High specific heat of the water. Option (c)
What is Specific heat?
The amount of heat required to increase the temperature of one gram of a substance by one degree Celsius is referred to as the substance's specific heat. Typically, calories or joules are used per gram and degree Celsius when referring to the units of specific heat.
The moderate temperature of islands has much to do with the water's high specific heat. The typical off-water is more significant than this clear land or soil. Due to this fact, water absorbs and releases eat more slowly. In comparison to the land.
Hence, the water has high specific heat.
To learn more about specific heat the link is given below:
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Answer:
That is false, Protons have a positive charge, Electrons have a negative charge, and Neutrons have a blank charge.
Explanation:
Answer:
Explanation:
Let the charge on proton be q .
energy gain by proton in a field having potential difference of V₀
= V₀ q
Due to gain of energy , its kinetic energy becomes 1/2 m v₀²
where m is mass and v₀ is velocity of proton
V₀ q = 1/2 m v₀²
In the second case , gain of energy in electrical field
= 2 V₀q , if v be the velocity gained in the second case
2 V₀q = 1/2 m v²
1/2 m v² = 2 V₀q = 2 x 1/2 m v₀²
mv² = 2 m v₀²
v = √2 v₀
Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. Therefore, there must be the same number of atoms of each element on each side of a chemical equation.
Hello!
Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:
← The circumference of the orbit
speed = orbital speed, we will solve for this later
time = period
Therefore:
Where 'r' is the orbital radius of the satellite.
First, let's solve for 'v' assuming a uniform orbit using the equation:
G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)
m = mass of the earth (5.98 × 10²⁴ kg)
r = radius of orbit (1.276 × 10⁷ m)
Plug in the givens:
Now, we can solve for the period:
