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ahrayia [7]
3 years ago
8

ASK YOUR TEACHER A baseball with a mass of 146 g is thrown horizontally with a speed of 40.6 m/s (91 mi/h) at a bat. The ball is

in contact with the bat for 1.05 ms and then travels straight back at a speed of 45.1 m/s (101 mi/h). Determine the average force exerted on the ball by the bat. Neglect the weight of the ball (it is much smaller than the force of the bat) and choose the direction of the incoming ball to be positive. (Indicate the direction with the sign of your answer.)
Physics
1 answer:
RideAnS [48]3 years ago
4 0

Answer:

Explanation:

mass of the ball = 146 g = 146 / 1000 = 0.146 kg

initial speed of the ball = 40.6 m/s

final speed of the ball = - 45.1 m/s

time of impact = 1.05 ms = 1.05 / 1000 = 0.00105 s

impulse, Ft = change in momentum = mv - mu = m (v-u)

F = m (v - u) / t = 0.146 kg ( -45.1 -40.6) / 0.00105 s = -11916.4 N

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3 years ago
After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r
mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

2 n t = (m + \dfrac{1}{2})\lambda

At m = 0 for the minimum thickness, so

2\times 1.45 \times t = (0 + 0.5)\times 600

2.9 \times t =300

t =\dfrac{300}{2.9}

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0
3 years ago
A thin, light wire 75.1 cm long having a circular cross section 0.555 mm in diameter has a 25.4 kg weight attached to it, causin
seraphim [82]

Answer:

(a) 3.23×10⁸ N/m²

(b)  1.46×10⁻³

(c) 2.21×10¹¹ N/m²

Explanation:

(a) Stress = Force/Area.

Stress = F/A................ Equation 1

But,

F = mg................. Equation 2

Where m = mass, and g = acceleration due to gravity

A = πd²/4................. Equation 3

d = diameter of the circular cross section.

Substitute equation 2 and equation 3 into equation 1

Stress = 4mg/πd²............. Equation 4

Given: m = 25.4 kg, d = 0.555 mm = 0.000555 m

Constant: g = 9.8 m/s², π = 3.142

Substitute these values into equation 4

Stress = 4(25.4)(9.8)/(3.142×0.00555²)

Stress = 995.68/(3.08×10⁻⁶)

Stress = 3.23×10⁸ N/m²

(b)

Strain = ΔL/L.............. Equation 5

Where ΔL = extension, L = Length.

Given: ΔL = 1.1 mm = 0.0011 m, L = 75.1 cm = 0.751 m

Substitute into equation 5

Strain = 0.0011/0.751

Strain = 1.46×10⁻³.

(c)

Young modulus = Stress/Strain

Young modulus = 3.23×10⁸/ 1.46×10⁻³

Young modulus = 2.21×10¹¹ N/m²

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Answer:

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What is the Weight of Earth?<br><br><br><br><br> Don't SpAm​
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