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NemiM [27]
3 years ago
10

Define specific heat capacity

Physics
2 answers:
sleet_krkn [62]3 years ago
7 0

Answer:you can just look this up yknow?

Explanation:

kirza4 [7]3 years ago
4 0
The heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).
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Which atoms have the highest electronegativity?
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Answer:

Note that there is little variation among the transition metals. Electronegativities generally decrease from top to bottom within a group due to the larger atomic size. Of the main group elements, fluorine has the highest electronegativity (EN = 4.0) and cesium the lowest (EN = 0.79).

3 0
3 years ago
Which statement best describes perigee?
algol13
<span>
A. The closet point in the Moon's orbit to Earth . . . . . perigee

B. The farthest point in the Moon's orbit to Earth . . . . . apogee

C. The Sun's orbit that is closest to the Moon . . . . . a meaningless description

D. The closest point in Earth's orbit of the Sun . . . . . perihelion

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4 years ago
Which letter (A, B, or C) shows where you should apply the most effort to lift the stone?
sergij07 [2.7K]
A becuz its at da it dont got no wa
8 0
3 years ago
A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

5 0
3 years ago
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