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2 years ago

Raising 100 grams of water from 40 to 60 °C (the specific heat capacity of water is 1

Physics

2 answers:

asambeis [7]2 years ago

6 0

Answer:i think it is c

ANTONII [103]2 years ago

5 0

C. 2000 calories.

Explanation/calculation:

Specific heat capacity = calories / mass * (final temperature - initial temperature)

1 = calories / 100 * (60 - 40)
1 = calories / 100 * 20
1 * (100 * 20) = calories
1 * 2000 = calories
2000 = calories

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The mass of a string is 7.7 × 10-3 kg, and it is stretched so that the tension in it is 190 N. A transverse wave traveling on th

Scilla [17]

Length of the strings = 2.33 m

3 0

2 years ago

A laser beam with a frequency of 180 Hz forms an 8 m standing wave with 10 nodes.

DIA [1.3K]

Answer:33

Explanation:

F = frequency

N = Node count

w = wave lenght

v = wave velocity

L = distance wave traveled

First find wave length of laser

w = (2/(N))*(L)

w = (2/(10))*(8)

w = 1.6

then using (w), find velocity

V = (w)(F)

V = (1.6)*(108)

V = 288

Plug in V and the new frequency to solve for new node count

F = NV/2L

(600) = (N)*(288) / 2 * (8)

(N) = 33.33

there are 33 nodes

8 0

2 years ago

A charge of 1.5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How m

belka [17]

Answer:

Energy stored in the capacitor is $U=2.7\times 10^{-5}\ J$

Explanation:

It is given that,

Charge, $q=1.5\ \mu C=1.5\times 10^{-6}\ C$

Potential difference, V = 36 V

We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :

$U=\dfrac{1}{2}q\times V$

$U=\dfrac{1}{2}\times 1.5\times 10^{-6}\times 36$

U = 0.000027 J

$U=2.7\times 10^{-5}\ J$

So, the potential energy is stored in the capacitor is $U=2.7\times 10^{-5}\ J$ . Hence, this is the required solution.

4 0

2 years ago

Read 2 more answers

How many types of quarks are there?<br> a. 2<br> b. 4<br> c. 6<br> d. 8

Nostrana [21]

There are 6 types of quarks.

Answer: C. 6

8 0

2 years ago

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A balloon filled with helium gas at 20°C occupies 4.91 L at 1.00 atm. The balloon is immersed in liquid nitrogen at -196°C, whil

mrs_skeptik [129]

Answer:

0.25 L

Explanation:

$P_1$ = Initial pressure = 1 atm

$T_1$ = Initial Temperature = 20 °C

$V_1$ = Initial volume = 4.91 L

$P_2$ = Final pressure = 5.2 atm

$T_2$ = Final Temperature = -196 °C

$V_2$ = Final volume

From ideal gas law we have

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow V_2=\dfrac{P_1V_1T_2}{T_1P_2}\\\Rightarrow P_2=\dfrac{1\times 4.91(273.15-196)}{(20+273.15)\times 5.2}\\\Rightarrow V_2=0.24849\ L\approx 0.25\ L$

The pressure experienced by the balloon is 0.25 L

7 0

2 years ago