A wave in the ocean has a wavelength of 2 m and a frequency of 0’5 Hz. What is the speed of this wave?

A period of one year on Earth is the time it takes Earth to

Kazeer [188]

D make one revolution around the sun hope this helps

7 0

2 years ago

If the atomic mass of Sodium-18 is 18.02597 u, what is the binding energy?

Klio2033 [76]

Answer:

<h3>The binding energy of sodium Na=<em>5.407791×10⁹J</em></h3>

Explanation:

<h3>Greetings !</h3>

Binding energy, amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Binding energy is especially applicable to subatomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals.

<h2>Formula : Eb=(Δm)c²</h2><h3>where:Eb= binding energy</h3><h3> .Δm= mass defect(kg)</h3><h3> c= speed of light 3.00×10⁸ms¯¹</h3><h2 /><h3><u>Given</u><u> </u><u>values</u></h3>

m= 18.02597
c=3.00×10⁸ms¯¹

<h3><u>required </u><u>value</u></h3>

Eb=?

<h3><u>Solution:</u></h3>

Eb=(Δm)c²
Eb=(18.02597)*(3.00*10⁸ms¯¹
Eb=5.407791*10⁹J

8 0

1 year ago

A wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same m

Nina [5.8K]

Answer: $0.05\ mm$

Explanation:

Given

Cross-sectional area of wire $A_1=4\ mm^2$

Extension of wire $\delta l=0.1\ mm$

Extension in a wire is given by

$\Rightarrow \delta l=\dfrac{FL}{AE}$

where, $E=\text{Youngs modulus}$

$\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)$

for same force, length and material

$\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)$

Divide (i) and (ii)

$\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm$

5 0

2 years ago

Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at

Stolb23 [73]

Answer:

a. Fx = -8.089 N b. Fy = 3.525 N c. 8.824 N d. 336.45°

Explanation:

Since F₁ = 9.2 N and acts at 57° above the negative axis in the second quadrant, its x-component is -F₁cos57° and its y- component is F₁sin57°

Since F₁ = 5.2 N and acts at 53.7° below the negative axis in the third quadrant, its x-component is -F₂cos53.7° and its y- component is -F₂sin53.7°

Part A

What is the x component Fx of the resultant force?

The x component of the resultant force Fx = -F₁cos57° + -F₂cos53.7° = -9.2cos57° + (-5.2cos53.7°) = (-5.011 - 3.078) N = -8.089 N

Part B

What is the y component Fy of the resultant force?

The y component Fy of the resultant force = F₁sin57° + -(F₂sin53.7°) = 9.2sin57° - 5.2sin53.7° = (7.716 - 4.191) N = 3.525 N

Part C

What is the magnitude F of the resultant force?

The magnitude F of the resultant force = √(Fx² + Fy²)

F = √(-8.089² N + 3.525² N) = √65.432 + 12.426 = √77.858 = 8.824 N

Part D

What is the angle ? that the resultant force forms with the negative x axis?

The angle the resultant force makes with the negative x axis is given by

θ = tan⁻¹(Fy/Fx) = tan⁻¹(3.525/-8.089) = tan⁻¹-0.4358 = -23.55°.

To measure it from the negative x axis, we add 360. So, our angle = 360 -23.55 = 336.45°

7 0

2 years ago

An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where ther

jeka94

Answer:

T = 188.5 s, correct is C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

initial instant. Before the crash

L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

L_f = I₀ w + m r v

L₀ = L_f

m r v₀ = I₀ w + m r v

angular and linear velocity are related

v = r w

w = v / r

m r v₀ = I₀ v / r + m r v

m r v₀ = (I₀ / r + mr) v

v = $\frac{m}{\frac{I_o}{r} +mr} \ r v_o$

let's calculate

v = $\frac{0.020}{\frac{1.4}{0.6 } + 0.020 \ 0.6 } \ 0.6 \ 4$

v = $\frac{0.020}{2.345} \ 2.4$

v = 0.02 m / s

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

v = x / T

T = x / v

the distance of a circle with radius r = 0.6 m

x = 2π r

we substitute

T = 2π r / v

let's calculate

T = 2π 0.6/0.02

T = 188.5 s

reduce

t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is C

6 0

2 years ago