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wariber [46]
3 years ago
11

While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the

basket. His 74-kg body decelerated from 7.6 m/s to 0 m/s in 0.16 seconds.
a. Determine the force acting upon Logan's body.
b. If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds. Determine what the force on his body would have been for such an abrupt collision.
Physics
1 answer:
olga2289 [7]3 years ago
6 0

Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

Mass M = 74kg

Initial velocity U = 7.6 m/s

Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

Change in momentum = 74×7.6 + 74×7.6

Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s

F = 1124.8/0.0080 = 140600 N

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Zinaida [17]
<h2>Let us find the efficiency : Ans = 0.6</h2>

Explanation:

we know :

efficiency = output/input

We also know that :

output = m x g x h

where :

m = mass of body

g = acceleration due to gravity

h = height of body from floor

Thus, output = 0.6 x 10 x 1.2 = 7.2J

Similarly ,input = 0.6 x 10 x 2 = 12J

Thus efficiency = 7.2/12 = 0.6

4 0
3 years ago
A rectangular field is to be enclosed on four sides with a fence. Fencing costs $2 per foot for two opposite sides, and $3 per f
borishaifa [10]
<h2>Dimension for cheap enclose = 32.45 ft x 23.52 ft</h2>

Explanation:

Area of rectangular field, A = 830 ft²

Length = l

Width = w

So we have

                 l x w = 830

                    l=\frac{830}{w}

Fencing costs $2 per foot for two opposite sides, and $3 per foot for the other two sides.

            Cost for fencing, C = 2 x 2 x w + 3 x 2 x l = 4 w + 6 l

             C=4w+6\times \frac{830}{w}

For minimum cost we have derivative is zero

           dC=4\times dw-6\times \frac{830}{w^2}\times dw\\\\0=4\times dw-6\times \frac{830}{w^2}\times dw\\\\w^2=1245\\\\w=32.45ft\\\\lw=830\\\\l\times 32.45=830\\\\l=23.52ft

Dimension for cheap enclose = 32.45 ft x 23.52 ft

8 0
3 years ago
"As you sit in a fishing boat, you notice that 12 waves pass the boat every 45 s. If the distance from one crest to the next is
Alchen [17]

Answer:

2.4564 m/s

Explanation:

Number of waves in 1 second

\dfrac{12}{45}=0.267\ Hz

This is the frequency 0.267 Hz

The distance mentioned in the question is the

\lambda = Wavelength = 9.2 m

The speed of wave is

v=f\lambda

\Rightarrow v=0.267\times 9.2

\Rightarrow v=2.4564\ m/s

The speed of the wave is 2.4564 m/s

4 0
3 years ago
a 1 m length of string is wrapped around a solid disk, of mass .25 kg and a radius of .30m, mounted on a frictionless axle. the
Nezavi [6.7K]

Answer:

60 rad/s

Explanation:

∑τ = Iα

Fr = Iα

For a solid disc, I = ½ mr².

Fr = ½ mr² α

α = 2F / (mr)

α = 2 (20 N) / (0.25 kg × 0.30 m)

α = 533.33 rad/s²

The arc length is 1 m, so the angle is:

s = rθ

1 m = 0.30 m θ

θ = 3.33 rad

Use constant acceleration equation to find ω.

ω² = ω₀² + 2αΔθ

ω² = (0 rad/s)² + 2 (533.33 rad/s²) (3.33 rad)

ω = 59.6 rad/s

Rounding to one significant figure, the angular velocity is 60 rad/s.

8 0
3 years ago
The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1
telo118 [61]
<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life h of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: h_{B-13}=5(10)^{-10}s=0.0000000005s

beryllium-15: h_{B-15}=2(10)^{-7}s=0.0000002s

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

h_{B-15}=X.h_{B-13}

Where X is the amount we want to find:

X=\frac{h_{B-15}}{h_{B-13}}

X=\frac{2(10)^{-7}s}{5(10)^{-10}s}

Finally:

X=400

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0
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