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Sauron [17]
3 years ago
13

Use Hooke's Law, which states that the distance a spring stretches (or compresses) from its natural, or equilibrium, length vari

es directly as the applied force on the spring. A force of 265 newtons stretches a spring 0.15 meter. (a) What force stretches the spring 0.4 meter
Physics
1 answer:
Phantasy [73]3 years ago
6 0

Answer:

706.68 N

Explanation:

By Hooke's law,

F = ke

k=\dfrac{F}{e}

Using the values in the question,

k=\dfrac{265\text{ N}}{0.15 \text{ m}}=1766.7\text{ N/m}

When e = 0.4 m,

F = 1766.7\text{ N/m}\times0.4\text{ m}=706.68\text{ N}

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hydroelectric energy is based on the blank energy of water being converted to blank energy as waterfalls and pushes against the
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3 years ago
Read 2 more answers
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.5 m>s. Two seconds later
ivanzaharov [21]

Answer:

4.28 s

Explanation:

after two seconds (2 s) His friends is

d = 3.5 m/s x 2 s = 7 meter ahead.

in this state, a bicylist start from initial velocity vo = 0 m/s and accelerat 2.4 m/s²

then, when bicylist reach his friend

t friend = t bicyclist = t

d bicylist = d friend + d

-------

d friend = 3.5 . t

d bicylist = vo . t + ½ a t²

d friend + d = vo . t + ½ a t²

3.5 t + 7 = 0 . t + ½ . 2.4 . t²

3.5 t + 7 = 1.2 t²

0 = 1.2 t² - 3.5 t - 7

t = -1.363 and t = 4.28

take the positive one

6 0
3 years ago
3. Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If
Daniel [21]

Answer:

Object 2 has the larger drag coefficient

Explanation:

The drag force, D, is given by the equation:

D = 0.5 c \rho A v^2

Object 1 has twice the diameter of object 2.

If d_2 = d

d_1 = 2d

Area of object 2, A_2 = \frac{\pi d^2 }{4}

Area of object 1:

A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2

Since all other parameters are still the same except the drag coefficient:

For object 1:

D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2

For object 2:

D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2

Since the drag force for the two objects are the same:

0.5 c_1 \rho (\pi d^2) v^2 = 0.5 c_2 \rho (\pi d^2/4) v^2\\4c_1 = c_2

Obviously from the equation above, c₂ is larger than c₁, this means that object 2 has the larger drag coefficient

8 0
3 years ago
In class we described the tidal forces that are responsible for raising and lowering thewater level near the shore of the ocean.
Sloan [31]

The newton's law of universal gravitation used to describe how a particle attracts every other particle in the universe.

The equation is given by,

F= G\frac{m_1 m_2}{r_2}

Where,

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m_1, m_2 = masses of the objects

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For our problem we have defined that,

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