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3 years ago

Some social skills address a personal ability to control impulsive behaviors and to express self-motivation.True or fals++e

Physics

1 answer:

hichkok12 [17]3 years ago

5 0

The answer would be true because social skills always help out in public and private settings.

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The volume of water in the Pacific Ocean is about 7.00 × 108 km3. The density of seawater is about 1030 kg/m3. For the sake of t

oee [108]

The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.

By definition the gravitational potential energy is given by,

$PE=\frac{GMm}{r}$

Where,

m = Mass of Moon

G = Gravitational Universal Constant

M = Mass of Ocean

r = Radius

First we calculate the mass through the ratio given by density.

$m = \rho V$

$m = (1030Kg/m^3)(7*10^8m^3)$

$m = 7.210*10^{11}Kg$

PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon

Now we define the radius at the most distant point

$r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m$

Then the potential energy at this point would be,

$PE_1 = \frac{GMm}{r_1}$

$PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}$

$PE_1 = 9.05*10^{15}J$

PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.

At the nearest point we perform the same as the previous process, we calculate the radius

$r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m$

The we calculate the Potential gravitational energy,

$PE_2 = \frac{GMm}{r_2}$

$PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}$

$PE_2 = 9.361*10^{15}J$

7 0

3 years ago

True or false the number of times a machine multipies effort force is a mechanical atlantis

lesantik [10]

Answer: i think its true

Explanation:

7 0

3 years ago

1.What type of sediment forms from minerals that crystallize from seawater?

nikdorinn [45]

The correct answers are as follows:
<span>1) hydrogenous sediment

2)sand and gravel

3) They rapidly break down at surface temperatures and pressures.</span>

7 0

3 years ago

Read 2 more answers

A 30 g bullet moving a horizontal velocity of 500 m/s comes to a stop 12 cm within a solid wall. (a) what is the change in the b

Sidana [21]

M = 30 g = 0.03 kg, the mass of the bullet
v = 500 m/s, the velocity of the bullet

By definition, the KE (kinetic energy) of the bullet is
KE = (1/2)*m*v²
= 0.5*(0.03 kg)*(500 m/s)² = 3750 J
Because the bullet comes to rest, the change in mechanical energy is 3750 J.

The work done by the wall to stop the bullet in 12 cm is
W = (1/2)*(F N)*(0.12 m) = 0.06F J

If energy losses in the form of heat or sound waves are ignored, then
W = KE.
That is,
0.06F = 3750
F = 62500 N = 62.5 kN

Answer:
(a) 3750 J
(b) 62.5 kN

7 0

4 years ago

A girl and her bicycle have a total mads of 42 kg. At the top of the hill her speed is 4 m/s. The hill is 14.3m high and 112m lo

frosja888 [35]

Answer:

13.78 m/s

Explanation:

Given that:

The mass of the girl & her bicycle = 42 kg

The speed at the top of the hilll= 4 m/s

The height of the hill = 14.3 m

The length of the hill (distance) = 112 m

The frictional force = 20 N

To find the speed at the bottom of the hill; we need to carry out the following processes.

The workdone by gravity = mass × acceleration due to gravity × Δh

= 42 × 9.8 × ( 14.3 - 0 )

= 5885.88 joules

The workdone by the friction = Force × distance = - 20 × 112 (since she is riding down the hill)

The workdone by the friction = -2240 Joules

The initial Kinetic friction = 1/2 mv²

= 1/2 × 42 × 4²

= 336 Joules

The final kinetic energy = Initial Kinetic energy + total work

The final kinetic energy = (336 + 5885.88 - 2240) Joules

The final kinetic energy = 3981.88 Joules

Using the final kinetic energy = 1/2 mv²

3981.88 = 1/2 × 42 × v²

3981.88 = 21 v²

v² = 3981.88/21

v² =189.61

v = $\sqrt{189.61}$

v = 13.78 m/s

Therefore, the speed at the bottom = 13.78 m/s

4 0

3 years ago