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Blababa [14]
2 years ago
13

What common laboratory measuring device will we likely always use in experiments that measure enthalpy changes?

Physics
1 answer:
adelina 88 [10]2 years ago
5 0
The most common measuring device to be used in measuring enthalpy changes is the thermometer. 
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You are given two rectangular blocks of shiny metal, Block A and Block B, and are asked to determine which one will float in a b
vladimir2022 [97]

Answer:

Explanation:

Volume of block A = 10 x 6 x 1 = 60 cm³

Mass of block A = 630 g

density of mass A = mass / density

= 630 / 60 = 10.5g / cm³

Volume of block B = 5 x 5 x 3 = 75 cm³

Mass of block A = 604 g

density of mass A = mass / density

= 604 / 75 = 8.05 g / cm³

Since density of both A and B are less than that of mercury , both will float in mercury.

7 0
2 years ago
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the jun
AlexFokin [52]

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

i_1+i_3=i_2                 (1)

i1 = 0.40 A

i2 = 0.75 A

you solve the equation i3 from the equation (1):

i_3=i_2-i_1=0.75A-0.40A=0.35A

Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

3 0
2 years ago
Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu
Anit [1.1K]
A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>

m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... y direction

Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 =  625N^2 =>

Net force = √625 = 25N

F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2

Answer: 5.00 m/s^2

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal. </span>

<span>m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... 60 degress above x direction

Components of F2

F2,x = F2*cos(60) = 15N / 2 = 7.5N

F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N


Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N

Total force in y = F2,y = 13.0 N

Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =

= 30.42N

a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2

Answer: 6.08 m/s^2


</span>
8 0
1 year ago
Read 2 more answers
Calculate the weight of a 4.5 kg rabbit.
solniwko [45]
The correct answer is: 13900589.
3 0
2 years ago
One of the Lady Spartans was falling to the ground after
dem82 [27]

Answer:

Explanation:

Given that,

A lady falling has a final velocity of 4m/s

v = 4m/s

Mass of the lady is 60kg.

m = 60kg

Using conservation of energy, the potential energy of the body from the point where the lady is dropping is converted to the final kinetic energy of the lady.

Therefore,

P.E = K.E(final) = ½mv²

P.E = ½ × 60 × 4²

P.E = 480 J.

0 0
2 years ago
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