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madreJ [45]
3 years ago
8

Any help please, physics is not my strong suit?

Physics
2 answers:
frez [133]3 years ago
5 0

Answer:

2000mg = 2g

5L = 5000mL

16cm = 160mm

Mashcka [7]3 years ago
5 0

1) kg

2) m

3) g

4) mL

5) mm

6) L

7) km

8) cm

9) mg

10) 2g

11) 104000m

12) 4.8m

13) 5600g

14) 8cm

15) 5000mL

16) 0.198kg

17) 0.075L

18) 0.15m

19) 560cm

20) 16mm

21) 25km

22) 6500mg

23) 66mm

24) 0.12g

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radar wave is transmitted and later reflected off an aircraft and recieved 1.4×10^3 sec after being sent out. how far is the air
lutik1710 [3]

I think there's a typo because the answer I'm getting is very large.

This is what I'm getting

--------------------------------------

c = speed of light

c = 3.0 x 10^8 m/sec approximately

This is roughly 300 million meters per second

The time it takes the signal to reach the aircraft and come back is 1.4 x 10^3 seconds. Half of this time period is going one direction (say from the radar station to the aircraft), so (1.4 x 10^3)/2 = 7.0 x 10^2 seconds is spent going in this one direction.

distance = rate*time

d = r*t

d = (3.0 x 10^8) * (7.0 x 10^2)

d = (3.0*7.0) x (10^8*10^2)

d = 21.0 x 10^(8+2)

d = 21.0 x 10^10

d = (2.1 x 10^1) * 10^10

d = 2.1 x (10^1*10^10)

d = 2.1 x 10^11 meters

d = 210,000,000,000 meters (this is 210 billion meters; equivalent to roughly 130,487,950 miles)

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3 years ago
What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k =
Doss [256]
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the car starts from rest at s=0 and increases its speed at at=4m/s^2. Determine the time when the magnitude of acceleration beco
dem82 [27]

Answer:

<em>Time =  5 seconds</em>

<em>Distance = 50 meters</em>

Explanation:

<u>Constantly Accelerated Motion</u>

When the velocity of a moving object changes at a constant rate, called acceleration, the velocity changes in same amounts in the same times. The question has a mistake when asking when the acceleration is 20 m/s. If the acceleration is constant, the only variable that can change to that value is the velocity. The equation to calculate the speed is

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\displaystyle s=v_o.t+\frac{gt^2}{2}

Given the object starts from rest, vo=0 and vf=20 m/s at a=4\ m/s^2. We compute t

\displaystyle t=\frac{v_f-v_o}{a}=\frac{20-0}{4}

\boxed{t=5\ sec}

Now we compute s

\displaystyle s=0+\frac{4\times 5^2}{2}

\boxed{s=50\ m}

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