Please help!!! Thank you!!!!!

Estimate the water volume in the graduated cylinder to the nearest 0.1 mL.

earnstyle [38]

Answer:

With a 50-mL graduated cylinder, read and record the volume to the nearest 0.1 mL. The 10-mL graduated cylinder scale is read to the nearest 0.01 mL and the 500-mL graduated cylinder scale is read to the nearest milliliter (1 mL). A buret is a scaled cylindrical tube attached to a stopcock, or valve.

5 0

3 years ago

Physics help please! Offering 75 pts!

Nataliya [291]

Answer:

A

Explanation:

i got it from 2021 edge

5 0

2 years ago

Read 2 more answers

What trends were seen in Mendeleev’s periodic table?

katrin [286]

He arranged his periodic table by each elements atomic mass

3 0

2 years ago

Read 2 more answers

A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s

klemol [59]

Given Information:

Resistance = R = 14 Ω

Inductance = L = 2.3 H

voltage = V = 100 V

time = t = 0.13 s

Required Information:

(a) energy is being stored in the magnetic field

(b) thermal energy is appearing in the resistance

(c) energy is being delivered by the battery?

Answer:

(a) energy is being stored in the magnetic field ≈ 219 watts

(b) thermal energy is appearing in the resistance ≈ 267 watts

(c) energy is being delivered by the battery ≈ 481 watts

Explanation:

The energy stored in the inductor is given by

$U = \frac{1}{2} Li^{2}$

The rate at which the energy is being stored in the inductor is given by

$\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1$

The current through the RL circuit is given by

$i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })$

Where τ is the the time constant and is given by

$\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16$

$i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}$

Therefore, eq. 1 becomes

$\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})$

At t = 0.13 seconds

$\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts$

(b) thermal energy is appearing in the resistance

The thermal energy is given by

$P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts$

(c) energy is being delivered by the battery?

The energy delivered by battery is

$P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts$

4 0

3 years ago

A spring has a spring constant of 20 N/m. How much potential energy is stored in the spring as it stretched 0.20 m

makkiz [27]

The elastic potential energy (Ep) is given by $Ep = \frac{1}{2}*k*x^2$

Data:
Ep = ? (Joule)
k = 20 N/m
x (displacement) = 0.20 m

Solving:
$Ep = \frac{1}{2}*k*x^2$
$Ep = \frac{1}{2}*20*0.20^2$
$Ep = \frac{20*0.04}{2}$
$Ep = \frac{0.8}{2}$
$\boxed{\boxed{Ep = 0.4\:J}}\end{array}}\qquad\quad\checkmark$

7 0

2 years ago