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harina [27]
3 years ago
14

Q: The picture above shows a frying pan. Why are frying pans usually made out of metal? 

Physics
2 answers:
Elena-2011 [213]3 years ago
5 0
A.good conduter of heat
Natasha_Volkova [10]3 years ago
4 0
A. Because metal is a good conductor of heat.

If a frying pan wasn't a good conductor of heat it's purpose of cooking things wouldn't really be resolved. Without heat it couldn't cook well. I think lol that's my way of thinking.



Hope I helped :)
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An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place
anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

mu=(M+m)v

v=\frac{m}{M+m}\times u

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

\frac{(M+m)v^2}{2}=(M+m)gh

here h=L(1-\cos \theta ) from diagram

therefore

v=\sqrt{2gL(1-\cos \theta )}

initial velocity in terms of v

u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}

For first case \theta =6.8^{\circ}

u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}

for second case \theta =11.4^{\circ}

u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}

Therefore \frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}

\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}

\frac{u_1}{u_2}=1.084

i.e.\frac{v_1}{v_2}=1.084

4 0
3 years ago
Please help this is really important thanks
Juliette [100K]
1. Giga is the largest
2. Stem-and-leaf
6 0
3 years ago
Capacitors C1 = 6.45 µF and C2 = 2.50 µF are charged as a parallel combination across a 250 V battery. The capacitors are discon
denpristay [2]

Answer:

For C1, Q =  1.6125×10⁻³ C

For C2, Q =  6.25×10⁻⁴ C

Explanation:

Note: Since the capacitors are connected in parallel, The voltage across each of them is equal.

From the question,

Q = CV........................ Equation 1

Where Q = Charge on the capacitor, V = Voltage across the capacitor, C = Capacitance of the capacitor.

For the first capacitor,

Q = C1V............. Equation 2

Where C1 = 6.45 μF= 6.45×10⁻⁶ F, V = 250 V

Substitute into equation 2

Q = (6.45×10⁻⁶ )(250)

Q = 1.6125×10⁻³ C.

For the the second capacitor,

Q = C2V............. Equation 3

Given: C2 = 2.50 μF = 2.5×10⁻⁶ F, V = 250 V

Q = (2.5×10⁻⁶ )(250)

Q = 6.25×10⁻⁴ C

4 0
3 years ago
3. Study 1 differed from Study 2 in which of the follow-
elena-s [515]
I believe the correct answer is C
4 0
2 years ago
The magnitude of a negative of a vector is negative. O True O False
Paha777 [63]

Answer:

The given statement is false.

Explanation:

For any negative vector

\overrightarrow{r}=-x\widehat{i}-y\widehat{j}

The magnitude of the vector is given by

|r|=\sqrt{(-x)^{2}+(-y)^{2}}\\\\|r|=\sqrt{x^2+y^2}

As we know that square root of any quantity cannot be negative thus we conclude that the right hand term in the above expression cannot be negative hence we conclude that magnitude of any vector cannot be negative.

8 0
3 years ago
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