# Which one of these are true about scale models? a. A map may have a scale model with the proportion of one centimeter to one kilometer. b. A scale model is always smaller than the object it represents, c. A model may be built to different scales.

Option A is true

Explanation:

For option A, it's true because a map that has a scale model with the proportion of one centimeter to one kilometer is known as verbal scale which is a type of scale.

For option B, it's not true because though a scale model is most times always smaller than the object it represents, there are sometimes when the scale model is an enlarged view/representation of a small object.

For option C, it's true because there is no one scale that is confined to just a model. A model can use different scales to depict an object.

## Related Questions

What displacement do I have if I travel at 10 m/s E for 10 s? A. 1 m E B. 1 m C. 100 m D. 100 m E Scalar quantities include what 2 things? A. Number and direction B. Numbers and units C. Units and directions D. Size and direction What measures distance in a car? A. Odometer B. Pressure gauge C. Speedometer D. Steering wheel What displacement do I have if I travel 10 m E, then 6 m W, then 12 m E? A. 28 m E B. 16 m E C. 16 m D. 28 m

The displacement is 100 m to the east.

### Explanation:

The displacement can be calculated using the formula:

Displacement = Velocity × Time

In this case, the velocity is 10 m/s to the east and the time is 10 seconds.

So, Displacement = 10 m/s × 10 s = 100 m to the east.

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A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and the density of helium is 0.179 kg/m3. (a) Draw a force diagram for the balloon. (Submit a file with a maximum size of 1 MB.) (b) Calculate the buoyant force acting on the balloon. (Give your answer to at least three significant figures.) 4159 N (c) Find the net force on the balloon. 1524 N Determine whether the balloon will rise or fall after it is released. The balloon will (d) What maximum additional mass can the balloon support in equilibrium? 155 kg (e) What happens to the balloon if the mass of the load is less than the value calculated in part (d)? The balloon and its load will remain stationary. The balloon and its load will accelerate downward. The balloon and its load will accelerate upward. (f) What limits the height to which the balloon can rise?

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

where is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

where is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

is the air density

is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

(c) 1524 N

The mass of the helium gas inside the balloon is

where is the helium density; so we the total mass of the balloon+helium gas inside is

So now we can find the weight of the balloon:

And so, the net force on the balloon is

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

where m' is the additional mass. Re-arranging the equation for m', we find

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

The physics involved in the functioning of helium balloons is based on buoyancy and Archimedes' Principle. The forces at play include the force due to gravity, the buoyant force and the net force, which determines the motion of the balloon. The balloon's height limit is determined by the decrease in air density with altitude.

### Explanation:

The several parts of this question are related to the principles of buoyancy and Archimedes' Principle. First, regarding the force diagram for the balloon (part a), it would show two primary forces. The force due to gravity (Fg) acting downwards and the buoyant force (Fb) acting upwards, which is a result of the displacement of air by the balloon. The net force mentioned in part (c) is calculated as the difference between these two forces.

Calculating the buoyant force (part b) involves multiplying the volume of the balloon by the density of the air and the acceleration due to gravity (Fb = V * ρ_air * g). For the net force on the balloon (part c), this is calculated by subtracting the weight of the balloon from the buoyant force (F_net = Fb - Fg). If the net force is positive, the balloon will rise, if it's negative, the balloon will fall, and if it is zero, the balloon will remain stationary.

The maximum additional mass the balloon can support in equilibrium (part d) is calculated using the net force divided by gravity. If the mass of the load is less than this value (part e), the balloon and its load will accelerate upward.

Lastly, the limit to the height to which the balloon can rise (part f) is determined by the decreasing density of the air as the balloon ascends. The buoyant force reduces as the balloon rises because the air density is lower at higher altitudes.

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A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound produced by the pipe, in SI units, is closest to:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

hence, the frequency of the third overtone is 102 Hz

A 3.00N rock is thrown vertically into the air from the ground. At h=15.0m, v=25m/s upward. Use the work-energy theorem to find the initial speed of the rock.a. 3m/s
b. 30.3 m/s
c. None of the above

so initial speed of the rock is 30.32 m/s

correct answer is b. 30.3 m/s

Explanation:

given data

h = 15.0m

v = 25m/s

weight of the rock m = 3.00N

solution

we use here work-energy theorem that is express as here

work = change in the kinetic energy    ..............................1

so it can be written as

work = force × distance     ...................2

and

KE is express as

K.E = 0.5 × m × v²

and it can be written as

F × d = 0.5 × m × (vf)² - (vi)²      ......................3

here

m is mass and vi and vf is initial and final velocity

F = mg = m  (-9.8)  , d = 15 m and v{f} = 25 m/s

so put value in equation 3 we get

m  (-9.8) × 15 = 0.5 × m × (25)² - (vi)²

solve it we get

(vi)² =  919

vi = 30.32 m/s

so initial speed of the rock is 30.32 m/s

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine provides anet thrust of 0.795 N perpendicular tothe tethering wire.(a) Find the torque the net thrust producesabout the center of the circle.
N·m

(b) Find the angular acceleration of the airplane when it is inlevel flight.

(c) Find the linear acceleration of the airplane tangent to itsflight path.
m/s2

(a) 24.6 Nm

The torque produced by the net thrust about the center of the circle is given by:

where

F is the magnitude of the thrust

r is the radius of the wire

Here we have

F = 0.795 N

r = 30.9 m

Therefore, the torque produced is

(b)

The equivalent of Newton's second law for a rotational motion is

where

is the torque

I is the moment of inertia

is the angular acceleration

If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is

And so we can solve the previous equation to find the angular acceleration:

(c)

The linear acceleration (tangential acceleration) in a rotational motion is given by

where in this problem we have

is the angular acceleration

r = 30.9 m is the radius

Substituting the values, we find

WILL MARK BRAINLIEST PLS HELPPP -- Which of Newton’s Laws explains why the satellite would collide with the moon if gravity is “turned off?”picture attached