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Alona [7]
3 years ago
15

*NEED ANSWER STAT*

Physics
2 answers:
tresset_1 [31]3 years ago
6 0
Becomes a +1 ion for this
icang [17]3 years ago
6 0
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A 1.00 kg block of ice, at -25.0°C, is warmed by 35 kJ of energy. What is the final temperature of the ice?
ahrayia [7]

Answer:

-8.4°C

Explanation:

From the principle of heat capacity.

The heat sustain by an object is given as;

H = m× c× (T2-T1)

Where H is heat transferred

m is mass of substance

T2-T1 is the temperature change from starting to final temperature T2.

c- is the specific heat capacity of ice .

Note : specific heat capacity is an intrinsic capacity of a substance which is the energy substained on a unit mass of a substance on a unit temperature change.

Hence ; 35= 1× c× ( T2-(-25))

35= c× ( T2+25)

35 =2.108×( T2+25)

( T2+25)= 35/2.108= 16.60°{ approximated to 2 decimal place}

T2= 16.60-25= -8.40°C

C, specific heat capacity of ice is =2.108 kJ/kgK{you can google that}

6 0
2 years ago
Which statements did both Aristotle and Ptolemy assume? Check all that apply.
tensa zangetsu [6.8K]

Bodies in space traveled in circles.

The planets revolved around the Earth.

6 0
2 years ago
Read 2 more answers
Your friend tells you that a shiny element has to be a metal. Do you agreed? Explain your reasoning
Svetradugi [14.3K]

no i dont agree, because a glass bowl is shiny but its not a metal. just because some metals are shiny doesnt make all metals shiny

3 0
3 years ago
Read 2 more answers
A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th
Stells [14]

Answer:

The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     V=\frac{V_{o}}{\kappa}

The capacitance is modified as:

                                              C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}

The stored energy without the dielectric is

                                               PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}

The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

5 0
3 years ago
An object with a non-zero speed must be _________.
pogonyaev
It would be either A or C if its still moving and not stopping
8 0
3 years ago
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