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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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12*8A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 13 feet. The ball is started in
max2010maxim [7]

Answer:

See attached pictures.

Explanation:

See attachments for explanation.

6 0
2 years ago
A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later
UNO [17]

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity.  If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

                (64.2 m/s)  -  [ (9.8 m/s² ) x (1.5 sec) ] 

            =  (64.2 m/s)  -       [      14.7 m/s      ]

            =             49.5 m/s  .  (upward)

7 0
2 years ago
3) 4 electrons are placed - one electron per corner - at the corners of a square of side 1 meter. One fixed proton is placed in
Eduardwww [97]

Explanation:

3

i believe that they are all going at 3.2 meters each, I did 4 times 0.8

4 0
2 years ago
Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3600 Hz . To do this, the car a
Alex17521 [72]

Answer:

The capacitance and the inductance can choose for a car-alarm circuit are

C = 215.27 μF

L = 9.078 μH

Explanation:

V =12.0 V, E = 1.55*10^2 J, f = 3600 Hz

To determine the capacitance can use the equation

U_c= \frac{1}{2}*C*V^2

Solve to C'

C = \frac{U_c*2}{V^2}=\frac{1.55x10^2J*2}{12.0^2V}

C=215.27 uF

To find the inductance can use the frequency of the circuit

f = \frac{1}{2\pi* \sqrt{C*L} }

Solve to L'

L = \frac{1}{4\pi^2*f^2*C}=\frac{1}{4\pi^2*3600^2*215.27 uF}}

L = 9.078 uH

6 0
2 years ago
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
MArishka [77]

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

6 0
2 years ago
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