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4 years ago

Please select the word from the list that best fits the definition. The movement of a magnetic pole away from the actual pole

Physics

1 answer:

elena55 [62]4 years ago

7 0

Answer:

Magnetosphere

Explanation:

It's the magnetosphere simply because a magnetic declination is not on abject, it's something that happens.

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If the Moon rises at a given location at 6:00 p.m. today, about what time will it rise tomorrow night?

Lena [83]

Answer:

6:50 PM

Explanation:

The Moon continually moves on ahead in its orbit while the Earth rotates. So 24 hours after, the Earth has rotated back around to the same place it was the night before, while the Moon has gone on ahead. assume of it like the second hand on an analog watch, it’s going around the face of the clock much faster than the minute hand, but each time the second hand goes around, the minute hand has moved, and so it takes an extra second to line back up with the minute hand. Because the Moon has moved 13 degrees or so since its last moon rise, it’s going to take another hour or so for the Earth to catch back up to the Moon’s new location, delaying the Moon's rising above your horizons by ~50 minutes each day.

3 0

3 years ago

Sheila lives in the seashore community of Beachview. At 6:00 a.m. one morning, she noticed that the local ocean water was at hig

Serhud [2]

The tide was high because of the sun just now coming up at noon it when the sun begins to settle causing the ocean to form at low tide .

7 0

3 years ago

Read 2 more answers

A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately

Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car $V_{c}$ = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

$V_{f}$ = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

$V_{f}$ = $V_{i}$ + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁ = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁ will be

$d_{c}$ = $V_{c}$ × t₁

we substitute

$d_{c}$ = 22.352 × 7

$d_{c}$ = 156.46 m

now distance between polive car and speeding car

Δd = $d_{c}$ - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( $V_{f}$ - $V_{c}$ )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = $V_{f}$ × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b) how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0

3 years ago

One end of a 7-cm-long spring is attached to the ceiling. When a 5.4 kg mass is hung from the other end, the spring is stretched

mash [69]

Answer:

2.63 cm

Explanation:

Hooke's law gives that the force F is equal to cy where c is spring constant and x is extension

Making c the subject of the formula then

$c=\frac {F}{y}$

Since F is gm but taking the given mass to be F

$c=\frac {5.4 kg}{4.3 cm}=1.2558139534883720930232558139534883720930$

By substitution now considering F to be 3.3 kg

$y=\frac {3.3 kg}{1.2558139534883720930232558139534883720930}=2.6277777777777 cm\approx 2.63 cm$

8 0

4 years ago

A charge of 2.00 μC flows onto the plates of a capacitor when it is connected to a 12.0-V potential source. What is the minimum

uysha [10]

To develop this problem we will apply the concept of energy conservation. For which the work carried out must be equivalent to the potential energy stored on the capacitor. We will start by finding the capacitance to later be able to calculate the energy and therefore the work in the capacitor

$C = \frac{Q}{V}$