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Tems11 [23]
2 years ago
13

What is the golden rule of lightning safety? (A.Don't stand under trees). (B.Avoid using electronics,)(C.Stay avvay from open sp

aces.) (D. Head for cover)​
Physics
1 answer:
IrinaK [193]2 years ago
6 0
The answer is d (head for cover)
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What is the anomalous expansivity of water
lora16 [44]

Answer:

The anomalous expansion of water is an abnormal property of water whereby it expands instead of contracting when the temperature goes from 4o C to 0o C, and it becomes less dense. The density is maximum at 4 degree centigrade and decreases below that temperature as shown in graph.

Explanation:do you want me to explain it more??

6 0
2 years ago
A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. its speed is 40 m/s when i
Alenkinab [10]

Given that,

Initial velocity , Vi = 0

Final velocity , Vf = 40 m/s

Acceleration due to gravity , a = 9.81 m/s²

Distance can be calculated as,

2as = Vf² - Vi²

2 * 9.81 *s = 40² - 0²

s = 81.55 m

For half height, that is, s = 40.77m

Vf= ??

2as = Vf² - Vi²

2 * 9.81 * 40.77 = Vf² - 0²

Vf² = 800

Vf = 28.28 m/s

Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.  

5 0
3 years ago
Which statement accurately describes a sample of water during parts a and c of the heating curve
vivado [14]

Answer:

A and C is about 12 cm away from each other.

Explanation:

5 0
3 years ago
Read 2 more answers
The strength of an electromagnet can be altered by
denpristay [2]
The strength of an electromagnet can be altered by increasing the number of coils around the core. The more times the coil is wrapped, the stronger the electromagnet is.

Your answer is: B) Increasing the number of coils around the core 

Have an amazing day and stay hopeful!

3 0
3 years ago
Read 2 more answers
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
3 years ago
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