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PilotLPTM [1.2K]
4 years ago
13

What is true about the force between charges?

Physics
2 answers:
shusha [124]4 years ago
5 0

Answer:

The statements A and D are true

The statements B and C are false

Explanation:

The force between charges can be explained by the Coulomb's Law.

According to Coulomb:

1 - Like charges always repel each other

2- Unlike or opposite charges always attrack each other

3- Force between 2 charges is given by:

F=k\frac{q_1q_2}{r^2}

where F is the force between 2 charges and r is the distance between 2 charges.

We can see that Force F is inversely proportional to the distance r

Which means that when F increase, r decreases and when F decreases, r increases

DedPeter [7]4 years ago
5 0

Answer: opposite charges attracts each other & like charges repel each other

Explanation:

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malfutka [58]
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3 0
3 years ago
An electric field of 1.27 kV/m and a magnetic field of 0.490 T act on a moving electron to produce no net force. If the fields a
lesantik [10]

Answer:

v = 2591.83 m/s

Explanation:

Given that,

The electric field is 1.27 kV/m and the magnetic field is 0.49 T. We need to find the electron's speed if the fields are perpendicular to each other. The magnetic force is balanced by the electric force such that,

qE=qvB\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.27\times 10^3}{0.49}\\\\v=2591.83\ m/s

So, the speed of the electron is 2591.83 m/s.

8 0
3 years ago
Convert Gravitational constant (G) = 6.67×10^-11 Nm²kg^-2 to cm³ g ^-1 s^-2.
GaryK [48]

Answer:

6.67×10⁻⁸ cm³/g/s²

Explanation:

6.67×10⁻¹¹ Nm²/kg²

= 6.67×10⁻¹¹ (kg m/s²) m²/kg²

= 6.67×10⁻¹¹ m³/kg/s²

= 6.67×10⁻¹¹ m³/kg/s² × (100 cm/m)³ × (1 kg / 1000 g)

= 6.67×10⁻⁸ cm³/g/s²

3 0
3 years ago
Brad has three beakers of water. Each beaker contains 500 mL of water. The temperature of the water in the first beaker is 40°C.
dexar [7]

Explanation:

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7 0
3 years ago
Design a tension member and slip-critical splice to carry a factored load of 500 kips. Please use a wide-flange section for the
Svetlanka [38]

Answer:

Kindly check the explanation section.

Explanation:

For the design we are asked for in this question/problem there is the need for us to calculate or determine the strength in fracture and that of the yield. Also, we need to calculate for the block shear strength.

From the question, we have that the factored load = 500kips. Also, note that the tension splice must not slip.

Also, the shear force are resisted by friction, that is to say shear resistance = 1.13 × Tb × Ns.

Assuming our db = 3/4 inches, then the slip critical resistance to shear service load = 18ksi(refer to AISC manual for the table).

If db = 7/8 inches, then the shear force resistance for n bolt = 10.2kips, n > 49.6.

The yielding strength = 0.9 × Aj × Fhb= 736 kips > 500

The fracture strength = .75 × Ah × Fhb = 309 kips.

The bearing strength of 7/8 inches bolt at the edge hole and other holes = 46 kips and 102 kips.

4 0
4 years ago
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