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2 years ago

CAN YOU EXPLAIN?

Physics

1 answer:

Anni [7]2 years ago

4 0

<h2>Answer:</h2><h2><u>this</u><u> this is an image of this is helpful to you please thank you</u></h2>

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What is population education?<br>

Leviafan [203]

Answer:

Is all about people – how many of us there are, how we shape the world around us, and how we interact with each other.

Explanation:

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3 years ago

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COULD SOMEONE PLS DO THESE FOR ME <br> I’LL LOVE U FOREVER AND EVER

sladkih [1.3K]

I can help you do them for yourself :)

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3 years ago

The acceleration due to gravity, g , is constant at sea level on the Earth's surface. However, the acceleration decreases as an

blsea [12.9K]

Answer:

g = g₀ [1- 2 h / Re + 3 (h / Re)²]

Explanation:

The law of universal gravitation is

F = G m Me / Re²

Where g is the universal gravitational constant, m and Me are the mass of the body and the Earth, respectively and R is the distance between them

F = G Me /Re² m

We call gravity acceleration a

g₀ = G Me / Re².

When the body is at a height h above the surface the distance is

R = Re + h

Therefore the attractive force is

F = G Me m / (Re + h)²

Let's take Re's common factor

F = G Me / Re² m / (1+ h / Re)²

As Re has a value of 6.37 10⁶ m and the height of the body in general is less than 10⁴ m, the h / Re term is very small, so we can perform a series expansion

(1+ h / Re)⁻² = 1 -2 h / Re + 6/2 (h / Re) 2 + ...

Let's replace

F = G Me /Re² m [1- 2 h / Re + 3 (h / Re)²]

F = g₀ m [1- 2 h / Re + 3 (h / Re)²]

If we call the force of attraction at height

m g =g₀ m [1- 2 h / Re + 3 (h / Re)²]

g = g₀ [1- 2 h / Re + 3 (h / Re)²]

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3 years ago

Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s

Ne4ueva [31]

The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
$F=qvB \sin \theta$
where $\theta$ is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is $q=2.7 \mu C=2.7 \cdot 10^{-6}C$ . In this case, v and B are perpendicular, so $\theta=90^{\circ}$ , therefore we have:
$B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T$

2) In this second case, the angle between v and B is $\theta=55^{\circ}$ . The charge is now $q=42.0 \mu C=42.0 \cdot 10^{-6}C$ , and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
$F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N$

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2 years ago

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What unit is used for speed?<br> 1. m/s<br> 2. S<br> 3. m/s2<br> 4. m

bixtya [17]

Answer:

The unit of speed is m/s.

7 0

2 years ago