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lara [203]
4 years ago
12

At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force t

hat the pole exerts back on the athlete is given by F(x) = (140 N/m )x - (190 N/m2 )x2 acting over a distance of 0.19 m .How much work is done on the athlete?
Physics
1 answer:
Salsk061 [2.6K]4 years ago
5 0

Answer:

The work done on the athlete is approximately 2.09 J

Explanation:

From the definition of the work done by a variable force:

\displaystyle{\int_{x_i}^{x_j}F(x)dx}

and substituting with the function of our problem:

\displaystyle{\int_{0}^{0.19}(140x-190x^2)dx\approx2.09\mathrm{J}}

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A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with rig
aleksandr82 [10.1K]

Answer:

The rate of change of the shadow length of a person is 2.692 ft/s

Solution:

As per the question:

Height of a person, H = 20 ft

Height of a person, h = 7 ft

Rate = 5 ft/s

Now,

From Fig.1:

b = person's distance from the lamp post

a = shadow length

Also, from the similarity of the triangles, we can write:

\frac{a + b}{20} = \frac{a}{7}

a = \frac{7}{13}b

Differentiating the above eqn w.r.t t:

\frac{da}{dt} = \frac{7}{13}.\frac{db}{dt}

Now, we know that:

Rate = \frac{db}{dt} = 5\ ft/s

Thus

\frac{da}{dt} = \frac{7}{13}.\times 5 = 2.692\ ft/s

5 0
3 years ago
consider a solid sphere and a solid disk wiht the same radius and the same mass. explain why the solid disk has a greater moment
andriy [413]

Answer:

Moment of inertia of the solid sphere:

I

s

=

2

5

M

R

2

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.

.

.

.

.

.

.

.

.

.

(

1

)

Is=25MR2...........(1)

Here, the mass of the sphere is

M

M

4 0
3 years ago
Hello please help i’ll give brainliest
Zarrin [17]

Answer:

C because it is in earth's mantle

3 0
3 years ago
A thin layer of oil of refractive index 1.22 is spread on the surface of water (n = 1.33), If the thickness of the oil is 275 nm
ANTONII [103]

Answer:

6.71×10⁻⁷ m

Explanation:

Using thin film constructive interference formula as:

<u>2×n×t = m×λ</u>

Where,

n is the refractive index of the refracted surface

t is the thickness of the surface

λ is the wavelength

If m =1

Then,

2×n×t = λ

Given that refractive index pf the oil is 1.22

Thickness of the oil = 275 nm

Also, 1 nm = 10⁻⁹ m

Thickness = 275×10⁻⁹ m

So,

Wavelength is :

<u>λ= 2×n×t = 2× 1.22 × 275×10⁻⁹ m = 6.71×10⁻⁷ m</u>

8 0
4 years ago
What is the centripetal acceleration of a small laboratory centrifuge in which the tip of the test tube is moving at 19.0 meters
Artist 52 [7]

Answer:3.61(10)^{3} \frac{m}{s^{2}}

The centripetal acceleration a_{c} of an object moving in a uniform circular path is given by the following equation:

a_{c}=\frac{V^{2}}{r}

Where:

V=19m/s is the velocity

r=10cm=0.1m is the radius of the circle

a_{c}=\frac{(19m/s)^{2}}{0.1m}

a_{c}=3610m/s^{2}=3.61(10)^{3}m/s^{2}

8 0
3 years ago
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