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Schach [20]
3 years ago
6

________ skids occur when the brakes are applied so hard that the front or rear wheels lost traction. Cornering

Physics
2 answers:
pochemuha3 years ago
8 0

Answer:

Braking skid

Explanation:

When you apply brake very hard so that the rear wheel lost traction leads to the occurrence of Braking skid.

Skid occurs when the Tyre looses it grip from the road due to sudden brake or road being smooth and many more reasons.

Skid can be due to the worn off Tyre.

some time skidding can cause accident also so  one should use good Tyre for their vehicle.

mihalych1998 [28]3 years ago
7 0

Answer:

Braking skid

Explanation:

A skid occurs when a vehicle is moving on the road and the tires of the vehicle lose their hold on the roadway.

Braking skid occurs when the brakes are applied suddenly and the braking is very hard that it results in the lose of hold of the rear wheels and the rear wheels lose grasp over the road resulting in the skid of the vehicle.

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Answer:

0.36 A.

Explanation:

We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:

Resistor 1 (R₁) = 35 Ω

Resistor 2 (R₂) = 20 Ω

Equivalent Resistance (Rₑq) =?

Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:

Rₑq = (R₁ × R₂) / (R₁ + R₂)

Rₑq = (35 × 20) / (35 + 20)

Rₑq = 700 / 55

Rₑq = 12.73 Ω

Next, we shall determine the total resistance in the circuit. This can be obtained as follow:

Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω

Resistor 3 (R₃) = 15 Ω

Total resistance (R) in the circuit =?

R = Rₑq + R₃ (they are in series connection)

R = 12.73 + 15

R = 27.73 Ω

Finally, we shall determine the current. This can be obtained as follow:

Total resistance (R) = 27.73 Ω

Voltage (V) = 10 V

Current (I) =?

V = IR

10 = I × 27.73

Divide both side by 27.73

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Therefore, the current is 0.36 A.

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3 years ago
When a planet orbits the Sun, the
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A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from
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Answer:

E=35921.96N/C

Explanation:

From the question we are told that:

Radius r=0.321mm

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