Explain why if air gets in brake system would reduce the efficiency of the brakes

How to make an uncharged particle positively charged

AlekseyPX

Answer:

If a neutral atom gains electrons, then it will become negatively charged. If a neutral atom loses electrons, then it become positively charged.

6 0

3 years ago

If the magnetic force is 3.5 × 10–2 N, how fast is the charge moving?

Sveta_85 [38]

Answer:

The velocity of charge is 1.1×10⁴

Option(D)

Explanation:

F = q(v×B)

3.5×10-²= (8.4×10-⁴)*(v)*(6.7×10-³)*sin35

v = (3.5)/(8.4×6.7×0.57) × 10⁵

v = 3.5/32.08 × 10⁵

v = 0.109 × 10⁵

v ≈ 1.1 × 10⁴ m/s

6 0

3 years ago

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A 2.43ug particle moves at 1.97 x 108 m/s. What is its momentum?

Ira Lisetskai [31]

Answer:

0.48 kgm/s

Explanation:

$m$ = mass of the particle = 2.43 μg = 2.43 x 10⁻⁶ x 10⁻³ kg = 2.43 x 10⁻⁹ kg

$v$ = velocity of the particle = 1.97 x 10⁸ m/s

$p$ = momentum of the particle

momentum of the particle is given as

$p = m v$

inserting the values

$p = (2.43\times 10^{-9})(1.97\times 10^{8})$

$p = 0.48$ kgm/s

4 0

3 years ago

Melting ice caps can result in _____.

IceJOKER [234]

Melting of ice caps ... causing them to shrink in coverage .. can result in ...

-- wet ground

-- more farming opportunities, but less skiing

-- higher sea level

-- coastal flooding

-- extinction of plant and animal species

-- more icebergs floating around

-- lower average albedo, higher solar absorption, widespread warming

4 0

3 years ago

Read 2 more answers

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.

Explain why if air gets in brake system would reduce the efficiency of the brakes​

Explain why if air gets in brake system would reduce the efficiency of the brakes