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kompoz [17]
1 year ago
5

Which has the lowest heat capacity? (values of heat capacities and calculations are unnecessary).

Physics
1 answer:
Brrunno [24]1 year ago
5 0

The correct answer is Metals.

Generally, the specific heat of metals is low. Very high specific heat exists in water.A physical feature of matter known as heat capacity or thermal capacity is the quantity of heat that must be applied to an object in order to cause a unit change in temperature. Heat capacity is measured in joules per kelvin (J/K), the SI unit. A broad property is heat capacity. Use the following equation to determine heat capacity: heat capacity = E / T, where E is the quantity of delivered heat energy and T is the change in temperature. The formula would be as follows, for instance, if it takes 2,000 Joules of energy to raise a block's temperature by 5 degrees Celsius: 2,000 Joules per °C is the heat capacity.

Learn more about heat capacity here :-

brainly.com/question/13499849

#SPJ4

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There is fluctuation in time measured by a pendulum clock why give reasons​
natulia [17]

A major source of error in pendulum clocks is thermal expansion; the pendulum rod changes in length slightly with changes in temperature, causing changes in the rate of the clock. An increase in temperature causes the rod to expand, making the pendulum longer, so its period increases and the clock loses time.

3 0
3 years ago
What is the experimental group(s)?
Vinil7 [7]
<h3>I think it B The group(s) that gets the special treatment.</h3><h3 /><h3>I hope this is correct.</h3><h3 /><h3 /><h3 />
5 0
3 years ago
A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h
noname [10]
The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 =  0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s
7 0
3 years ago
1. Why did Asteroid 2019 OK cause concern among scientists? * The asteroid was not detected until it was extremely close to Eart
AveGali [126]

Answer:

<u>The asteroid was not detected until it was extremely close to Earth. </u>

Explanation:

According to data from NASA, the Asteroid named 'Astriod 2019 OK', was detected when it was extremely close to earth with just about an estimated distance of  73,000 kilometers  (45,000 miles) from the Earth.

Scientists were concerned at the proximity of this space object to the Earth before it was discovered, and it brought about a cause of concern that since it was not extremely large (estimated 57 to 130 meters wide) it creates a potential for other smaller asteroids to escape detection and struck the earth.

7 0
2 years ago
Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y
WITCHER [35]

Question:

A point charge of -2.14uC  is located in the center of a spherical cavity of radius 6.55cm  inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm  from the center of the cavity.  

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 3.65\times 10^5N/C

Explanation:

A point charge ,q = -2.14\times 10^{-6} C is located in the center of a spherical cavity of radius , r =6.55\times 10^{-2}  m inside an insulating spherical charged solid.  

The charge density in the solid , d = 7.35 \times 10^{-4}C/m^3.

Distance from the center of the cavity,R =9.5\times 10^{-2 }m

Volume of shell of charge= V  =(\frac{4\pi}{3})[ R^3 - r^3 ]

Charge on the shell ,Q = V \times d'

Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d

Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}

Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}

Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}

Q =1.7745 \times 10^{-6 }C

Electric field at 9.5\times 10^{-2}m due to shellE1  = \frac{k Q}{R^2}

E1 =  \frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}

E1 =1.769\times 10^6 N/C

Electric field at  9.5\times 10^{-2} due to 'q' at center E2 = \frac{kq}{R^2}

E2 =\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}

E2 =2.134\times 10^6 N/C

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

=[  2.134  - 1.769 ]\times 10^6

= 3.65\times 10^5 N/C

8 0
3 years ago
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