Answer:
F = 85696.5 N = 85.69 KN
Explanation:
In this scenario, we apply Newton's Second Law:
where,
F = Upthrust = ?
m = mass of space craft = 5000 kg
g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)
g = 15.0093 m/s²
a = acceleration required = 2.13 m/s²
Therefore,
<u>F = 85696.5 N = 85.69 KN</u>
The two substances that are mostly likely examples of covalent bonding are Sucrose and Ethanol.
<h3 /><h3 /><h3>What is a covalent Bond?</h3>
- A covalent bond is a type of chemical bond that involves the sharing of pairs of electron between atoms.
Examples of compounds with covalent bond include the following;
- Distilled water
- Sucrose
- Ethanol
Olive oil is a mixture not a compound
Sodium Chloride & Potassium lodide are examples of ionic bond.
Thus, the two substances that are mostly likely examples of covalent bonding are Sucrose and Ethanol.
Learn more about covalent bonds here: brainly.com/question/12732708
Answer:
the group of units suggested by the international convention of scientists in 1960 AD to make similarties in meseurment all over the world is called SI units
Answer:
the correct one is D,
Ultraviolet, x-ray, gamma ray
Explanation:
Electromagnetism radiation are waves of energy that is expressed by the Planck relationship
E = h f
where h is the plank constant and f the frequency of the radiation.
Also the speed of light is
c = λ f
we substitute
E = h c /λ
therefore to damage the cells of the body radiation of appreciable energy is needed
microwave radiation has an energy of 10⁻⁵ eV
infrared radiation E = 10⁻² eV
visible radiation E = 1 to 3 eV
radiation Uv E = 3 to 6 eV
X-ray E = 10 eV
gamma rays E = 10 5 eV
therefore we see that the high energy radiation is gamma rays, x-rays and ultraviolet light.
When checking the answers, the correct one is D
Answer:
a. 2v₀/a b. 2v₀/a
Explanation:
a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.
Since the dragster starts from rest with an acceleration, a, using
s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster
s' = 0t + 1/2at²
s' = 1/2at²
Since the distance moved by me and the dragster must be the same,
s = s'
v₀t. = 1/2at²
v₀t. - 1/2at² = 0
t(v₀ - 1/2at) = 0
t= 0 or v₀ - 1/2at = 0
t= 0 or v₀ = 1/2at
t= 0 or t = 2v₀/a
So the maximum time tmax = 2v₀/a
b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s = v₀(2v₀/a)
= 2v₀/a
