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Molodets [167]
3 years ago
10

What is isostasy dependent on a balance between

Physics
2 answers:
Scorpion4ik [409]3 years ago
7 0

Answer: Option (A) and (D)

Explanation: Isostacy is the gravitational state of equilibrium in which a block sinks down due to the addition of materials and also the same block rises up if materials are eroded.

          In terms of our earth, the gravitational force is responsible for the sinking or down-pushing of the continents, and the bouyancy forces in the mantle is responsible for the uplifting of the continents.

Salsk061 [2.6K]3 years ago
5 0
I think its D hope i helped?\
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Greg drew a diagram to compare two of the fundamental forces.
Mashcka [7]

Answer:

X: Always attractive

Y: Infinite range

Z: Attractive or repulsive

ANSWER IS C

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3 years ago
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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
5 0
3 years ago
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If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is t
Aleksandr-060686 [28]

Answer:

The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Explanation:

Given;

first magnetic field at first distance, B₁ = 2.50 mT

first distance, r₁ = 12.6 cm = 0.126 m

Second magnetic field at Second distance, B₂ = ?

Second distance, r₂ = ?

Magnetic field for a straight wire is given as;

B = \frac{\mu I}{2 \pi r}

Where:

μ is permeability

B is magnetic field

I is current flowing in the wire

r distance to the wire

Let \ \frac{\mu I}{2\pi}  \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

7 0
3 years ago
HELPPP
AlekseyPX
Friction is caused by the uneven surfaces of touching objects
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2 years ago
How far from the base of the platform does she land?
seropon [69]

When Janet leaves the platform, she's moving horizontally at 1.92 m/s.  We assume that there's no air resistance, and gravity has no effect on horizontal motion.  There's no horizontal force acting on Janet to make her move horizontally any faster or slower than 1.92 m/s.

She's in the air for 1.1 second before she hits the water.

Moving horizontally at 1.92 m/s for 1.1 second, she sails out away from the platform

(1.92 m/s) x (1.1 sec) = <em>2.112 meters</em>

3 0
3 years ago
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