Ask question

Ask question

All categories

3 years ago

15

Focusing on irelevant information can negatively impact the ability to reason and problem solve effectively.

1 answer:

My name is Ann [436]3 years ago

3 0

Answer:

true

Explanation:

You might be interested in

If we're interested in knowing the rate at which light energy is recieved by a unit of area on a particular surface, we're reall

beks73 [17]

Illluminance is the measurement of photometric power. That means, illuminance is the rate of photometric flux that is received by a surface per area. It is usually expressed as a unit of W/m^2. Thus, from the choices, the answer we're looking for is illuminance<span>.</span>

5 0

4 years ago

Read 2 more answers

Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's

mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

$T^2\alpha R^3\\T^2=kR^3.......................(1)$

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

$\frac{T^2}{R^3}=k.......................(2)$

Let the orbital period of the earth be $T_e$ and its mean distance of from the sun be $R_e$ .

Also let the orbital period of the planet be $T_p$ and its mean distance from the sun be $R_p$ .

Equation (2) therefore implies the following;

$\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)$

We make the period of the planet $T_p$ the subject of formula as follows;

$T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)$

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

$R_p=16R_e...............(5)$

Substituting equation (5) into (4), we obtain the following;

$T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}$

$R_e^3$ cancels out and we are left with the following;

$T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)$

Recall that the orbital period of the earth is about 365.25 days, hence;

$T_p=64*365.25\\T_p=23376days$

4 0

3 years ago

You want to place a mirror at a blind turn on the staircase in your house. Which would be best suited for this purpose?

Eduardwww [97]

You can't see beyond a blind turn, so a mirror would allow you to see around the corner.

6 0

3 years ago

Is this equation balanced? 2HCl + CaCO3 → CaCl2 + H2O + CO2

Serggg [28]

Answer:

Yes it is

Explanation:

the sum moles at the left side equals the sum of moles at the right side

7 0

3 years ago

A 17926-lb truck enters an emergency exit ramp at a speed of 75.6 ft/s. It travels for 6.4 s before its speed is reduced to 30.3

I am Lyosha [343]

Answer:

$F_{braking}=337299 pdl$

Explanation:

Impulse-Momentum relation:

$I=\Delta p\\ F_{total}*t=m(v_{f}-v{o})$

$F_{total}=-F_{braking}+mgsin{\theta}$

We solve the equations in order to find the braking force:

$F_{braking}=m(v_{o}-v{f})/t+mgsin{\theta}=17926(75.6-30.3)/6.4+17926*32.17*sin21.4=337299 pdl$

6 0

3 years ago