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Ksivusya [100]
3 years ago
7

At which of the following points does a roller coaster have the most potential energy? As it is going down a hill. At the top of

the hill. As it is climbing a hill. At the bottom of the hill.
Physics
1 answer:
vova2212 [387]3 years ago
6 0

Answer:

At the top of the hill.

Explanation:

As the roller coaster goes up the hill, kinetic energy (K.E) decreases, gravitational potential energy (G.P.E) increases .

As it reach the top of the hill, K.E becomes zero and G.P.E reaches <em>m</em><em>a</em><em>x</em><em>i</em><em>m</em><em>u</em><em>m</em> .

As it goes down the hill, K.E starts to increase and G.P.E decrease .

At the bottom of the hill, K.E reaches <em>maximum</em> and G.P.E becomes zero .

(Correct me it I am wrong)

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The weight of air in a column 1-m2 in cross section that extends from sea level to the top of the atmosphere is?
Sauron [17]

From sea level to the top of the atmosphere, a column of air with a 1-m2 cross section weighs 101,000 N.

To find the answer we have to know about the pressure.

<h3>What is Pressure?</h3>
  • The pressure at a point on a surface is the thrust acting per unit area around that point.

                    Pressure, P=\frac{Thrust}{Area}=\frac{F}{A}

  • A particular mass of air is contained in a column that rises from the ocean to the top of the atmosphere. Then,

                   P=P_a=1.013*10^5 pascal

Pa is the atmospheric pressure at the sea level.

  • By combining both the equations, we get the weight of air,

                                       \frac{F}{A}=P_a

              F=W=P_a*A=1.013*10^5*1=101300N

Thus, we can conclude that, the weight of air in a column 1-m2 in cross section that extends from sea level to the top of the atmosphere is 101300N.

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3 0
1 year ago
Which of these is one of six most common elements found in the human body??
Lady bird [3.3K]

Answer: Oxygen

Explanation: Oxygen is the most common element in the human body. It makes up 65% of the body.

6 0
2 years ago
A billion years ago, Earth and its moon were just 200000 kilometers apart. Express this distance in meters.​
Luda [366]

Answer:

The  value is  x =  200000000 \  m

Explanation:

From the question we are told that

  The  distance between the earth and the moon is  d  =  200000 \  km

Generally ,

     1 \  km  =  1000 \  m

      200000 \  km \to x \  m

=>    x = \frac{200000 *  1000}{1 }

=>   x =  200000000 \  m

           

4 0
2 years ago
15 points! An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remai
alexandr1967 [171]

The alpha particle is emitted at 4235 m/s

Explanation:

We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:

p_i = p_f\\ Mu=m_1 v_1 + m_2 v_2 =  

where:  

M =222u is the mass of the original nucleus

v=420 m/s is the initial velocity of the nucleus

m_1 = 4 u is the mass of the alpha particle

v_1 is the final velocity of the alpha particle

m_2 = 222u-4u = 218 u is the mass of the daughter nucleus

v_2 = 350 m/s is the final velocity of the nucleus

Solving for v_1, we  find the final velocity of the alpha particle:

v_1 = \frac{Mu-m_2 v_2}{m_1}=\frac{(222)(420)-(218)(350)}{4}=4235 m/s

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4 0
3 years ago
A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while
dem82 [27]

12.8 rad

Explanation:

The angular displacement \theta through which the wheel turned can be determined from the equation below:

\omega^2 = \omega_0^2 + 2\alpha\theta (1)

where

\omega_0 = 0

\omega = 34.7\:\text{rad/s}

\alpha = 47.0\:\text{rad/s}^2

Using these values, we can solve for \theta from Eqn(1) as follows:

2\alpha\theta = \omega^2 - \omega_0^2

or

\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}

\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}

\:\:\:\:= 12.8\:\text{rad}

7 0
2 years ago
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