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4 years ago

What is the total amount of heat in a substance

1 answer:

7 0

heat capacity: The capability of a substance to absorb heat energy; the amount of heat required to raise the temperature of one mole or gram of a substance by one degree Celsius without any change of phase.

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Beyond what point must an object be squeezed for it to be a black hole? A : Schwarzschild Radius. B : Crackpoint. C : Event Hori

7nadin3 [17]

Answer:

A:Schwarzschild radius

8 0

3 years ago

Read 2 more answers

What would be the Elastic Potential Energy (EPE) stored in a spring with a constant k = 200 N/m that is pulled to stretch 0.45m?

harkovskaia [24]

Answer:

During the summer, oxygen is supplied by surface winds. M/NPT 16 to 26 0.14 to 0.45 m 1/ 77.80 $ 28 Phone: 418 81-1 Fax: 418 81-2882 SZ5-2025 SZ22-200 SZ2-250 SZ24-00 12.5 x 60 12.5 x 60 12.5 x 60 8 x 10 10 x 15 10 x 20 15 x HI-950 can take measures between 0.1 C to 999.9 C/ F. K, J & T-type: The HI-9551

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3 years ago

A 1.0 kg object moving at 4.5 m/s has a wavelength of:

Lisa [10]

By wave particle duality.

Wavelength , λ = h / mv

where h = Planck's constant = 6.63 * 10⁻³⁴ Js, m = mass in kg, v = velocity in m/s.
m = 1kg, v = 4.5 m/s

λ = h / mv

λ = (6.63 * 10⁻³⁴) /(1*4.5)

λ ≈ 1.473 * 10⁻³⁴ m

Option D.

7 0

3 years ago

The focal length for a spherical convex mirror is –20 cm. What is its radius of curvature?

Svetradugi [14.3K]

Answer:

40 cm

Explanation:

The focus of a spherical convex mirror is the point at which the rays of an object converge in the infinite, it is also the point at which an object must be placed so that its image is formed in the infinite. The distance from the focus to the origin is called the focal length and is called f. It is related to the radius of the mirror, R, according to:

$f=-\frac{R}{2}$

rewriting for R:

$R=-2f=-2*-20cm=40 cm$ .

4 0

3 years ago

One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.

Zepler [3.9K]

Answer:

The answer is " $\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}$ "

Explanation:

Given:

$\to v=1\ liter= 10^{-3} \ m^3\\\\\to w= 9.6 \ N\\$

calculation:

$Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978$

4 0

3 years ago