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3 years ago

1.

Physics

2 answers:

Jet001 [13]3 years ago

4 0

Answer:

1. high frequency and high energy.

2. short wavelengths and high frequencies.

3. less energy and long wavelengths.

4. high frequencies and high energy

5. X-rays

6. infrared waves

7. radio waves.

8. Gamma Rays

9. Microwaves

10. infrared waves

sweet [91]3 years ago

3 0

Answer: 1: High frequency and high energy. 2: Short wavelengths and high frequencies. 3: Less energy and long wavelengths. 4: High frequencies and high energy. 5: X-rays. 6: Infrared waves. 7: Radio waves. 8: Gamma Rays. 9: Microwaves. 10: Infrared waves.

Explanation: 1: High frequency and high energy. 2: Short wavelengths and high frequencies. 3: Less energy and long wavelengths. 4: High frequencies and high energy. 5: X-rays. 6: Infrared waves. 7: Radio waves. 8: Gamma Rays. 9: Microwaves. 10: Infrared waves.

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vfiekz [6]

Answer:

Current, I = 0.0011 A

Explanation:

It is given that,

Diameter of rod, d = 2.56 cm

Radius of rod, r = 1.28 cm = 0.0128 m

The resistivity of the pure silicon, $\rho=2300\ \Omega-m$

Length of rod, l = 20 cm = 0.2 m

Voltage, $V=1\times 10^3\ V$

The resistivity of the rod is given by :

$R=\rho\dfrac{L}{A}$

$R=2300\ \Omega-m\dfrac{0.2\ m}{\pi (0.0128\ m)^2}$

R = 893692.30 ohms

Current flowing in the rod is calculated using Ohm's law as :

V = I R

$I=\dfrac{V}{R}$

$I=\dfrac{10^3\ V}{893692.30\ \Omega}$

I = 0.0011 A

So, the current flowing in the rod is 0.0011 A. Hence, this is the required solution.

6 0

3 years ago

The glass in a window is 35 inches wide and 20 inches tall, and standard atmospheric pressure is 14.7 pounds per square inch. Wh

iogann1982 [59]

Answer:103 pounds

Explanation:

Given

width of window $b=35 in.$

height of window $h=20 in.$

standard atmospheric pressure $P_{outside}=14.7 psi$

Also $P_{inside}=1.01P_{outside}$

Thus Net Force on the window will be the algebraic sum of Force due to outside and inside Pressure .

$F_{net}=(P_{inside}-P_{outside})\cdot A$

$F_{net}=P_{outside}(1.01-1)\times 35\times 20$

$F_{net}=14.7\times 0.01\times 35\times 20$

$F_{net}=102.9\ pounds\approx 103\ pounds$

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3 years ago

A person observes a firework display for A safe distance of .750 km. Assuming that sound travels at 340 m/s in air what is the t

WINSTONCH [101]

Answer:

t = 2.2 s

Explanation:

Given that,

A person observes a firework display for A safe distance of 0.750 km.

d = 750 m

The speed of sound in air, v = 340 m/s

We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.

$v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{750\ m}{340\ m/s}\\\\t=2.2\ s$

So, the required time is 2.2 seconds.

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3 years ago

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3 years ago