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g100num [7]
2 years ago
13

A piece of a metal alloy with a mass of 114 g was placed into a graduated cylinder that contained 25.0 mL of water, raising the

water level to 42.5 mL. What is the density of the metal?
Physics
1 answer:
Minchanka [31]2 years ago
8 0

Answer:

6.51 g/c.c

Explanation:

mass, m = 114 g

initial volume, V1 = 25 mL

final volume, V2 = 42.5 mL

Volume of the metal piece, V = V2 - V1 = 42.5 - 25 = 17.5 mL

1 mL = 1 c.c

So, Volume of metal, V = 17.5 c.c.

Let the density of the metal is d.

density = mass / volume

d = 114 / 17.5 = 6.51 g/c.c

Thus, the density of metal is 6.51 g/c.c.

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What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22
SIZIF [17.4K]

Answer:

5.45\times 10^{-4} W

Explanation:

T_{r} = Temperature of the room = 22.0 °C = 22 + 273 = 295 K

T_{s} = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K

A = Surface area = 1.50 m²

\epsilon = emissivity = 0.97

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴

Rate of heat transfer is given as

R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})

R = (0.97)(5.67\times 10^{-8}) (1.50) ((306)^{2} - (295)^{2})

R = 5.45\times 10^{-4} W

3 0
2 years ago
A water wave traveling in a straight line on a lake is described by the equation:y(x,t)=(2.75cm)cos(0.410rad/cm x+6.20rad/s t)Wh
Georgia [21]

Answer:

A) The wave equation is defined as

y(x,t) = A\cos(kx + \omega t)=0.0275\cos(0.0041x + 6.2t)\\

Using the wave equation we can deduce the wave number and the angular velocity. k = 0.0041 and ω = 6.2.

The time it takes for one complete wave pattern to go past a fisherman is period.

\omega = 2\pi f\\ f = 1/ T

T = 1.01 s.

The horizontal distance the wave crest traveled in one period is

\lambda = 2\pi / k = 2\pi / 0.0041 = 1.53\times 10^3~m

y(x = \lambda,t = T) = 0.0275\cos(0.0041*1.53*\10^3 + 6.2*1.01) = 0.0275~m

B) The wave number, k = 0.0041 . The number of waves per second is the frequency, so f = 0.987.

C) A wave crest travels past the fisherman with the following speed

v = \lambda f = 1.53\times 10^3 * 0.987 = 1.51\times 10^3~m/s

The maximum speed of the cork floater can be calculated as follows.

The velocity of the wave crest is the derivative of the position with respect to time.

v(x,t) = \frac{dy(x,t)}{dt} = -(6.2\times 0.0275)\sin(0.0041x + 6.2t)

The maximum velocity can be found by setting the derivative of the velocity to zero.

\frac{dv_y(x,t)}{dt} = -(6.2)^2(0.0275)\cos(0.0041*1.53\times 10^3 + 6.2t) = 0

In order this to be zero, cosine term must be equal to zero.

0.0041*1.53\times 10^3 + 6.2t = 5\pi /2\\t = 0.255~s

The reason that cosine term is set to be 5π/2 is that time cannot be zero. For π/2 and 3π/2, t<0.

v(x=\lambda, t = 0.255) = -(6.2\times0.0275)\sin(0.0041\times 1.53\times 10^3 + 6.2\times 0.255) = -0.17~m/s

6 0
2 years ago
A proposed space elevator would consist of a cable stretching from the earth's surface to a satellite, orbiting far in space, th
NISA [10]

To solve this problem we will apply the concepts related to energy conservation. Here we will use the conservation between the potential gravitational energy and the kinetic energy to determine the velocity of this escape. The gravitational potential energy can be expressed as,

PE= \frac{GMm}{d}

The kinetic energy can be written as,

KE= \frac{1}{2} mv^2

Where,

G = 6.67*10^{-11}m^3/kg\cdot s^2Gravitational Universal Constant

m = 5.972*10^{24}kg Mass of Earth

h = 56*10^6m  Height

r = 6.378*10^6m Radius of Earth

From the conservation of energy:

\frac{1}{2} mv^2 = \frac{GMm}{d}

Rearranging to find the velocity,

v = \sqrt{\frac{2Gm}{d}} \rightarrow  Escape velocity at a certain height from the earth

If the height of the satellite from the earth is h, then the total distance would be the radius of the earth and the eight,

d = r+h

v = \sqrt{\frac{2Gm}{r+h}}

Replacing the values we have that

v = \frac{2(6.67*10^{-11})(5.972*10^{24})}{6.378*10^6+56*10^6}

v = 3.6km/s

Therefore the escape velocity is 3.6km/s

3 0
2 years ago
In a harbor, you can see sea waves traveling around the edges of small stationary boats. Why does this happen?
faust18 [17]
Below are the choices that can be found in the other sources:

A. diffraction 
<span>B. refraction </span>
<span>C. reflection </span>
<span>D. transmission
</span>
The answer is diffraction. It means that <span>the process by which a beam of light or other system of waves is spread out as a result of passing through a narrow aperture or across an edge, typically accompanied by interference between the wave forms produced.</span>
8 0
2 years ago
To measure moderately low pressures, oil with a density of 9.0 x 102 kg/m3 is used in place of mercury in a barometer. A change
Yuki888 [10]

Answer:

Δ P =  13.24 Pa

Explanation:

Given that

Density of oil ,ρ₁ = 9 x 10² kg/m³

We know that density for mercury ,ρ₂  = 13.6 x 10³ kg/m³

The change in the height of column ,Δh = 1.5 mm

The pressure given as

P = ρ g h

Change in the pressure

Δ P =  ρ₁ g Δh

Now by putting the values

Δ P =  9 x 10² x 9.81 x 1.5 x 10⁻³    Pa

Δ P =  13.24 Pa

Therefor the change in the pressure will be 13.24 Pa.  

       

3 0
2 years ago
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