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prohojiy [21]
3 years ago
14

How does the electric force between two charged particles change if particle's charge is reduced by a factor of 3?

Physics
2 answers:
MAVERICK [17]3 years ago
3 0

The force between two charges is proportional to the product of the charges.

If only one of the charges is reduced by a factor of 3, then the force is reduced by a factor of 3.

If both charges are reduced by a factor of 3, then the force is reduced by a factor of 9.

mamaluj [8]3 years ago
3 0

Answer: It is reduced by a factor of 3.

Explanation:

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Answer:

Speed of both blocks after collision is 2 m/s

Explanation:

It is given that,

Mass of both blocks, m₁ = m₂ = 1 kg

Velocity of first block, u₁ = 3 m/s

Velocity of other block, u₂ = 1 m/s

Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)v

v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

v=\dfrac{1\ kg\times 3\ m/s+1\ kg\times 1\ m/s}{2\ kg}

v = 2 m/s

Hence, their speed after collision is 2 m/s.

7 0
3 years ago
Click on the boxes below to indicate the number of electrons or protons in each labeled position for a zinc atom (z=30, a=65). P
g100num [7]

Answer:

Name: Zinc

Symbol: Zn

Atomic Number: 30

Atomic Mass: 65.39 amu

Melting Point: 419.58 °C (692.73 K, 787.24396 °F)

Boiling Point: 907.0 °C (1180.15 K, 1664.6 °F)

Number of Protons/Electrons: 30

Number of Neutrons: 35

Classification: Transition metal

Crystal Structure: Hexagonal

Density at 293 K: 7.133 g/cm3

Color: bluish

('lil long, sorry)

7 0
2 years ago
Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3
Scilla [17]

Explanation:

Given that,

Angle by the normal to the slip α= 60°

Angle by the slip direction with the tensile axis β= 35°

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Applied stress = 12 MPa

We need to calculate the shear stress applied at the slip plane

Using formula of shear stress

\tau=\sigma\cos\alpha\cos\beta

Put the value into the formula

\tau=12\cos60\times\cos35

\tau=4.91\ MPa

Since, the shear stress applied at the slip plane is less than the critical resolved shear stress

So, The crystal will not yield.

Now, We need to calculate the applied stress necessary for the crystal to yield

Using formula of stress

\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}

Put the value into the formula

\sigma=\dfrac{6.2}{\cos60\cos35}

\sigma=15.13\ MPa

Hence, This is the required solution.

3 0
3 years ago
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STatiana [176]

Answer:

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Explanation:

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4 0
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