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3 years ago

After you enlarge a map, which one of the following scale remains correct?

Physics

1 answer:

9966 [12]3 years ago

6 0

Answer:

None

Explanation:

An scale is the factor by which actual features on ground are enlarged or reduced for representing on a plane. There are different kinds of scales:

Verbal scale use of words to represent scale information on the map. The distance or linear units are used for depicting this scale on the map. For example: 1 inch = 1 Kilo meter.
Fractional scale uses the numbers or values for showing the scale instead of words. As the name says, it is represented using a fraction or ratio. Example: 1: 10,000 or 1/10,000
In large scale more details are shown in a map, however, less area coverage will be shown in a single map as the scale is large and more details are given. Example: 1:500
Small scale is exactly opposite to the large scale, less details are shown as magnification is not enough, however a large amount of area can be shown in a single map. Example: 1:25,000
A graphic scale is a bar that has been calibrated to show map distances. On maps that have been reduced or enlarged the original ratio and written scales are incorrect, since the relationship between map distance and real world distance has been altered, graphic scale is enlarged or reduced to the same extent as the map, this makes it the right option.

I hope you find this information useful and interesting! Good luck!

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Calculate the phase angle (in radians) for a circuit with a maximum voltage of 12 V and w-50 Hz. The voltage source is connected

Vinvika [58]

Answer:

The phase angle is 0.0180 rad.

(c) is correct option.

Explanation:

Given that,

Voltage = 12 V

Angular velocity = 50 Hz

Capacitance $C= 20\times10^{-2}\ F$

Inductance $L=20\times10^{-3}\ H$

Resistance $R= 50\ Omega$

We need to calculate the impedance

Using formula of impedance

$z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}$

$z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}$

$z=50.00$

We need to calculate the phase angle

Using formula of phase angle

$\theta=\cos^{-1}(\dfrac{R}{z})$

$\theta=\cos^{-1}(\dfrac{50}{50.00})$

$\theta=0.0180\ rad$

Hence, The phase angle is 0.0180 rad.

3 0

3 years ago

The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various

Schach [20]

Answer:

$q = -461532.5 \ C$

Explanation:

From the question we are told that

The electric filed is $E = 102 \ N/C$

Generally according to Gauss law

=> $E A = \frac{q}{\epsilon_o }$

Given that the electric field is pointing downward , the equation become

$- E A = \frac{q}{\epsilon_o }$

Here $q$ is the excess charge on the surface of the earth

$A$ is the surface area of the of the earth which is mathematically represented as

$A = 4\pi r^2$

Where r is the radius of the earth which has a value $r = 6.3781*10^6 m$

substituting values

$A = 4 * 3.142 * (6.3781*10^6 \ m)^2$

$A =5.1128 *10^{14} \ m^2$

$q = -E * A * \epsilon _o$

Here $\epsilon_o$ s the permitivity of free space with value

$\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}$

$q = -461532.5 \ C$

6 0

3 years ago

A car veers off course and runs straight into a brick wall. This is an example

ValentinkaMS [17]

Short time large force

4 0

3 years ago

A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa

Lesechka [4]

At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
$U_i= \frac{1}{2} ka^2$
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
$K_i =0$
And the total energy of the system is
$E_i = U_i+K= \frac{1}{2}ka^2$

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
$U_f = 0$
while the mass is moving at speed v, and therefore the kinetic energy is
$K_f = \frac{1}{2} mv^2$
And the total energy is
$E_f = U_f + K_f = \frac{1}{2} mv^2$

For the law of conservation of energy, the total energy must be conserved, therefore $E_i = E_f$ . So we can write
$\frac{1}{2} ka^2 = \frac{1}{2}mv^2$
that we can solve to find an expression for v:
$v= \sqrt{ \frac{ka^2}{m} }$

6 0

3 years ago

A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The

USPshnik [31]

Answer:

a) see attached, a = g sin θ

c) v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

Wₓ = m a

W sin θ = m a

a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

θ' = 9/2 θ

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

Em₀ = mg h = mg L (1-cos tea)

Lowest point

Emf = K = ½ m v²

Em₀ = Emf

g L (1-cos θ) = v² / 2

v = √(2gL (1-cos θ))

4 0

3 years ago