The specific heat of the unknown substance with a mass of 0.158kg is 0.5478 J/g°C
HOW TO CALCULATE SPECIFIC HEAT CAPACITY:
The specific heat capacity of a substance can be calculated using the following formula:
Q = m × c × ∆T
Where;
- Q = quantity of heat absorbed (J)
- c = specific heat capacity (4.18 J/g°C)
- m = mass of substance
- ∆T = change in temperature (°C)
According to this question, 2,510.0 J of heat is required to heat the 0.158kg substance from 32.0°C to 61.0°C. The specific heat capacity can be calculated:
2510 = 158 × c × (61°C - 32°C)
2510 = 4582c
c = 2510 ÷ 4582
c = 0.5478 J/g°C
Therefore, the specific heat capacity of the unknown substance that has a mass of 0.158 kg is 0.5478 J/g°C.
Learn more about specific heat capacity at: brainly.com/question/2530523
Answer:
Answers at the bottom!!
Explanation:
For the first answer: Gamma ray
Gamma rays and x-rays consist of high-energy waves that can travel great distances at the speed of light and generally have a great ability to penetrate other materials. For that reason, gamma rays (such as from cobalt-60) are often used in medical applications to treat cancer and sterilize medical instruments.
For the second answer: They both have technological uses.
In my opinion I'm pretty sure it's A because we do use microwaves and x-rays as technological uses.
Hope this helps!!
Happy Holidays and Season Greeting!!
ii Feliz Navidad a todos !!
If you mean pattern designs it’s rocky bumpy and you can see the circle patterns
If you mean phases the here’s a picture
Answer:
Explanation:
The angular momentum of electron mvR = 6 x 10⁻²⁵ Js
Magnetic field B = 2.5 x 10⁻³ T
radius of circular path R = mv / Bq
where m is mass , v is velocity and q is charge on electron
R² = mvR / Bq
R² = 6 x 10⁻²⁵ / 2.5 x 10⁻³ x 1.6 x 10⁻¹⁹
= 1.5 x 10⁻³
R = 3.87 x 10⁻² m
mvR = 6 x 10⁻²⁵
v = 6 x 10⁻²⁵ / mR
= 6 x 10⁻²⁵ / 9.1 x 10⁻³¹ x 3.87 x 10⁻²
= .17 x 10⁸
= 17 x 10⁶ m/s
