Four +2 μC charges are placed at the positions (10 cm, 0 cm), (−10 cm, 0 cm), (0 cm, 10 cm), and (0 cm, −10 cm) such that they f
1 answer:
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Answer:
The coefficient of static friction between the puppy and the floor is 0.7273.
Explanation:
The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:
Where is the static friction force,
is the coefficient of static friction and
is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore,
, to make the puppy moving we need to use a force of 80 N, therefore,
, so we can solve for the coefficient as shown below:
The coefficient of static friction between the puppy and the floor is 0.7273.
Answer:
= 33.33 cm
Explanation:
Given:
When mass, =21 kg
distance travelled is = 140 cm
When mass, =5 kg
distance travelled is = ?
Hooke's law state that within elastic limit, when an external force is applied to a body, the body gets deformed and when the force is released the gets back to its original form.
Therefore according to the question,
= 33.33 cm
Distance travelled is 33.33 cm when mass is 5 kg.
Answer:
0.2631 N/C
Explanation:
Given that:
The radius of the wire r = 0.22 mm = 0.22 × 10⁻³ m
The radius of the thick wire r' = 0.55 mm = 0.55 × 10⁻³ m
The numbers of electrons passing through B, N = 6.0 × 10¹⁸ electrons
Electron mobility μ = 6.0 x 10-4 (m/s)/(N/C)
= 0.0006
The number of electron flow per second is calculated as follows:
The magnitude of the electric field is:
E =
E =
E =
E = 0.2631 N/C