A circuit is used in which process? A. electricity B. magnetism C. velocity D. gravity

WHY is it so important to include units when expressing numbers?

Natasha_Volkova [10]

Answer:

Without units, the results are unclear and it is hard to keep track of what each seperate measurement entails.

5 0

4 years ago

A 10-gram aluminum cube absorbs 677 joules when its temperature is increased from 50°c to 125°

kotegsom [21]

Answer : The specific heat of aluminum is, $0.90J/g^oC$

Solution : Given,

Heat absorbs = 677 J

Mass of the substance = 10 g

Final temperature = $125^oC$

Initial temperature = $50^oC$

Formula used :

$Q= m\times c\times \Delta T$

or,

$Q= m\times c\times (T_{final}-T_{initial})$

Q = heat absorbs

m = mass of the substance

c = heat capacity of aluminium

$T_{final}$ = final temperature

$T_{initial}$ = initial temperature

Now put all the given values in the above formula, we get the specific heat of aluminium.

$677g= (10g)\times c\times (125-50)^oC$

$c=0.9026J/g^oC=0.90J/g^oC$

Therefore, the specific heat of aluminum is, $0.90J/g^oC$

7 0

3 years ago

Read 2 more answers

An electronics technician wishes to construct a parallelplate capacitor using rutile (k=100) as the dielectric. The area of the

My name is Ann [436]

Given k=100, Area if plate A = 1.00 cm2

$= 1 x (10-2)2 m^{2}$

Thickness of rutile $ds = t = 1.00 mm$

$= 1 x 10-3 cm$

Without any dielectric medium capacitance of a parallel plate capacitor is c = oAd

When a dielectric is placed between capacitors plates then capacitance c1 = KoAd

Here $d = t = 1 x 10-1 cm$ , $o = 8.854 x 10-12 c^{2} m^{-2} N$

Now $c1 = 100 X 8.859 X 10^{-12} X 1 X (10-2)^{21} X 10^{-2}$

$= 8.859 X 10-1910-3= 8.85 X 10-11 F= 8.85 X 10-12 F$

$C1 = 88.5 X 10-12 F$

$C1 = 88.5 PF$

Correct Option: A

What purposes serve parallel plate capacitors?

The following are some uses for parallel plate capacitors:

This kind of capacitor is utilized in rechargeable energy systems such as batteries.
These capacitors are used in systems for dynamic digital memory.
Such capacitors are used in pulsed laser and radar circuits.

To learn more about Parallel plate capacitors, visit:

brainly.com/question/12733413

#SPJ4

5 0

2 years ago

Observe yourself breathing and count the number of times you inhale per second. During each breath you probably inhale 0.66 L of

Pavel [41]

To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

$PV=nRT$

Where,

P = Pressure

V = Volume

T = Temperature

n = amount of substance

R = Ideal gas constant

We start by calculating the volume of inhaled O_2 for it:

$V = 21\% * 0.66L$

$V = 0.1386L$

Our values are given as

P = 1atm

T=293K $R = 0.083145kJ*mol^{-1}K^{-1}$

Using the equation to find n, we have:

$PV=nRT$

$n = \frac{PV}{RT}$

$n = \frac{(1)(0.1386)}{(0.0821)(293)}$

$n = 5.761*10^{-3}mol$

Number of molecules would be found through Avogadro number, then

$\#Molecules = 5.761*10^{-3}*6.022*10^{23}$

$\#Molecules = 3.469*10^{21} molecules$

7 0

3 years ago

Two small frogs simultaneously leap straight up from a lily pad. frog a leaps with an initial velocity of 0.551 m/s, while frog

Setler [38]

Define
g = 9.8 m/s², acceleration due to gravity, positive downward.

Assume that wind resistance may be neglected.

Frog A:
u = 0.551 m/s, launch velocity, upward.
When the frog lands back on the pad, its vertical position is zero, and its vertical velocity will be 0.551 m/s downward.
If the time of flight is t, then
(0.551 m/s)*(t s) - 0.5*(9.8 m/s²)*(t s)² = 0
0.551t - 4.9t² = 0
t = 0, or t = 0.1124 s
t = 0 corresponds to launch, and t = 0.1124 s corresponds to landing.

Frog B:
Launch velocity is 1.75 m/s
When t = 0.1124 s, the position of the frog is
s = (1.75 m/s)(0.1124 s) - 0.5*(9.8 m/s²)*(0.1124 s)²
= 0.135 m
The velocity of frog B is
v = (1.75 m/s) - (9.8 m/s²)*(0.1124 s)
= 0.6485 m/s

Answer:
When frog A lands on the ground,
Frog B is 0.135 m above ground and its velocity is 0.649 m/s upward.

5 0

4 years ago