Answer:
98 kg
Mass is given as 10 kg. Therefore, Weight = 10 kg * 9.8 m/s^2. Weight = 98 kg.m/s^2. = 98 Newtons.
Explanation:
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Answer and Explanation:
a. An oxygen-filled balloon is not able to float in the air, because the oxygen inside the balloon is of the same density, that is, the same "weight" as the oxygen outside the balloon and present in the atmosphere. The balloon can only float if the gas inside it is less dense than atmospheric oxygen. Helium gas is less dense than atmospheric gas, so if a balloon is filled with helium gas, that balloon will be able to float because of the difference in density.
b. The ship is able to float in the water because its steel construction is hollow and full of air. This makes the average density of this ship less than the density of water, which makes the ship lighter than water and for this reason, this ship is able to float. In addition, the ship is partially immersed, allowing the weight of the ship on the water to counteract the buoyant force that the water promotes on the ship. Weight and buoyant are two opposing forces that keep the ship afloat.
Answer:
W = 145.8 [N]
Explanation:
To solve this problem we must remember that weight is defined as the product of mass by gravity, in this case lunar gravity.
W = m*g
where:
m = mass = 90 [kg]
g = gravity acceleration = 1.62 [kg/m²]
W = 90*1.62
W = 145.8 [N]
Answer:
CH4
Explanation:
CH4 is joined together by a covalent bond, aka a bond between two non-metals. Non-metals are found on the right side of the periodic table and include Carbon (C) and Hydrogen. Although Hydrogen is technically on the left side of the table, it has the characteristics of a non-metal. Futhermore, Ionic bonds generally are between an element on the right joined with an element on the left. This is because ionic bonds want charges that will cancel out to create a neutral molecule.
example: LiF
Li→ Li+
F→F-
(Li+)+(F-)=charges cancel out.
Answer:
electrons
Explanation:
The magnitude of the electric field outside an electrically charged sphere is given by the equation
where
k is the Coulomb's constant
Q is the charge stored on the sphere
r is the distance (from the centre of the sphere) at which the field is calculated
In this problem, the cloud is assumed to be a charged sphere, so we have:
is the maximum electric field strength tolerated by the air before breakdown occurs
is the radius of the sphere
Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:
Assuming that the cloud is negatively charged, then
And since the charge of one electron is
The number of excess electrons on the cloud is
