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3 years ago

13

Egocentrism exists during which of Piaget's stages?

1 answer:

Goryan [66]3 years ago

7 0

Answeri believe it is the first one but not sure

Explanation:

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Of all the planets in our solar system, Jupiter has the greatest gravitational strength. If a 1.5 kg pair of running shoes would

Andre45 [30]

Answer:

gₓ = 23.1 m/s²

Explanation:

The weight of an object is on the surface of earth is given by the following formula:

$W = mg$

where,

W = Weight of the object on surface of earth

m = mass of object

g = acceleration due to gravity on the surface of earth = strength of gravity on the surface of earth

Similarly, the weight of the object on Jupiter will be given as:

$W_{x} = mg_{x}$

where,

Wₓ = Weight of the object on surface of Jupiter = 34.665 N

m = mass of object = 1.5 kg

gₓ = acceleration due to gravity on the surface of Jupiter = strength of gravity on the surface of Jupiter = ?

Therefore,

$34.65 N = (1.5 kg)g_{x}$

$g_{x} = \frac{34.65 N}{1.5 kg}$

<u>gₓ = 23.1 m/s²</u>

7 0

3 years ago

It is easier to climb up a slanted slope than a vertical slope

V125BC [204]

IT IS EASIER TO CLIMB A SLANTED SLOPE

3 0

3 years ago

Read 2 more answers

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

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3 years ago

1. A car is slowing down at a constant rate from 40 m/s to come to a stop. The acceleration is - 5 m/s2. How

Oxana [17]

Answer:

Explanation:

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3 years ago

A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation. (a) Calculate the

Illusion [34]

angular vel to tangential vel

v=r omega

v = 56 x 100/60 x 2 pi

v = 56x5/3x6

v=560m/s as estimate

100 revs, 5.00m

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3 years ago

Read 2 more answers