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s2008m [1.1K]
3 years ago
5

Calculate the impulse (in kgm/s) received by a football when it's kicked in a kickoff. The ball has a mass of 0.27 kg and travel

s a horizontal distance of 68 m. The ball is kicked at an angle q of 20.5 with the horizontal and spends 2.276 seconds in the air.
Physics
1 answer:
Alona [7]3 years ago
7 0

Answer:

Impulse experienced by the ball = 8.61 kgm/s

Explanation:

Impulse is given as change in momentum according to Newton's second law of motion

Impulse = mv (assuming the ball goes to rest after being kicked)

Range of a projectile is given as

R = uₓt

uₓ = horizontal component of the velocity = u cos(θ) = u cos 20.5° = 0.9367 u

t = time of flight, given in the question as 2.276 s

R = 68 m

68 = 0.9367 u × 2.276 = 2.132 u

u = 31.9 m/s

Impulse = mv = 0.27 × 31.9 = 8.61 kgm/s

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From this diagram which of the following can you conclude
Volgvan

Answer: A

Explanation:

I say opposites attract

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3 years ago
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1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.
NikAS [45]

Answer:

1.

109.6 cm ,  - 1.74 , real

2.

1.5

Explanation:

1.

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{40} = \frac{1}{63} + \frac{1}{d}

d = 109.6 cm

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{-109.6}{63}

m = - 1.74

The image is real

2

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}

a = 10

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{- (- (a +5))}{a}

m = \frac{(5 + 10)}{10}

m = 1.5

6 0
3 years ago
You have two contentious friends, Chris and Pat, and you’ve quickly discovered that they need you to resolve arguments they have
pshichka [43]

Answer:

v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points,

Explanation:

To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.

             v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points, in the only case that it is so is when there is no acceleration.

Therefore neither of them is right.

3 0
3 years ago
. How does the area of a rectangle relate to the lengths of each side?
iragen [17]
Area is 2-dimensional which means it has a length and a width area is measured in square units such as square inches, square feet or square meters. To find the area of a rectangle multiply the length by the width.
3 0
3 years ago
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In Millikan's oil drop experiment an oil drop is at rest between two large plates separated by 1.3 cm when the potential differe
sweet-ann [11.9K]

Answer:

Mass of the oil drop, m=3.01\times 10^{-15}\ kg

Explanation:

Potential difference between the plates, V = 400 V

Separation between plates, d = 1.3 cm = 0.013 m

If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

qE=mg

m=\dfrac{qE}{g}

q=6e, e is the charge on electron

E is the electric field, E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m

m=\dfrac{6\times 1.6\times 10^{-19}\times 30769.23}{9.8}

m=3.01\times 10^{-15}\ kg

So, the mass of the oil drop is 3.01\times 10^{-15}\ kg. Hence, this is the required solution.

5 0
3 years ago
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