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2 years ago

Omar rubs a balloon with wool, giving the balloon a

Physics

1 answer:

CaHeK987 [17]2 years ago

3 0

Answer:

-8.9×10-14=-89-14=-103°C

25°C=2km

-103°C=? Less

=-103/25×2km

=-8.24km

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The human ear canal is, on average, 2.5cm long and aids in hearing by acting like a resonant cavity that is closed on one end an

Troyanec [42]

Answer:

$3400$ Hz

Explanation:

We know that

1 cm = 0.01 m

$L$ = Length of the human ear canal = 2.5 cm = 0.025 m

$V$ = Speed of sound = 340 ms⁻¹

$f$ = First resonant frequency

The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as

$f = \frac{(2n - 1)V}{4L}$

for first resonant frequency, we have n = 1

Inserting the values

$f = \frac{(2(1) - 1) 340}{4(0.025)}$

$f = \frac{340}{4(0.025)}$

$f = 3400$ Hz

4 0

3 years ago

IMPORTANT!!! DUE IN 5 MINS I NEED HELP PLEASE

Ket [755]

Answer:

684.5 is the weight on mars and 1813 on earth

Explanation:

185*3.7=684.5 185*9.8=1813 you multiply for earth by 9.8 because that's the gravity on earth and you multiply by 3.7 because that's the gravity on mars

6 0

2 years ago

4. Explain how states of matter change in regards to<br> a. Temperature-<br> b. Pressure-

Alina [70]

Temperature can change the state from solid to liquid causing it to melting, liquid to gas causing vaporization or a solid to a gas causing sublimation. Pressure alone cannot change the state of matter.

5 0

2 years ago

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

8 0

2 years ago

Substances released into the air are known as

irakobra [83]

Your answer is Emissions

3 0

2 years ago

Read 2 more answers