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Savatey [412]
3 years ago
13

A toddler drops his pacifier from the Royal Gorge Bridge in Colorado. If the pacifier took 7.71 s to fall how fast was it travel

ing when it hit the water?
Physics
1 answer:
Pavel [41]3 years ago
3 0

Answer:

75.56 m/s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 0 m/s

Time (t) = 7.71 s

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) =..?

Thus, we can obtain the velocity (v) with which the pacifier hit the water as follow:

v = u + gt

v = 0 + (9.8 × 7.71)

v = 0 + 75.558

v = 75.558 ≈ 75.56 m/s

Therefore, velocity with which the pacifier hit the water is 75.56 m/s

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A magnet attracts a piece of iron. The iron can then attract another piece of iron. On the basis of domain alignment, explain wh
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Answer :

The magnet produces a domain alignment that allows the iron become magnetic such that it is attracted to the original magnet.the magnetic iron can then create the same effect on another piece of iron.

Explanation:

the piece of magnet is brought closer to the iron where the domains of the iron align including the magnetic poles.

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4 years ago
What are antigens that would exsist on red blood cells of a person was type AB- blood
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AB blood would lead to an ABO incompatibility reaction. Hope it helped! (Asked a friend for you to know :D)
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3 years ago
Which type of rock can only form below earths surface?
Inessa [10]

Answer: Igneous it forms because of magma but magma is under the earths surface so its Igneous

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7 0
2 years ago
Read 2 more answers
34)You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then
velikii [3]

Answer:

F_H_n=230.04 N

The Required  horizontal force is 230.04N

Explanation:

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

F_H_n=F_f\\F_h=mg*u

where:

F_{Hn} is the required Force

u is the friction coefficient

m is the mass

g is gravitational acceleration=9.8m/s^2

200=mg*u                         Eq (1)

Now, weight increases by 42% and friction coefficient decreases by 19%

New weight=(1.42*m*g) and new friction coefficient=0.81u

F_H=(1.42m*g*.81u)          Eq (2)

Divide Eq(2) and Eq (1)

\frac{F_H_n}{200}=\frac{1.42m*g*0.81u}{m*g*u}\\F_H_n=1.42*0.81*200\\F_H_n=230.04 N

The Required  horizontal force is 230.04N

4 0
4 years ago
A sled with a mass of 25.0 kg rests on a horizontal sheet of essentially frictionless ice. It is attached by a 5.00-m rope to a
Tpy6a [65]

Answer:

The value is  F  =  34.3 \  N

Explanation:

From the question we are told that

    The mass of the shed is  m  =  25 \  kg

    The length of the rope is  l = 5.0 \  m

    The angular speed of the shed is w = 5 rev/minute = \frac{2* \pi * 5}{60 }= 0.524 \  rad/sec

Generally the force exerted on the shed is mathematically represented as

     F  =  m  *  w^2 *  l

=>  F  =  25  *  0.524^2 *  5

=>  F  =  34.3 \  N    

7 0
4 years ago
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