Egocentrism exists during which of Piaget's stages?

An archer shoots an arrow with a mass of 45.0 grams from bow pulled

Sladkaya [172]

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

3 0

3 years ago

Three cars (car F, car G, and car H) are moving with the same speed and slam on their brakes. The most massive car is car F, and

Crazy boy [7]

To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.

From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by

$v_f^2=v_i^2+2ax$

Where,

$v_f =$ Final velocity

$v_i =$ Initial Velocity

a = Acceleration

x = Displacement

Acceleration can be expressed in terms of the drag coefficient by means of

$F_f = \mu_k (mg) \rightarrow$ Frictional Force

$F = ma \rightarrow$ Force by Newton's second Law

Where,

m = mass

a= acceleration

$\mu_k =$ Kinetic frictional coefficient

g = Gravity

Equating both equation we have that

$F_f = F$

$\mu_k mg=ma$

$a = \mu_k g$

Therefore,

$v_f^2=v_i^2+2ax$

$0=v_i^2+2(\mu_k g)x$

Re-arrange to find x,

$x = \frac{v_i^2}{2(-\mu_k g)}$

The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.

3 0

3 years ago

A satellite is in a circular orbit 8200 km above the Earth’s surface; i.e., it moves on a circular path under the influence of n

andreyandreev [35.5K]

Answer:

5.23km/s

Explanation:

Given

Radius of Earth = 6.37 * 10^6 m

Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m

Gravity Acceleration on Satellite Altitude = 1.87965m/s²

For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.

Centripetal Acceleration = V²/R

So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R

Make V the subject of formula

A = V²/R

V² = AR

V = √AR

Where R = (radius of earth) + (altitude of satellite)

R = 6.37 * 10^6 + 8.2 * 10^6

R = 14.57 * 10^6m

A = 1.87965m/s²

V = √(1.87965 * 14.57x10^6)

V = √27386500.5

V = 5233.211299001789

V = 5233.2113 m/s ------- Approximated

V = 5.23km/s

7 0

3 years ago

1. Water boils at 100°C at sea level. If the water in this experiment did not boil at 100°C, what could be the reason?

umka2103 [35]

Answer:

It may not be at the sea level

Explanation:

The reason here is water only boils at sea level. This means that if you move water to a different height, say top of a mountain, the boiling temperature of water would change. This is due to the pressure drop at high place. The drop of pressure would make it harder to transform water liquid to gas, thus requiring more temperature.

8 0

3 years ago

Read 2 more answers

: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

e-lub [12.9K]

Answer:

$x_1 = 3.74m$