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3 years ago

How much heat is contained in 100 kg of water at 60.0 °C?

Physics

1 answer:

yulyashka [42]3 years ago

8 0

Answer:

so how much heat is there at 0 C? That's zero. But for every degree above that you have 4.184 J. You take it from there. Remember q = mc*delta T.

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On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3, length 88.8 cm and diameter 2.30 cm fro

vichka [17]

Answer:

w = 28.25 N

Explanation:

To do this, we need to use two expressions.

First, to calculate the weight of any object, we use the 2nd law of newton. In this case, the weight is:

w = m*g (1)

However we do not have the mass of the rod. We need to calculate that. To calculate the mass, we'll use the expression of density which is:

d = m/V

From here, we solve for mass:

m = d * V (2)

Finally, we can know the volume of the rod, because is cylindrical, therefore, the volume of a cylinder is:

V = π * r² * h (3)

So, in resume, we need to solve for the volume of the rod, then, the mass ans finally the weight. Let's calculate the volume of the rod, converting the units of centimeter to meters, just dividing by 100:

diameter = 2.3 cm ---> radius = 2.3/2 = 1.15 cm -----> 0.0115 m

Length or height = 88.8 cm ----> 0.888 m

Replacing in (3):

V = π * (0.0115)² * 0.888

V = 3.69x10⁻⁴ m

Now, let's use (2) to calculate the mass:

m = 7800 * 3.69x10⁻⁴

m = 2.88 kg

Finally for the weight, we'll use expression (1):

w = 2.88 * 9.81

<h2>w = 28.25 N</h2><h2>And this is the weight of the rod.</h2>

4 0

3 years ago

Calculate the minimum thickness (in nm) of an oil slick on water that appears blue when illuminated by white light perpendicular

mr_godi [17]

Answer:

The minimum thickness = 83.92 nm

Explanation:

The relation between the wavelength in a particular medium and refractive index $\lambda_n = \frac{ \lambda }{n}$

where ;

$\lambda$ = wavelength of the light in vacuum

n = refractive index of medium with respect to vacuum

For one phase change :

$2t = \frac{\lambda_n}{2}\\\\where \ \lambda_n = \frac{\lambda}{n}\\\\Then \ \\\\2t = \frac{\lambda}{2n}\\\\t = \frac{\lambda_n}{4n}$

Replacing 1.43 for n and 480 nm for λ; we have:

$t = \frac{480}{4(1.43)}$

t = 83.92 nm

Thus; the minimum thickness = 83.92 nm

4 0

3 years ago

Fusion and fission are both processes that involve which fundamental force? to ask? A. gravitational B. weak nuclear C. electrom

marin [14]

Fusion & Fission are both processes that involve D. Strong Nuclear Fundamental forces
"D". is the answer

5 0

3 years ago

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

3 years ago

Can anyone help me on the summer and winter Gizmo questions. Activities A and B?

Anuta_ua [19.1K]

Answer:

sure, do you have more detailed of the assignments...?

Explanation:

6 0

2 years ago