All categories

3 years ago

Fuerzas que se aplican al mover una caja

1 answer:

8 0

El peso de la caja, la fricción, si la superficie es áspera, así como la fuerza normal que se llama el "fuerza de reacción " en respuesta al peso de la caja.

You might be interested in

You stare at a bright red screen for so long that your red cones become saturated and no longer function. The red screen is then

adell [148]

Answer: you'll see cyan color on the screen

Explanation:

Saturating the red cone causes them to stop functioning, hence you can't perceive the red part of white light. White light is made up of three main colors which are blue, red and green. When one can no longer perceive the red part of light, one is left with the grean and blue part. The green and blue part of light will superimpose to give a cyan color.

5 0

3 years ago

Read 2 more answers

Which of the following stages follows a protostar?

goldfiish [28.3K]

The answer is a White Dwarf.

4 0

3 years ago

Which are examples of perfectly in elastic collisions

sweet [91]

Typical examples of inelastic collision are between cars, airlines, trains, etc.
For instance, when two trains collide, the kinetic energy of each train is transformed into heat, which explains why, most of the times, there is a fire after a collision. However, the momentum of the two trains that are involved in the collision remains unaffected. So, the trains collide with all their speed, maintaining their momentum, yet their kinetic energy is transformed into heat energy.
Another way to explain a train or a car collision is this: when the two trains or cars collide, they stick together while slowing down. They slow down because their kinetic energy is gradually lost. Still, they collide because they conserve their momentum.

8 0

3 years ago

Two bulbs are connected in parallel across a source of EMF = 8.0V with a negligible internal resistance. One bulb has a resistan

Sonja [21]

<h2>Answer:</h2>

(a) 3.18Ω

(b) 3.18Ω

<h2>Explanation:</h2>

Let the two bulbs be A and B

Given;

$R_{A}$ = Resistance in bulb A = 3.0Ω

$R_{B}$ = Resistance in bulb B = 2.5Ω

Since the two bulbs are connected in parallel;

i. their effective resistance ( $R_{X}$ ) is given by

$\frac{1}{R_{X}}$ = $\frac{1}{R_{A} }$ + $\frac{1}{R_{B} }$ ---------------(i)

Substitute the values of $R_{A}$ and $R_{B}$ into equation (i)

=> $\frac{1}{R_{X}}$ = $\frac{1}{3.0}$ + $\frac{1}{2.5}$

Solve for $R_{X}$

$R_{X}$ = 1.36Ω

ii. voltage (potential difference), V, across them is the same;

Therefore we can get the total current (I) that will flow through them if the voltage to be supplied is 2.4V.

Use the Ohm's law;

V = I x R -----------------(ii)

Where;

V = voltage across them = 2.4V

I = total current flowing through them

R = their effective resistance = $R_{X}$ = 1.36Ω

Substitute these values into equation (ii) as follows;

2.4 = I x 1.36

I = 2.4 / 1.36

I = 1.76A

(a) Now get the value of R

Since the voltage across the two bulbs is 2.4V out of the 8.0V supplied by the source, then the remaining (8.0 - 2.4 = 5.6)V will pass across the resistor R.

Also, since the two bulbs make a series connection with the resistor R, the same total current (I = 1.76A) that flows through these bulbs will flow through the resistor R.

Therefore, to get the value of R, we use the relation

V = I x R ------------------------------(iii)

Where;

V = voltage across the resistor = 5.6V

I = current through the resistor = 1.76A

<em>Substitute these values into equation (iii)</em>

=> 5.6 = 1.76 x R

=> R = 5.6 / 1.76

=> R = 3.18Ω

Therefore, the value of R to be chosen in order to supply each bulb with a voltage of 2.4V is 3.18Ω

(b) The potential difference and voltage across refer to the same thing. Therefore, the value of R that would make the potential difference across each of the bulbs be 2.4V is the same as the one calculated in (a) which is 3.18Ω

3 0

3 years ago

A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be li

statuscvo [17]

Answer:

37357 sec

or 622 min

or 10.4 hrs

Explanation:

GIVEN DATA:

Lifting weight 80 kg

1 cal = 4184 J

from information given in question we have

one lb fat consist of 3500 calories = 3500 x 4184 J

= 14.644 x 10^6 J

Energy burns in 1 lift = m g h

= 80 x 9.8 x 1 = 784 J

lifts required $= \frac{(14.644 x 10^6)}{784}$

= 18679

from the question,

1 lift in 2 sec.

so, total time = 18679 x 2 = 37357 sec

or 622 min

or 10.4 hrs

3 0

2 years ago