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3 years ago

Going around a ferris wheel without stopping is an example of _____ circular motion(apex)

Physics

2 answers:

Natasha_Volkova [10]3 years ago

5 0

Explanation:

Going around a Ferris wheel without stopping is an example of <u>uniform</u> circular motion. In a circular motion, the objects moves with constant speed but its velocity is continuously changing because of changing direction.

The force acting when the object when it is moving in a circular motion is called centripetal force. The direction of both force and acceleration is inwards.

Hence, "going around a Ferris wheel without stopping" is an example of uniform circular motion.

victus00 [196]3 years ago

3 0

going around a ferris wheel without stopping is an example of uniform circular motion (apex).

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A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f

MaRussiya [10]

To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

The total mass of the cars would be,

$m_T = 25(7.7*10^4)Kg$

$m_T = 1.925*10^6Kg$

Therefore the friction force at 29Km / h would be,

$f=250v$

$f= 250*29Km/h$

$f = 250*29*(\frac{1000m}{1km})(\frac{1h}{3600s})$

$f = 2013.889N$

In this way the tension exerts between first car and locomotive is,

$T=m_Ta+f$

$T=(1.925*10^6)(0.2)+2013.889$

$T= 3.8701*10^5N$

Therefore the tension in the coupling between the car and the locomotive is $3.87*10^5N$

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2 years ago

You use 35 J of energy to move an object 5 m. What is the weight of the object

Vlad [161]

Explanation:

35 ÷ 5 = 7kg.

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7 0

2 years ago

What pattern is seen on the moon from earth?

igomit [66]

If you mean pattern designs it’s rocky bumpy and you can see the circle patterns

If you mean phases the here’s a picture

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2 years ago

a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if

klio [65]

Answer:

Velocity of the two balls after collision: $60\; \rm m \cdot s^{-1}$ .

$100\; \rm J$ of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

Mass of the first ball: $100\; \rm g = 0.1\; \rm kg$ .
Mass of the second ball: $400\; \rm g = 0.4 \; \rm kg$ .

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass $m$ and velocity $v$ is: $p = m \cdot v$ .

Momentum of the two balls before collision:

First ball: $p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}$ .
Second ball: $p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}$ .
Sum: $10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1}$ given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be $30\; \rm kg \cdot m \cdot s^{-1}$ . The mass of the two balls, combined, is $0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg$ . Let the velocity of the two balls after the collision $v\; \rm m \cdot s^{-1}$ . (There's only one velocity because the collision had sticked the two balls to each other.)

Momentum after the collision from $p = m \cdot v$ : $(0.5\, v)\; \rm kg \cdot m \cdot s^{-1$ .
Momentum after the collision from the conservation of momentum: $30\; \rm kg \cdot m \cdot s^{-1}$ .

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about $v$ :

$0.5\, v = 30$ .

$v = 60$ .

In other words, the velocity of the two balls right after the collision should be $60\; \rm m \cdot s^{-1}$ .

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass $m$ and velocity $v$ is $\displaystyle \frac{1}{2}\, m \cdot v^{2}$ .

Kinetic energy before the collision:

First ball: $\displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J$ .
Second ball: $\displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J$ .
Sum: $500\; \rm J + 500\; \rm J = 1000\; \rm J$ .

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

Mass of the two balls, combined: $0.5\; \rm kg$ .
Velocity of the two balls right after the collision: $60\; \rm m\cdot s^{-1}$ .

Sum of the kinetic energies of the two balls right after the collision:

$\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J$ .

Therefore, $1000\; \rm J - 900\; \rm J = 100\; \rm J$ of kinetic energy would be lost during this collision.

7 0

3 years ago

Two charged spheres are 20cm apart and exert an attractive force of 6x10^-9 N on each other. What will be the force of attractio

Travka [436]

The force of attraction between 2 charged spheres can be explained by Coulomb's law,
It states the force of attraction is directly proportional to the magnitudes of the charges and inversely proportional to the square of the distance between the charges.

/ $F = \frac{k q_{1} q_{2} }{ r^{2} }$
where F - force of attraction/repulsion
q₁ and q₂ - charges of the 2 spheres
k - Coulomb's law constant
r - distance between the spheres

In the question given, the charges of the spheres remain constant in both instances, only distance changes. Therefore (kq₁q₂) = c which is a constant
then F = c / r²
first instance
6 x 10⁻⁹ N = c/ (20 cm)² ---1)
F = c/(10 cm)² --- 2)
2) / 1)
$\frac{F}{6* 10^{-9} } = \frac{400}{100}$
F = 6 x 10⁻⁹ x 4
F = 2.4 x 10⁻⁸ N

8 0

3 years ago