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Anna11 [10]
2 years ago
10

A 1.0 kg block is attached to an unstretched

Physics
1 answer:
KonstantinChe [14]2 years ago
4 0

Answer:

Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = 1 Nm.

Explanation:

Here, spring constant, k  = 50 N/m.

given block comes down eventually 0.2 m below.

here, g = 10 m/s.

let block be at a height h above the ground in figure 1.

⇒In figure 2, potential energy of the block-spring-Earth

system = m×g×(h - 0.2) + 1/2× k × x². where, x = change in spring length.

⇒ Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = (m×g×(h - 0.2)) - (1/2× k × x²)

              =  (1×10×0.2) - (1/2×50×0.2×0.2) = 1 Nm.

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Why is the following situation impossible? Two identical dust particles of mass 1.00 µg are floating in empty space, far from an
Igoryamba

Answer:

This is a conceptual problem so I will try my best to explain the impossible scenario. First of all the two dust particles ara virtually exempt from any external forces and at rest with respect to each other. This could theoretically happen even if it's difficult for that to happen. The problem is that each of the particles have an electric charge which are equal in magnitude and sign. Thus each particle should feel the presence of the other via a force. The forces felt by the particles are equal and opposite facing away from each other so both charges have a net acceleration according to Newton's second law because of the presence of a force in each particle:

a=\frac{F}{m}

Having seen Newton's second law it should be clear that the particles are actually moving away from each other and will not remain at rest with respect to each other. This is in contradiction with the last statement in the problem.

4 0
2 years ago
PLEASE HELP!!!!!! WILL GIVE BRAINLIEST TO WHOEVER ANSWERS WITH THE RIGHT ANSWER !!!!!!!! 
Solnce55 [7]

It would be B and D your welcome


7 0
2 years ago
Read 2 more answers
Pistons are fitted to two cylindrical chambers connected through a horizontal tube to form a hydraulic system. The piston chambe
Ivenika [448]

Answer:

order   d> a = e> c> b = f

Explanation:

Pascal's law states that a change in pressure is transmitted by a liquid, all points are transmitted regardless of the form

      P₁ = P₂

Using the definition of pressure

      F₁ / A₁ = F₂ / A₂

      F₂ = A₂ /A₁   F₁

Now we can examine the results

a) F1 = 4.0 N A1 = 0.9 m2 A2 = 1.8 m2

     F₂ = 1.8 / 0.9 4

     F₂a = 8 N

b) F1 = 2.0 N A1 = 0.9 m2 A2 = 0.45 m2

    F₂b = 0.45 / 0.9 2

    F₂b = 1 N

c) F1 2.0 N A1 = 1.8 m2 A2 = 3.6 m2

    F₂c = 3.6 / 1.8 2

    F₂c = 4 N

d) F1 = 4.0N A1 = 0.45 m2 A2 = 1.8 m2

    F₂d = 1.8 / 0.45 4.0

    F₂d = 16 m2

e) F1 = 4.0 N A1 = 0.45 m2 A2 = 0.9 m2

   F₂e = 0.9 / 0.45 4

   F₂e = 8 N

f) F1 = 2.0N A1 = 1.8 m2 A2 = 0.9 m2

   F₂f = 0.9 / 1.8 2.0

   F₂f = 1 N

Let's classify the structure from highest to lowest

F₂d> F₂a = F₂e> F₂c> F₂b = F₂f

I mean the combinations are

 d> a = e> c> b = f

6 0
2 years ago
A microwave oven operates with sinusoidal microwaves at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the
Degger [83]

Answer:

F = 2 × 10⁻³ N

Explanation:

Given:

frequency, f = 2400 MHz

Height, h = 25cm = 0.25 m

Area of the base, A = 30 cm x 30 cm = 900 cm² = 0.09 m²

Energy of the  microwave, E = 0.50 mJ = 0.5 x 10⁻³ J

Now, the time taken by the wave from top to the base, t = h/c

here, c is the speed of the light

thus,

t = 0.25/(3 x 10⁸) = 8.33 x 10⁻¹⁰ s

The radiation pressure P_r = Intensity/c

now, the intensity is given as:

I = Power/ area

also,

Power = Energy/ time = 0.5 x 10⁻³ J/8.33 x 10⁻¹⁰ s = 600000 W

thus,

I = 600000 W/ 0.09 m² = 6666666.6 W/m²

substituting the value in the formula for pressure due to radiation, we have

P_r = 6666666.6 W/m²/(3 x 10⁸)

also

pressure = Force/ area

thus,

Force/ area = 6666666.6 W/m²/(3 x 10⁸)

or

Force (F) = (6666666.6 W/m² × 0.09 m²)/(3 x 10⁸)

or

F = 2 × 10⁻³ N

6 0
2 years ago
Read 2 more answers
A vertical cylinder is divided into two parts by a movable piston of mass m. The piston and cylinder system is well insulated (t
Mekhanik [1.2K]

Answer:

Final temperature will be 438.076 K

Explanation:

We have given temperature T_1=323K

Volume V_1=V\ and\ V_2=\frac{V}{2}

As there is no heat transfer so this is an adiabatic process

For and adiabatic process TV^{\gamma -1}=constant

Here \gamma =1.4

So T_1V_1^{\gamma -1}=T_2V_2^{\gamma -1}

T_2=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}\times T_1

T_2=\left ( \frac{V}{\frac{V}{2}} \right )^{1.4 -1}\times 332=2^{0.4}\times 332=438.076K

4 0
2 years ago
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