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3 years ago

How are sky maps made?

Physics

2 answers:

Andre45 [30]3 years ago

5 0

A star chart or star map is a map of night sky

r-ruslan [8.4K]3 years ago

4 0

A star chart or star map is a map of the night sky Astronomers divide these into grids to use them more easil They are used to identify and locate astronomical objects such as stars constellations and galaxie it is in my text book

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Police radar equipment is used to detect the speed of objects. In one trial, the radar equipment records a stationary tree as tr

Brut [27]

Answer:

reliability

accuracy

Explanation:

If a reading of a measurement is consistently the same then the measurement is reliable.

If a reading of measurement is close the actual value of the measurement then the reading is accurate.

Here, a stationary tree shows reading 6 mph once and 0 mph another instant. So, neither the reading of a measurement is consistent not the reading of measurement is close the actual value.

Hence, the radar has problems in its reliability and accuracy

8 0

3 years ago

Which of the following is an example of how humans can increase biodiversity?

Leto [7]

Answer:

The following is an example of how humans can increase biodiversity

Provide Wildlife Corridors and Connections Between Green Spaces

Use Organic Maintenance Methods and Cut Back On Lawns

Use a Native Plant Palette and Plant Appropriately

Utilize Existing Green Space Connections

Be Mindful of Non-Native Predators

7 0

2 years ago

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

Sergeeva-Olga [200]

Answer:

$8.37*10^5$ rpm

Explanation:

Given that rotational kinetic energy = $4.66*10^9J$

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = $\frac{1}{2} mr ^2$

Rotational K.E is illustrated as $(K.E)_{rt} = \frac{1}{2} I \omega^2$

$\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }$

$\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }$

$\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }$

$\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }$

$\omega = 87636.04$

$\omega = 8.76*10^4 rad/s$

Since 1 rpm = $\frac{2 \pi}{60} rad/s$

$\omega = 8.76*10^4(\frac{60}{2 \pi})$

$\omega = 836518.38$

$\omega = 8.37 *10^5 rpm$

3 0

2 years ago

A hole is punched at A in a plastic sheet by applying a 660-N force P to end D of lever CD, which is rigidly attached to the sol

olasank [31]

Answer:

hello your question lacks some data and required diagram

G = 77 GPa, т all = 80 MPa

answer : required diameter = 252.65 * 10-^3 m

Explanation:

Given data :

force ( P ) = 660 -N force

displacement = 15 mm

G = 77 GPa

т all = 80 MPa

i) Determine the required diameter of shaft BC

considering the vertical displacement ( looking at handle DC from free body diagram )

D' = 0.3 sin∅ , where D = 0.015

hence ∅ = 2.8659°

calculate the torque acting at angle ∅ of CD on the shaft BC

Torque = 660 * 0.3 cos∅

= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N

hello attached is the remaining part of the solution

4 0

2 years ago

The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a

Degger [83]

Answer:

147.456077993 Hz

Explanation:

$f_0$ = Frequency of the sonar = 22 kHz

$v_w$ = Velocity of the whale = 4.95 m/s

v = Speed of sound in water = 1482 m/s

The difference in frequency is given by

$\Delta f=f_0\times\dfrac{2v_{w}}{v-v_w}\\\Rightarrow \Delta f=22000\times\dfrac{2\times 4.95}{1482-4.95}\\\Rightarrow \Delta f=147.456077993\ Hz$

The difference in frequency is 147.456077993 Hz

6 0

2 years ago