What does a 12 kg mass weigh on the Moon?

Tom’s father is 48 years old. He is not able to see nearby objects clearly. (ref: p.508-511)

marin [14]

Answer:

Explanation:

a.) hypermetropia is the reason for his problem

b.)Image is formed farther from the near point

c.) defect is corrected by convex lens

7 0

4 years ago

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9. A 30 cm ruler is found to have a center of mass of 15.6 cm. The percent error of the center of mass is _____, if the ruler is

grigory [225]

Answer:

3.85 percent

Explanation:

From the question,

Percentage error = (error/actual)×100................ Equation 1

Given: actual center of mass = 15 cm, error = 15.6-15 = 0.6 cm

Substitute these values into equation 1

Percentage error = (0.6/15.6)×100

Percentage error = 3.85 percent

Hence the percentage error of the uniform mass = 3.85 percent

4 0

3 years ago

A plane flies 446 km east from city A to city B in 43.0 min and then 939 km south from city B to city C in 1.10 h. For the total

NeTakaya

Answer:

(a) Magnitude is 1039 km

(b) Direction of the displacement is $64.59^{\circ}$ South of East

(c) Average velocity magnitude is 570.88 km

(d) The direction of average velocity is $64.59^{\circ}$ South of East

(e) Average speed is 759.34 km/h

Solution:

Distance moved from A to B in East direction, $\vec{AB} = 446 km$

Distance moved from B to C in South direction, $\vec{BC} = - 939 km$

Time taken to move from A to B, t = 43.0 min = 0.72 h

Time taken to move from B to C, t' = 1.10 h

Now,

(a) The magnitude of displacement of the plane is provided by AC as shown in fig 1 and can be given as:

$AC = \sqrt{(AB)^{2} + (BC)^{2}}$

$AC = \sqrt{(446)^{2} + (- 939)^{2}}$ = 1039 km

(b) Direction of the displacement is given by:

$tan\theta = \frac{\vec{BC}}{\vec{AB}}$

$\theta = tan^{- 1}(\frac{- 939}{\vec{446}}) = - 64.59^{\circ}$

$64.59^{\circ}$ South of East

(c) Magnitude of the average speed is given by:

$v_{avg} = \frac{AC}{t + t'}$

$v_{avg} = \frac{1039}{1.82} = 570.88 km/h$

(d) The direction of the average velocity is the same as that of the displacement, i.e., $64.59^{\circ}$ South of East.

(e) The average speed of the [plane is given by:

$v'_{avg} = \frac{Total\ Distance\ Traveled}{Total\ Time}$

$v'_{avg} = \frac{446 + 939}{1.10 + 0.72} = 759.34 km/h$

6 0

3 years ago

Arrange the following substances in order of decreasing magnitude of lattice energy.

kolbaska11 [484]

The formula is F = ( q1 * q2 ) / r ^ 2
where: q is the individual charges of each ion 
r is the distance between the nuclei 
The formula is not important but to explain the relationship between the atoms in the compounds and their lattice energy. 

From the formula we can first conclude that compounds of ions with greater charges will have a greater lattice energy. This is a direct relationship. 
For example, the compounds BaO and SrO, whose ions' charges are ( + 2 ) and ( - 2 ) respectively for each, will have greater lattice energies that the compounds NaF and KCl, whose ions' charges are ( + 1 ) and ( - 1 ) respectively for each. 

So Far: ( BaO and SrO ) > ( NaF and KCl ) 

The second part required you find the relative distance between the atoms of the compounds. Really, the lattice energy is stronger with smaller atoms, an indirect relationship. 
For example, in NaF the ions are smaller than the ions in KCl so it has a greater lattice energy. Because Sr is smaller than Ba, SrO has a greater lattice energy than BaO. 

Therefore: 
Answer: SrO > BaO > NaF > KCl

5 0

4 years ago

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Why is it that when riding in a car, you don't feel like you're moving?

UkoKoshka [18]

This may shock you:

We NEVER feel speed, velocity, or motion, as long as it's constant.

We only feel CHANGES in speed, velocity, or motion.

That means speeding up, slowing down, or changing direction.

As long as we're moving in a straight line at a constant speed, we don't feel anything.

6 0

3 years ago