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tigry1 [53]
3 years ago
8

4. What is the period of the microwaves in the above question?

Physics
1 answer:
Sidana [21]3 years ago
6 0

Answer:

The period of a wave is the time for a particle on a medium to make one complete vibrational cycle. Period, being a time, is measured in units of time such as seconds, hours, days or years. The period of orbit for the Earth around the Sun is approximately 365 days; it takes 365 days for the Earth to complete a cycle.

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A woman weighs 580 N. What is her mass?
mote1985 [20]

w = mg where g = 9.8 N/kg  

m = 580/9.8 = 59 kg

5 0
4 years ago
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Solve these problems on charges by finding out Q1 and Q2. Q1Q2= 4x10^6, Q1 + Q2 =8.00×10^-6​
inna [77]

Answer:

Answer:

Q_1 = 7Q

1

=7

Q_2 = 10Q

2

=10

Q_3 = 13.5Q

3

=13.5

Step-by-step explanation:

Given

5, 7, 7, 8, 10, 11, 12, 15, 17.

Required

Determine Q1, Q2 and Q3

The number of data is 9

Calculating Q1:

Q1 is calculated as:

Q_1 = \frac{1}{4}(N + 1)Q

1

=

4

1

(N+1)

Substitute 9 for N

Q_1 = \frac{1}{4}(9 + 1)Q

1

=

4

1

(9+1)

Q_1 = \frac{1}{4}*10Q

1

=

4

1

∗10

Q_1 = 2.5th\ itemQ

1

=2.5th item

This means that the Q1 is the mean of the 2nd and 3rd data.

So:

Q_1 = \frac{1}{2}(7+7)Q

1

=

2

1

(7+7)

Q_1 = \frac{1}{2}*14Q

1

=

2

1

∗14

Q_1 = 7Q

1

=7

Calculating Q2:

Q2 is calculated as:

Q_2 = \frac{1}{2}(N + 1)Q

2

=

2

1

(N+1)

Substitute 9 for N

Q_2 = \frac{1}{2}(9 + 1)Q

2

=

2

1

(9+1)

Q_2 = \frac{1}{2}*10Q

2

=

2

1

∗10

Q_2 = 5th\ itemQ

2

=5th item

Q_2 = 10Q

2

=10

Calculating Q3:

Q3 is calculated as:

Q_3 = \frac{3}{4}(N + 1)Q

3

=

4

3

(N+1)

Substitute 9 for N

Q_3 = \frac{3}{4}(9 + 1)Q

3

=

4

3

(9+1)

Q_3 = \frac{3}{4}*10Q

3

=

4

3

∗10

Q_3 = 7.5th\ itemQ

3

=7.5th item

This means that the Q3 is the mean of the 7th and 8th data.

So:

Q_3 = \frac{1}{2}(12+15)Q

3

=

2

1

(12+15)

Q_3 = \frac{1}{2}*27Q

3

=

2

1

∗27

Q_3 = 13.5Q

3

=13.5

4 0
3 years ago
If a body is moving with a constant velocity then it has ................. acceleration
MakcuM [25]

Answer:

accelerates

Explanation:

tell more please

5 0
2 years ago
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A cosmic-ray electron moves at perpendicular to earth’s magnetic field at an altitude where the field strength is. what is the r
Harlamova29_29 [7]

4.266 m is the radius of the circular path the electron follows.

Given

Speed of electron (v) = 7.5 × 10⁶ m/s

Earth's Magnetic Field (B) = 1 × 10⁻⁵ T

We already know that

Mass of electron (m) = 9.1 × 10⁻³¹ kg

Charge on electron (q) = 1.6 × 10⁻¹⁹ C

According to the formula

Radius of circular path(r) = mass on electron × speed/ Charge × Magnetic field

Radius of circular path(r) = m × v/q × B

Put the values into the formula

r = 9.1 × 10⁻³¹ × 7.5 × 10⁶/ 1.6 × 10⁻¹⁹ × 10⁻⁵

On solving, we get

r = 4.266 m

Hence, 4.266 m is the radius of the circular path the electron follows.

Learn more about magnetic field here brainly.com/question/26257705

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In this system, ___________________ energy from the sunlight causes the solar panel to create ___________________ energy which f
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solar, heat,heat, light

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