree) from a free ele

JulijaS [17]

Answer: $4.86(10)^{-12}m$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda'=500 nm=500(10)^{-9} m$ is the wavelength of the scattered photon

$\lambda_{o}$ is the wavelength of the incident photon

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=180\°$ the angle between incident phhoton and the scatered photon.

$\Delta \lambda=2.43(10)^{-12} m (1-cos(180\°))$ (2)

$\Delta \lambda=4.86(10)^{-12}m$ (3) This is the shift in wavelength

5 0

3 years ago

A ladder 7.60 m long leans against an exterior wall. If the ladder is inclined at an angle of 68.0° to the horizontal, what is t

nata0808 [166]

Answer:

2.85 m

Explanation:

From trigonometry,

Cosine = Adjacent/Hypotenuse

Assuming, The wall, the ladder and the ladder forms a right angle triangle as shown in fig 1, in the diagram attached below.

cos∅ = a/H....................... Equation 1

Where ∅ = Angle the ladder makes with the horizontal, a = The horizontal distance from the bottom of the ladder to the wall, H = The length of the ladder.

make a the subject of the equation

a = cos∅(H)..................... Equation 1

Given: ∅ = 68 °, H = 7.6 m.

Substitute into equation 2

a = cos(68)×7.6

a = 0.375×7.6

a = 2.85 m.

Hence the horizontal distance from the bottom of the ladder to the wall = 2.85 m

8 0

3 years ago

To determine an athlete's body fat, she is weighed first in air and then again while she's completely underwater. it is found th

il63 [147K]

The difference in weight is due to the displacement of water (the buoyancy of water is acting on the athlete thus giving her smaller weight).<span>

The amount of weight displaced or the amount of buoyant force is: </span>

Fb = 690 N - 48 N

Fb = 642 N

From newtons law, F = m*g. Using this formula, we can calculate for the mass of water displaced:

m of water displaced = 642N / 9.8m/s^2

m of water displaced = 65.5 kg

Assuming a water density of 1 kg/L, and using the formula volume = mass/density:

V of water displaced = 65.5kg / 1kg/L = 65.5 L

The volume of water displaced is equal to the volume of athlete. Therefore:

V of athlete = 65.5 L

The mass of athlete can also be calculated using, F = m*g

m of athlete = 690 N/ 9.8m/s^2

m of athlete = 70.41 kg

Knowing the volume and mass of athlete, her average density is therefore:

average density = 70.41 kg / 65.5 L

<span>average density = 1.07 kg/L = 1.07 g/mL</span>

5 0

3 years ago

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