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ryzh [129]
3 years ago
6

Adding or removing thermal energy to or from a substance does not always cause its temperature

Physics
2 answers:
julia-pushkina [17]3 years ago
6 0
Change in thermal energy not always cause it's temperature change. It is the situation when water reaches either at 0 C or 100 C then thermal energy doesn't cause change in temperature instead it changes the state of matter.

In short, Your Answer would be "True"

Hope this helps! 
cluponka [151]3 years ago
3 0
<span>Adding or removing thermal energy to or from a substance does not always cause its temperature
to change.

a)True

This is true in the case of boiling and melting point. That is when a change of state is involved.

For example in boiling the water gets to 100 degrees, if heat is continuously added, there is no change in temperature. The water simply changes state.
</span>
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A certain signal molecule S in heart tissue is degraded by two different biochemical pathways: when only Path 1 is active, the h
Misha Larkins [42]

Answer:

Half life of S = 3.76secs

Explanation:

The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.

For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564

for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615

Simply say = 1/3.9 + 1/104 = 0.266015

So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active

6 0
4 years ago
An electron in a photosynthetic pigment that is excited to a higher energy state generally has one of two fates. The first optio
wolverine [178]

Answer:

The excited electron itself can be passed to another molecule

Explanation:

4 0
3 years ago
Explain how a dense substance, such as metal, is able to float on a less dense substance, like water. Please Help! Remember you
wlad13 [49]
Archimedes and the principle of flotation ... "Cast iron sinks" ... unless it's a ship, of course is an old joke.
What has to happen is that supposing the object has a mass of 1kg, then it has to displace 1kg of water to float. However the volume of the object is decided by the object's density. Unless, that is, it is "hollowed out" or otherwise manipulated, such as is done in the hull of a metal ship ...
There's a thing called a "Eureka can" used in physics labs.

4 0
3 years ago
a 2kg mass is held 4m above the ground. what is the approximate potential energy of the mass with respect to the ground?
aleksley [76]

The potential energy of the object is 78.4 J

Explanation:

The gravitational potential energy (GPE) of an object is the energy possessed by the object due to its position in a gravitational field.

For an object near the Earth's surface, the GPE is given by:

GPE=mgh

where

m is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity near the Earth's surface

h is the height of the object relative to the surface

For the object in this problem,

m = 2 kg

h = 4 m

Substituting, we find its GPE:

GPE=(2)(9.8)(4)=78.4 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

6 0
4 years ago
"If the object has a speed of" -2.5 m/s at x = 0m, find its speed at x = 5.00 m and its speed at x = 15.0 m.
frez [133]

Answer:

Hello your question has some missing parts attached below is the missing part

A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below.

answer : speed at x = 5 m = 3.35 m/s

              speed at x = 15m  = 5.12 m/s

Explanation:

initial speed( x = 0 ) = 2.5 m/s

speed at x = 5.00 m = ?

speed at x = 15 m = ?

Determine speed at x = 5 m

First we will apply the expression for work-energy theorem

w = \frac{1}{2} m(v^2 - v_{0}^2 )  ----- ( 1 )

where : w = 7.50J, v_{0} = 2.5m/s , m = 3.0 kg ( in the expression of work )

input values into equation 1

7.50 = 1/2 (3 ) ( v^2 - 2.5^2 )

7.50 = 3/2 ( v^2 - 6.25 )

5 j/kg = v^2 - 6.25

∴ v = √11.25 = 3.35 m/s

Determine speed at X = 15

first we will determine the work done form x = 5 to x = 15

W = 7.5J + 15J + 7.5J = 30J,  v_{0} = 2.5m/s , m = 3.0 kg

w = \frac{1}{2} m(v^2 - v_{0}^2 ) --- ( 2 )

equation2 becomes

30J = 1/2 ( 3 ) ( v^2 - 6.25 )

30J = 3/2 ( V^2 - 6.25 )

20 J/kg = v^2 - 6.25

v = √26.25 = 5.12 m/s

5 0
3 years ago
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