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LiRa [457]
3 years ago
9

When you power an electromagnet with 3 batteries, what variable decreases as a result of removing one of those batteries.

Physics
1 answer:
Snezhnost [94]3 years ago
3 0

Answer:

The strength of the magnetic field decreases as a result of removing one of those batteries.

Hope this helps!

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A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met
faltersainse [42]

Answer:

83%

Explanation:

On the surface, the weight is:

W = GMm / R²

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.

In orbit, the weight is:

w = GMm / (R+h)²

where h is the height of the shuttle above the surface of the Earth.

The ratio is:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

The shuttle in orbit retains 83% of its weight on Earth.

4 0
4 years ago
Find the speed vfinal of the joined cars after the collision. mastering physics
Tanya [424]
<span>Px = 0 Py = 2mV second, Px = mVcosφ Py = –mVsinφ add the components Rx = mVcosφ Ry = 2mV – mVsinφ Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²) and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²) simplifying Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²) Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²) Vf = (V/3)âš((cosφ)² + (2 – sinφ)²) Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ)) Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ)) using the identity sin²(Ď)+cos²(Ď) = 1 Vf = (V/3)âš1 + 4 – 2sinφ) Vf = (V/3)âš(5 – 2sinφ)</span>
6 0
4 years ago
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σal
galben [10]
Hello

1) First of all, since we know the radius of the wire (r=1.2~mm=0.0012~m), we can calculate its cross-sectional area
A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2

2)  Then, we can calculate the current density J inside the wire. Since we know the current, I=3~A, and the area calculated at the previous step, we have
J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity \sigma=3.6\cdot10^7~ \frac{A}{Vm} of the aluminium, the electric field is given by
E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m

4 0
3 years ago
A flexible balloon contains 0.400 molmol of an unknown polyatomic gas. Initially the balloon containing the gas has a volume of
Nastasia [14]

Answer:

Mass released = 8.6 g

Given data:

Initial number of moles nitrogen= 0.950 mol

Initial volume = 25.5 L

Final mass of nitrogen released  = ?

Final volume = 17.3 L

Formula:

V₁/n₁  = V₂/n₂

25.5 L / 0.950 mol = 17.3 L/n₂

n₂ =  17.3 L× 0.950 mol/25.5 L

n₂ = 16.435 L.mol /25.5 L

n₂ = 0.644 mol

Initial mass of nitrogen:

Mass = number of moles × molar mass

Mass = 0.950 mol × 28 g/mol

Mass = 26.6 g

Final mass of nitrogen:

Mass = number of moles × molar mass

Mass = 0.644 mol × 28 g/mol

Mass = 18.0 g

Mass released = initial mass - final mass

Mass released = 26.6 g - 18.0 g

Mass released = 8.6 g

Read more on Brainly.com - brainly.com/question/15623698#readmore

7 0
3 years ago
"The greater the height of an object, the *BLANK* its gravitational potential energy."
-Dominant- [34]

Answer:

the greater the height of an object the *greater* its gravitational potential energy

greater

3 0
2 years ago
Read 2 more answers
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