The answer is 107 degrees. The geometric shape for ammonia is Trigonal Pyramidal, even though its electron geometry is “Tetrahedral”. This is because ammonia has a lone pair of electrons that occupy its space like the other 3 hydrogens in the geometric structure.
The answer 180 degrees. This is because of the linear geometric structure of carbon dioxide. The oxygen atom is on either side of the carbon atom, each is bound by a double covalent bond. All the atoms are involved in the bond and there are no one pair electrons.
The answer is tetrahedral geometry. This is because all the 4 valence electrons of the carbon are involved in a bond with a hydrogen atom. The angles in a tetrahedral geometric arrangement, such as in methane, is 109.5 degrees, where the hydrogen atoms are as far apart, from each other, as possible .
Answer:
The answer to your question is:
a) 2.7 m/s²
b) -3.6 m/s²
Explanation:
Data
mass of the toolbox = 3.2 kg
a = ?
F = 40 N and F = 20 N
g = 9.81 m/s²
Formula
Second law of motion = F = ma
a + g = F / m
a = F/m - g
a) a = 40/3.2 - 9.81
a = 2.69 ≈ 2.7 m/s² positive up
b) a = 20/ 3.2 - 9.81
a = 6.25 - 9.81
= - 3.56 ≈ - 3.6 m/s² negative down
Answer: They travel equal distances
Explanation:
Distance is amount of space covered from one point to another. It is measured in metres. Distance = speed x time taken
In this case,
- Since Tony and Tanya both run at the same speed: let speed = b
- Also, both run for the same amount of time: let time = c
Therefore, distance covered by Tony and Tanya = speed x time taken
Distance = b x c
Distance = bc (i.e Tony travelled a distance of bc, so also Tanya)
There are various ways to compute for power, and one of them is by using the voltage and current of a system. This formula is given by P = IV, where I is the current in amperes and V is voltage in volts. Using the given data above, the power in watts is calculated as follows:
P = 110(2.7) = 297 watts.
Answer: A
Explanation:
From the question, the given parameters are given.
Mass M = 30kg
Radius r = 2 m
Coefficient of static friction μ = 0.8
Coefficient of kinetic friction μ = 0.6
Kinetic friction Fk = μ × mg
Fk = 0.6 × 30 × 9.8
Fk = 176.4 N
The force acting on the merry go round is a centripetal force F.
F = MV^2/r
This force must be greater than or equal to the kinetic friction Fk. That is,
F = Fk
F = 176.4
Substitute F , M and r into the centripetal force formula above
176.4 = (30×V^2)/2
Cross multiply
352.8 = 30V^2
V^2 = 352.8/30
V = sqrt (11.76) m/s
V = 5.24 m/s
Therefore, the maximum speed of the merry go round before the child begins to slip is sqrt (12) m/s approximately
