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vazorg [7]
2 years ago
11

Nora tied a string around a tennis ball, and then she swung it in a circle in front of her to demonstrate a planet orbiting the

Sun. She explained that her hand represented the Sun, the ball represented a planet, and the string kept the ball from leaving the "orbit" around her hand. Why doesn't the string work in the same way as gravity? A. The string holding the ball in orbit is a contact force; gravity is a noncontact force. B. The string only pulls one way; gravity pulls both ways. C. The string keeps the distance between the bodies the same; gravity makes the distance vary. D. There is no difference between the two systems.
Physics
2 answers:
777dan777 [17]2 years ago
8 0

<em>Choice-A </em>is a true statement. The string holding the ball in orbit is a contact force, whereas gravity is a non-contact force.

Choice-C is half-true.  The string keeps the distance between the bodies constant, but gravity doesn't "MAKE" the distance vary.

shutvik [7]2 years ago
5 0

Answer:The string holding the ball in orbit is a contact force; gravity is a noncontact force.

Explanation:

You might be interested in
Function of a simple pendulum​
Misha Larkins [42]

Answer:

A pendulum is a mechanical machine that creates a repeating, oscillating motion. A pendulum of fixed length and mass (neglecting loss mechanisms like friction and assuming only small angles of oscillation) has a single, constant frequency. This can be useful for a great many things.

From a historical point of view, pendulums became important for time measurement. Simply counting the oscillations of the pendulum, or attaching the pendulum to a clockwork can help you track time. Making the pendulum in such a way that it holds its shape and dimensions (in changing temperature etc.) and using mechanisms that counteract damping due to friction led to the creation of some of the first very accurate all-weather clocks.

Pendulums were/are also important for musicians, where mechanical metronomes are used to provide a notion of rhythm by clicking at a set frequency.

The Foucault pendulum demonstrated that the Earth is, indeed, spinning around its axis. It is a pendulum that is free to swing in any planar angle. The initial swing impacts an angular momentum in a given angle to the pendulum. Due to the conservation of angular momentum, even though the Earth is spinning underneath the pendulum during the day-night cycle, the pendulum will keep its original plane of oscillation. For us, observers on Earth, it will appear that the plane of oscillation of the pendulum slowly revolves during the day.

Apart from that, in physics a pendulum is one of the most, if not the most important physical system. The reason is this - a mathematical pendulum, when swung under small angles, can be reasonably well approximated by a harmonic oscillator. A harmonic oscillator is a physical system with a returning force present that scales linearly with the displacement. Or, in other words, it is a physical system that exhibits a parabolic potential energy.

A physical system will always try to minimize its potential energy (you can accept this as a definition, or think about it and arrive at the same conclusion). So, in the low-energy world around us, nearly everything is very close to the local minimum of the potential energy. Given any shape of the potential energy ‘landscape’, close to the minima we can use Taylor expansion to approximate the real potential energy by a sum of polynomial functions or powers of the displacement. The 0th power of anything is a constant and due to the free choice of zero point energy it doesn’t affect the physical evolution of the system. The 1st power term is, near the minimum, zero from definition. Imagine a marble in a bowl. It doesn’t matter if the bowl is on the ground or on the table, or even on top of a building (0th term of the Taylor expansion is irrelevant). The 1st order term corresponds to a slanted plane. The bottom of the bowl is symmetric, though. If you could find a slanted plane at the bottom of the bowl that would approximate the shape of the bowl well, then simply moving in the direction of the slanted plane down would lead you even deeper, which would mean that the true bottom of the bowl is in that direction, which is a contradiction since we started at the bottom of the bowl already. In other words, in the vicinity of the minimum we can set the linear, 1st order term to be equal to zero. The next term in the expansion is the 2nd order or harmonic term, a quadratic polynomial. This is the harmonic potential. Every higher term will be smaller than this quadratic term, since we are very close to the minimum and thus the displacement is a small number and taking increasingly higher powers of a small number leads to an even smaller number.

This means that most of the physical phenomena around us can be, reasonable well, described by using the same approach as is needed to describe a pendulum! And if this is not enough, we simply need to look at the next term in the expansion of the potential of a pendulum and use that! That’s why each and every physics students solves dozens of variations of pendulums, oscillators, oscillating circuits, vibrating strings, quantum harmonic oscillators, etc.; and why most of undergraduate physics revolves in one way or another around pendulums.

Explanation:

7 0
2 years ago
An object is placed at O ona number line. It moves 3 units to the right, then 4 units to the left, and then 6 units to
kvv77 [185]

Answer:

You have a displacement of 5 units to the right.

Explanation:

First you go three to the right which lands on the 3 mark. Then you move it 4 to the left which substracts 4, landing the object at -1. Finally you move 6 to the right, and you finish at marker 5. Since displacement is not total distance but just final distance from the start point directly to end point, it is only a displacement of 5.

6 0
2 years ago
Read 2 more answers
On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m
Nastasia [14]

Answer:

F = 1.489*10^{-7}  N

Explanation: Weight of space probes on earth is given by:W= m*g

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 \frac{m}{s^{2} })

Therefore,

m_{1} = \frac{14500}{9.81}

m_{1} = 1478.08  kg

Similarly,

m_{2} = \frac{4800}{9.81}

m_{2} = 489.29  kg

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

F =  \frac{Gm_{1} m_{2}}{R^{2} }

G= gravitational constant (6.67 * 10^{-11} m^{3} kg^{-1} s^{-2})

m_{1} , m_{2}= masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

F = 1.489*10^{-7}  N

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.

7 0
3 years ago
(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i
murzikaleks [220]

Answer:

a) t = -\frac{ln(2)}{k}

b) See the proof below

A(t) = A_o 2^{-\frac{t}{T}}

c) t = 3T \frac{ln(2)}{ln(2)}= 3T

Explanation:

Part a

For this case we have the following differential equation:

\frac{dA}{dt}= kA

With the initial condition A(0) = A_o

We can rewrite the differential equation like this:

\frac{dA}{A} =k dt

And if we integrate both sides we got:

ln |A|= kt + c_1

Where c_1 is a constant. If we apply exponential for both sides we got:

A = e^{kt} e^c = C e^{kt}

Using the initial condition A(0) = A_o we got:

A_o = C

So then our solution for the differential equation is given by:

A(t) = A_o e^{kt}

For the half life we know that we need to find the value of t for where we have A(t) = \frac{1}{2} A_o if we use this condition we have:

\frac{1}{2} A_o = A_o e^{kt}

\frac{1}{2} = e^{kt}

Applying natural log we have this:

ln (\frac{1}{2}) = kt

And then the value of t would be:

t = \frac{ln (1/2)}{k}

And using the fact that ln(1/2) = -ln(2) we have this:

t = -\frac{ln(2)}{k}

Part b

For this case we need to show that the solution on part a can be written as:

A(t) = A_o 2^{-t/T}

For this case we have the following model:

A(t) = A_o e^{kt}

If we replace the value of k obtained from part a we got:

k = -\frac{ln(2)}{T}

A(t) = A_o e^{-\frac{ln(2)}{T} t}

And we can rewrite this expression like this:

A(t) = A_o e^{ln(2) (-\frac{t}{T})}

And we can cancel the exponential with the natural log and we have this:

A(t) = A_o 2^{-\frac{t}{T}}

Part c

For this case we want to find the value of t when we have remaining \frac{A_o}{8}

So we can use the following equation:

\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}

Simplifying we got:

\frac{1}{8} = 2^{-\frac{t}{T}}

We can apply natural log on both sides and we got:

ln(\frac{1}{8}) = -\frac{t}{T} ln(2)

And if we solve for t we got:

t = T \frac{ln(8)}{ln(2)}

We can rewrite this expression like this:

t = T \frac{ln(2^3)}{ln(2)}

Using properties of natural logs we got:

t = 3T \frac{ln(2)}{ln(2)}= 3T

8 0
3 years ago
Bob is pushing a box across the floor at a constant speed of 1.7 m/s, applying a horizontal force whose magnitude is 70 N. Alice
OlgaM077 [116]

Answer:

a) 70 N, b) b. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.

Explanation:

a) A constant speed means that magnitude of friction force is equal to the magnitude of the external force. The friction force is directly proportional to the normal force, which is equal to the weight of the box. Therefore, the magnitude of the force is 70 N.

b) Alice used initially a greater force to accelerate the box up to needed speed and later reduced the external force to keep speed constant. The right choice is option b.

3 0
2 years ago
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