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3 years ago

What is another way to describe the vector below?

Physics

2 answers:

nignag [31]3 years ago

8 0

Answer:

-40 feet to the left

Explanation:

opposite of 40 feet from right is -40 to left

IrinaVladis [17]3 years ago

8 0

Answer:

B. -40 feet to the left

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A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be in a s

Greeley [361]

Answer:

$a = 4.76 cm$

Explanation:

As we know by the law of diffraction through circular aperture

$a sin\theta = 1.22 \lambda$

here we know that

a = diameter of the aperture

$\theta$ = diffraction angle

$\lambda$ = wavelength

now we have

$a = \frac{1.22 \lambda}{sin\theta}$

Now in order to find the wavelength we know

$\lambda = \frac{c}{f}$

$\lambda = \frac{343}{9100}$

$\lambda = 0.0376 m$

now we have

$a = \frac{1.22(0.0376)}{sin75}$

$a = \frac{0.046}{0.966}$

$a = 0.0476 m= 4.76 cm$

6 0

3 years ago

Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit.

Effectus [21]

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, $\lambda=608\ nm=608\times 10^{-9}\ m$

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

$y=\dfrac{mD\lambda}{a}$

where

a = width of the slit

$a=\dfrac{mD\lambda}{y}$

$a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}$

a = 0.000167 m

$a=1.67\times 10^{-4}\ m$

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

7 0

3 years ago

A pirate ship shoots a cannon towards an oncoming ship. The cannon ball has a mass of 25 kg and the ship has a mass of 2500 kg.

mars1129 [50]

Answer:

25 m/s in the opposite direction with the ship recoil velocity.

Explanation:

Assume the ship recoil velocity and velocity of the cannon ball aligns. By the law of momentum conservation, the momentum is conserved before and after the shooting. Before the shooting, the total momentum is 0 due to system is at rest. Therefore, the total momentum after the shooting must also be 0:

$m_sv_s + m_bv_b = 0$

where $m_s = 2500 kg, m_b = 25 kg$ are masses of the ship and ball respectively. $v_s = 0.25 m/s, v_b$ are the velocities of the ship and ball respectively, after the shooting.