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2 years ago

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A metal bar has a volume of 32 cm3. The mass of the bar is 256 g. What is the density of the metal? A. 290 g/cm3 6 B. 8,200 g/cm

C. 8.0 g/cm3 O D. 220 g/cm

1 answer:

lina2011 [118]2 years ago

7 0

Answer:

espera que te sirva

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A 3.00 kg block moving 2.09 m/s

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Ixchelt burns her tongue when she takes a sip of hot coffee from her mug. Which part of this example represents heat?

bekas [8.4K]

Answer:

Explanation:

"The thermal energy moving from her coffee to the tongue" represent the heat.

Here coffee is at high temperature while tongue is at low temperature, when Ixchelt tongue make contact with coffee then thermal energy of coffee is absorbed by tongue and tongue gets burned.

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Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13

lyudmila [28]

Answer:

$y = 0.14 Cos\left ( 2.512x-34.66t \right )$

Explanation:

wavelength, λ = 2.5 m

speed, v = 13.8 m/s

Amplitude, A = 0.14 m

The general equation of the transverse harmonic wave which is travelling right is given by

$y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right )$

where, Ф is phase

At t = 0, x = 0 , y = 0.14 m

0.14 = 0.14 Sin Ф

Ф = π/2

So, the equation is

$y = 0.14Sin\left ( \frac{2\pi }{2.5 } (x - 13.8t)+\frac{\pi }{2} \right )$

$y = 0.14 Cos\left ( 2.512x-34.66t \right )$

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3 years ago

Which is a more objective measurement, sound intensity or loudness, and why? 1. sound intensity is exactly same as loudness. 2.

fgiga [73]

The correct answer is:
<span>2. sound intensity is a more objective and physical attribute of a sound wave because loudness can vary from person to person

indeed, sound intensity is a measurable quantity, and so it is objective, while loudness is the subjective perception of the sound level, so it varies from person to person.</span>

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3 years ago

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A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit

Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

$d_1=4.7\times 10^{10} m$

$v_i=9.5\times 10^4 m/s$

$d_2=6\times 10^{12} m$

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}$

$\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f$

$v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})$

$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

$v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

$v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}$

Where mass of sun= $M=1.98\times 10^{30} kg$

$G=6.7\times 10^{-11}$

$v_f=58515.9 m/s$

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3 years ago