All categories

2 years ago

To heat the water in the tank from 50°C to 58°C the immersion heater transfers

Physics

1 answer:

LenKa [72]2 years ago

4 0

Let's see

∆T=58-50=8°C

$\\ \rm\Rrightarrow Q=mc\Delta T$

$\\ \rm\Rrightarrow 4032=4200(8)m$

$\\ \rm\Rrightarrow 4032/8=4200m$

$\\ \rm\Rrightarrow 504=4200m$

$\\ \rm\Rrightarrow m=0.12kg$

$\\ \rm\Rrightarrow m=120g$

You might be interested in

WILL GIVE BRAINLIEST! Briefly describe the three phases of self-regulation. _PSYCHOLOGY_

Zinaida [17]

Phase 1. Forethought/preaction—This phase precedes the actual performance; sets the stage for action; maps out the tasks to minimize the unknown; and helps to develop a positive mindset. Realistic expectations can make the task more appealing. Goals must be set as specific outcomes, arranged in order from short-term to long-term. We have to ask students to consider the following:

When will they start?Where will they do the work?How will they get started?What conditions will help or hinder their learning activities are a part of this phase?

Phase 2. Performance control—This phase involves processes during learning and the active attempt to utilize specific strategies to help a student become more successful.

We have to ask students to consider the following:

Are students accomplishing what they hoped to do?Are they being distracted?Is this taking more time than they thought?Under what conditions do they accomplish the most?What questions can they ask themselves while they are working?How can they encourage themselves to keep working (including self-talk—come on, get your work done so you can watch that television show or read your magazine!)

Phase 3. Self-reflection—This phase involves reflection after the performance, a self-evaluation of outcomes compared to goals.

We have to ask students to consider the following:

Did they accomplish what they planned to do?Were they distracted and how did they get back to work?Did they plan enough time or did they need more time than they thought?Under what conditions did they accomplish the most work.
Hope this helps!!!!!

6 0

3 years ago

Read 2 more answers

Using an elastic cord, the astronaut in the black shirt pulls the astronaut with the red shirt back towards him before releasing

pishuonlain [190]

Answer:

50 T-shirt Design Ideas That Won't Wear Out - 99Designshttps://99designs.com › blog › creative-inspiration › t-s...

Whatever message your t-shirt needs to send, we've got 50 t-shirt design ... Illustrated chicken astronaut t-shirt ... of the challenges associated with a t-shirt for a business is getting people to wear it. ... shirt is a great way to create memorabilia that participants can look back on ... Graphic tees are what shirts were made for.

Explanation:

8 0

2 years ago

A 0.600 m long pendulum is used to determine the acceleration due to gravity on a distant plane. If 20 oscillations are complete

katrin2010 [14]

Answer:

7.50 m/s^2

Explanation:

The period of a pendulum is given by:

$T=2\pi \sqrt{\frac{L}{g}}$ (1)

where

L = 0.600 m is the length of the pendulum

g = ? is the acceleration due to gravity

In this problem, we can find the period T. In fact, the frequency is equal to the number of oscillations per second, so:

$f=\frac{N}{t}=\frac{20}{35.5 s}=0.563 Hz$

And the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{0.563 Hz}=1.776 s$

And by using this into eq.(1), we can find the value of g:

$g=\frac{4 \pi^2 L}{T^2}=\frac{4 \pi^2 (0.600 m)}{(1.776 s)^2}=7.50 m/s^2$

6 0

3 years ago

A diagram of a closed circuit with power source on the left labeled 45 V, two resistors on the top labeled 2.0 Ohms and 3.0 Ohms

CaHeK987 [17]

Answer:

Explanation:

ANSWER KEY

6 0

3 years ago

Read 2 more answers

A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an

olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = $\pi r^2$

V = Potential applied = 2 mV

k = Dielectric constant = 40

$\epsilon_0$ = Electric constant = $8.854\times 10^{-12}\ \text{F/m}$

Capacitance is given by

$C=\dfrac{k\epsilon_0A}{d}$

Charge is given by

$Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}$

Number of electron is given by

$n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}$

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

6 0

2 years ago