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3 years ago

If a water wave completes one cycle in 2 seconds, what is the period of the wave?

2 answers:

8 0

The definition of the period of a wave is the time that it takes for the wave to complete one cycle. As stated in the problem, the certain water wave completes one cycle in 2 seconds. From this information alone, we can conclude that the period of the wave is indeed, 2 seconds.

Alex3 years ago

3 0

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You are climbing in the High Sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of t

s344n2d4d5 [400]

Answer:

Height with sound ignored = Gravity x Time taken = 9.8 x 8.60 = 84.28 meters

Time taken by the sound = 84.28/330 = 0.255 seconds

Height with sound involved = (84.28 x 0.255) + 84.28 = 105.80 meters

a. Underestimated

5 0

3 years ago

Two objects attract each other gravitationally with a force of 2.5 x 10^-10N when they are 0.25 m apart. Their total mass is 4.0

In-s [12.5K]

Answer:

M = 3.9406 kg and m = 0.0594 kg

Explanation:

The gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically it is expressed as follows:

Fg = (G×M×m)/r² Formula (1)

Where:

Fg is the gravitational force (N)

G is the universal gravitation constant, G = 6.67 × 10⁻¹¹ (N×m²)/kg²

M and m are the masses of the bodies that interact (kg).

r is the distance that separates them (m).

Known Data

Fg = 2.5 × 10⁻¹⁰ N

r = 0.25 m

G = 6.67 × 10⁻¹¹ (N×m²)/kg²

Problem development

We propose 2 equations

M + m = 4kg

M = 4 - m equation (1)

We replace in formula (1)

2.5 × 10⁻¹⁰ = (6.67 × 10⁻¹¹ × M × m)/(0.25)²

2.5 × 10⁻¹⁰ × (0.25)² = (6.67 × 10⁻¹¹ × M × m)

(2.5 × 10⁻¹⁰ × (0.25)²)/(6.67 × 10⁻¹¹) = M × m

M × m = 0.234 equation (2)

We replace M = 4 - m in equation (2)

(4 - m) × m = 0.234

4m - m² = 0.234

m² - 4m + 0.234 = 0 (quadratic equation)

We apply the formula for the quadratic equation and obtain 2 values for m that meet the conditions:

m = 3.9406 kg or m = 0.0594 kg

We replace m in equation (1)

M = 4 - 3.9406 = 0.0594 kg or M = 4 - 0.0594 = 3.9406

To meet the condition that M + m must give 4 kg, one mass must be equal 3.9406 and the other must equal 0.0594, then:

M = 3.9406 kg and m = 0.0594 kg

6 0

4 years ago

In 1991 at smith college, in massachusetts, ferdie adoboe ran 1.00 × 102 m backward in 13.6 s. suppose it takes adoboe 2.00 s to

Alik [6]

as it is given that it covers a total distance 1 * 10^2 m

total time taken by it = 13.6 s

now the average speed is given as ratio of total distance and total time

$v = \frac{d}{t}$

$v = \frac{1* 10^2 }{13.6}$

$v = 7.35 m/s$

so the average speed will be 7.35 m/s

now if it starts from rest and achieve the final speed as 7.35 m/s

now we can use kinematics

$v_f = v_i + at$

$7.35 = 0 + a* 2$

$a = 3.68 m/s^2$

so its acceleration will be 3.68 m/s^2

4 0

4 years ago

Millikan is doing his oil drop experiment. He has a droplet with radius 1.6 µm suspended motionless in a uniform electric field

swat32

Answer:

The charge on the droplet is $3.106\times10^{-16}\ C$ .

Yes, quantization of charge is obeyed within experimental error.

Explanation:

Given that,

Radius = 1.6μm

Electric field = 46 N/C

Density of oil = 0.085 g/cm³

We need to calculate the charge on the droplet

Using formula of force

$F= qE$

$mg=qE$

$V\times\rho\times g=qE$

$q=\dfrac{V\times\rho}{E}$

$q=\dfrac{\dfrac{4}{3}\pi\times r^3\times\rho\times g}{E}$

Put the value into the formula

$q=\dfrac{\dfrac{4}{3}\times\pi\times(1.6\times10^{-6})^3\times85\times9.8}{46}$

$q=3.106\times10^{-16}\ C$

We need to calculate the quantization of charge

Using formula of quantization

$n = \dfrac{q}{e}$

Put the value into the formula

$n=\dfrac{3.106\times10^{-16}}{1.6\times10^{-19}}$

$n=1941.25$

Yes, quantization of charge is obeyed within experimental error.

Hence, The charge on the droplet is $3.106\times10^{-16}\ C$ .

Yes, quantization of charge is obeyed within experimental error.

7 0

3 years ago

Read 2 more answers

Would thermal energy be greater at 0°C or 48°F?

Alik [6]

If you know 0 C= 32 F, it is greater at 48 F

5 0

4 years ago