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3 years ago

Do you think a baseball curves better at the top of a high mountain or down on a flat plain

Physics

1 answer:

Serhud [2]3 years ago

4 0

A baseball will curve better on the flat plain if it is higher than sea level but low elevation.

Hope this helped!

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In a discussion person A is talking 1.2 dB louder than person B, and person C is talking 3.2 dB louder than person A. What is th

liberstina [14]

Answer: 3.84dB

Explanation:

Since person A is talking 1.2dB louder than B, we will have

A = 1.2B... (1)

Similarly, person C is talking 3.2 dB louder than person A, we have

C = 3.2A... (2)

From equation 1, B = A/1.2... (3)

To get the ratio of the sound intensity of person C to the sound intensity of person B, we will divide equation 2 by 3 to give

C/B = 3.2A/{A/1.2}

C/B = 3.2A×1.2/A

C/B = 3.2×1.2

C/B = 3.84dB

3 0

4 years ago

Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum

asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

6 0

3 years ago

A projectile fired up into the air at an angle has a range of 235 m and a flight time of 47 s.

madam [21]

<h2>Answer:5 $ms^{-1}$ ,133.6 $m$ ,51.18 $ms^{-1}$ </h2>

Explanation:

Let $v_{x}$ , $v_{y}$ be the horizontal and vertical components of velocity.

Question a:

Horizontal component of velocity is the ratio of range and time of flight.

So,horizontal component of velocity is $\frac{235}{47}=5ms^{-1}$

So, $v_{x}=5ms^{-1}$

Question b:

Time of flight= $\frac{2v_{y}}{g}$

So, $v_{y}=\frac{47\times 9.8}{2}=51.18ms^{-1}$

Maximum height is given by $\frac{v_{y}^{2}}{2g}$

So,maximum height is $\frac{51.18^{2}}{2\times 9.8}=133.6m$

Question c:

The vertical velocity is already calculated in Question b.

$v_{y}=51.18ms^{-1}$

7 0

3 years ago

The x vector component of a displacement vector has a magnitude of 146 m and points along the negative x axis. The y vector comp

larisa86 [58]

Answer:

a) the magnitude of r is 184.62

b) the direction is 37.74° south of the negative x-axis

Explanation:

Given the data in the question;

as illustrated in the image blow;

To find the the magnitude of r, we will use the Pythagoras theorem

r² = y² + x²

r = √( y² + x²)

we substitute

r = √((-113)² + (-146)²)

r = √(12769 + 21316 )

r = √(34085 )

r = 184.62

Therefore, the magnitude of r is 184.62

To find its direction, we need to find ∅

from SOH CAH TOA

tan = opposite / adjacent

tan∅ = -113 / -146

tan∅ = 0.77397

∅ = tan⁻¹( 0.77397 )

∅ = 37.74°

Therefore, the direction is 37.74° south of the negative x-axis

7 0

3 years ago

Ninais ba ni floranteng mapalapit ang kanyang loob kay adolfo

luda_lava [24]

Answer:

Explanation:

8 0

2 years ago