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icang [17]
3 years ago
11

Exit

Physics
1 answer:
Mila [183]3 years ago
3 0
The gravitation field strength of Planet X is 10 times that of Earth's, as the object weight has increased 10 times. This is only possible if the planet is more massive or more dense (smaller diameter).

Therefore, the answer is C.
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How many miles of MgCI2 are there in 339g of compound?
harkovskaia [24]

I'm going to answer this by using rounded numbers for the atomic masses. You need only go back and put the numbers in from your periodic table. My answers will be close, but not what you should get.

Find the Molar Mass of MgCl2

Mg =                   24 grams

2Cl = 2 * 35.5 = 71 grams

Total  =              95 grams

Find the mols in 339 grams

1 mole = 95 grams

x mol = 339

Solve

339 = 95x   Divide by 95

339/95 = x

x = 3.67 mols  Answer

4 0
3 years ago
Difference between moment and displacement​
harkovskaia [24]

<u>Displacement</u> is the difference between final position and initial position.

<u>Momentum</u> is the quantity of motion contained by an object.

  • It is the product of <em><u>mass and velocity.</u></em>
7 0
3 years ago
Elements in which family are most likely to have properties associated with
Elena-2011 [213]

Answer:  As with all metals, the alkali metals are malleable, ductile, and are good conductors of heat and electricity. The alkali metals are softer than most other metals.

Alkaline earth metals

The alkaline earth elements are metallic elements found in the second group of the periodic table

Explanation:

5 0
3 years ago
A permanent magnet has a magnetic dipole moment of 0.160 A · m^2. The magnet is in the presence of an external uniform magnetic
Elena L [17]

Answer:

the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

Explanation:

The torque is given by :

\bar {N} = \bar {m} * \bar {B}

where ;

m = 0.160 A.m²

B = 0.0800 T

θ = 35°

So the magnitude of the torque N = mBsinθ

N = (0.160)(0.0800)(sin 35°)

N = 0.007341

N = 7.34×10⁻³ Nm

Hence, the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

b) The potential energy \bar{U} = \bar{-m} * \bar{B}

U = -mBcosθ

U = (- 0.160)(0.0800)(cos 45)

U = -0.010485

U = -1.0485 ×10⁻² J

Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

6 0
3 years ago
Because the Triple Beam Balance is zeroed before being put into use, it measures
pickupchik [31]

Answer:

B

Explanation:

8 0
3 years ago
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