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coldgirl [10]
3 years ago
14

Technician A says that OAT type coolant should be changed every 2 years. Technician B says that IAT type coolant should be chang

ed every 2 years. Which technician is correct?
Physics
2 answers:
julia-pushkina [17]3 years ago
7 0

Answer:

Technician B would be right

Explanation:

Andru [333]3 years ago
5 0

Answer:

Technician B is correct i-e IAT type coolant should be changed every 2 years.

Explanation:

Coolant is one of the most important fluid for the car as it prevents the engine from overheating during summers and freezing during winters. What it does is that it actually lowers the freezing point during summers and increases the boiling point of the system during summers. There are different types of coolant used by different cars. Among which, two of them are:

1) OAT i-e Organic Acid Technology

2) IAT i-e Inorganic Additive Technology  

IAT type coolants are usually used by older cars and generally should be usually replaced every 2 years. While, the other type which is the OAT type coolant should be replaced every 5 years.

Thus, Technician B is correct i-e IAT type coolant should be changed every 2 years.

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Answer: 0.4

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8 0
3 years ago
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A tennis player swings her 1000 g racket with a speed of 10 m/s. She hits a 60 g tennis ball that was approaching her at a speed
Mashutka [201]

(a)

consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.

m = mass of the tennis ball = 60 g = 0.060 kg

v₀ = initial velocity of the tennis ball before being hit by racket = 20 m/s

v = final velocity of the tennis ball after being hit by racket = - 39 m/s

ΔP = change in momentum of the ball

change in momentum of the ball is given as

ΔP = m (v - v₀)

inserting the above values

ΔP = (0.060) (- 39 - 20)

ΔP = - 3.54 kgm/s

hence , magnitude of change in momentum : 3.54 kgm/s

7 0
3 years ago
I am ion with 17 protons, 18 neutrons, and 18
gtnhenbr [62]

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7 0
3 years ago
The period of a 440 Hertz sound wave is __ seconds.
snow_lady [41]
The formula relating frequency and period of a wave is simply frequency = 1 / time. Therefore, by rearranging we also see that time = 1 / frequency. The period of a 440 Hertz sound wave is 1 / 440 seconds. The speed of the wave can also be determined very simply using the formula speed = wavelength x frequency. 
8 0
4 years ago
Two asteroids collide and stick together. The first asteroid has mass of 1.50 × 104 kg andis initially moving at 0.77 × 103 m/s.
KengaRu [80]

Answer:

Magnitude 900m/s, direction 12.8° respect to the velocity of the first asteroid.

Explanation:

This is a perfectly inelastic collision, because the two asteroids stick together at the end. That means that the kinetic energy doesn't conserves, but the linear momentum does. But, since the velocities of the asteroids have different directions, we have to break down them in components. For convenience, we will take the direction of the first asteroid as x-axis, and its perpendicular direction  (in the plane of the two velocity vectors) as y-axis. So, we have that:

p_{1ox}+p_{2ox}=p_{fx}\\\\p_{2oy}=p_{fy}

And, since p=mv, we get:

m_1v_{1o}+m_2v_{2o}\cos\theta=(m_1+m_2)v_{fx}\\\\m_2v_{2o}\sin\theta=(m_1+m_2)v_{fy}

Solving for v_fx and v_fy, and calculating their values, we get:

v_{fx}=\frac{m_1v_{1o}+m_2v_{2o}\cos\theta}{m_1+m_2}\\\\\implies v_{fx}=\frac{(1.50*10^{4}kg)(0.77*10^{3}m/s)+(2.00*10^{4}kg)(1.02*10^{3}m/s)\cos20\°}{1.50*10^{4}kg+2.00*10^{4}kg}=878m/s\\\\\\v_{fy}=\frac{m_2v_{2o}\sin\theta}{m_1+m_2}\\\\\implies v_{fy}=\frac{(2.00*10^{4}kg)(1.02*10^{3}m/s)\sin20\°}{1.50*10^{4}kg+2.00*10^{4}kg}=199m/s

Now, the final speed can be calculated using the Pythagorean Theorem:

v_f=\sqrt{v_{fx}^{2}+v_{fy}^{2}} \\\\\implies v_f=\sqrt{(878m/s)^{2}+(199m/s)^{2}}=900m/s

And the direction \beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°can be obtained using trigonometry:

\beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°

That means that the final velocity of the two asteroids has a magnitude of 900m/s and a direction of 12.8° with respect to the velocity of the first asteroid.

7 0
3 years ago
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