According to Binet, mental age relates to chronological age because __________

A backpack has a mass of 8 kg. It is lifted and given 54.9 J of gravitational potential energy. How high is it lifted? Accelerat

sweet [91]

Potencial Energy=hma
Where
Potencial Energy =E= 54.9J
h=?
m=8kg
a=9.8m/s^2
You need to know that 1 J=1(kgm^2)/s^2

Isolate h=E/(ma)
h=(54.9)/(8*9.8)

8 0

3 years ago

Read 2 more answers

An object is removed from a room where the temperature is 69 degrees and is taken outside, where the air temperature is 30 degre

Yuliya22 [10]

Answer:

The temperature of the object at any time t, T(t) is given as

T = T∞ + (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ

Explanation:

Let T be the temperature of the object at any time

T∞ be the temperature outside = 30°

T₀ be the initial temperature of the object in the room = 69°

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the object = Rate of Heat gain by the outside air

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 30) = (69 - 30)e⁻ᵏᵗ

(T - 30) = 39 e⁻ᵏᵗ

At 1 minute, T = 52°

52 - 30 = 39 e⁻ᵏᵗ

22/39 = e⁻ᵏᵗ

- kt = In (22/39) = In (0.564)

- k(1) = - 0.5725

k = 0.5725 /min

(T - T∞) = (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ

T = T∞ + (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ

4 0

3 years ago

Suppose that she pushes on the sphere tangent to its surface with a steady force of F = 75 N and that the pressured water provid

PolarNik [594]

Answer:

The time taken to rotate the sphere one time is, t = 22 s

Explanation:

Given data,

The mass of the sphere, m = 8200 kg

The radius of the sphere, r = 90 cm

= .9 m

The force applied by the girl, F = 75 N

The moment of inertia of the sphere is,

I = 2/5 mr²

= (2/5) 8200 x (.9)²

= 2657 kg·m²

The torque,

τ = I α

75 x 0.9 = 2657 x α

α = 0.0254 rad/s²

The angular displacement,

θ = ½αt²

2π = ½ x 0.0254 rad/s² x t²

t = 22 s

Hence, the time taken to rotate the sphere one time is, t = 22 s

8 0

3 years ago

A 6 kg table has a net force of 15 N acting on it. What is the magnitude of the acceleration of the table?

beks73 [17]

Answer:

2.5m/s2

Explanation:

F=ma

a=F/m

a=15/6

a=2.5m/s2

8 0

3 years ago

A potential difference of 300 volts is applied to a 2.0-µf capacitor and an 8.0-µf capacitor connected in series. (a) What are t

zubka84 [21]

Answer:

a) Q1=Q2=480μC V1=240V V2=60V

b) Q1=96μC Q2=384μC V1=V2=48V

c) Q1=Q2=0C V1=V2=0V

Explanation:

Let C1 = 2μC and C2=8μC

For part (a) of this problem, we know that charge in a series circuit, is the same in C1 and C2. Having this in mind, we can calculate equivalent capacitance first:

$C=\frac{1}{\frac{1}{C1} +\frac{1}{C2} } = 1.6\mu C$

$Q_{T} = V_{T}*C_{T} = 480\mu C$

$V1 = \frac{Q1}{C1}=240V$

$V2 = \frac{Q2}{C2}=60V$

For part (b), the capacitors are in parallel now. In this condition, the voltage is the same for both capacitors:

$V1=V2$ So, $\frac{Q1}{C1} = \frac{Q2}{C2}$

Total charge is the same calculated for part (a), so:

$\frac{Qt - Q2}{C1} = \frac{Q2}{C2}$ Solving for Q2:

Q2 = 384μC Q1 = 96μC.

Therefore:

V1=V2=48V

For part (c), both capacitors would discharge, since their total voltage of 300V would by applied to a wire (R=0Ω). There would flow a huge amount of current for a short period of time, and capacitors would be completely discharged. Q1=Q2=0C V1=V2=0V

3 0

3 years ago

According to Binet, mental age relates to chronological age because ___________.