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2 years ago

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Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal ener

gy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.

1 answer:

AVprozaik [17]2 years ago

8 0

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

= 1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

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A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

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It is important that your muscles are very warm when doing this type of stretching.

MaRussiya [10]

Answer:

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3 years ago

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A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t

Anastaziya [24]

Answer:

The gauge pressure is $P_g = 2058 \ P_a$

Explanation:

From the question we are told that

The height of the water contained is $h_w = 30 \ cm = 0.3 \ m$

The height of liquid in the cylinder is $h_t = 40 \ cm = 0.4 \ m$

At the bottom of the cylinder the gauge pressure is mathematically represented as

$P_g = P_w + P_o$

Where $P_w$ is the pressure of water which is mathematically represented as

$P_w = \rho_w * g * h_w$

Now $\rho_w$ is the density of water with a constant values of $\rho_w = 1000 \ kg /m^3$

substituting values

$P_w = 1000 * 9.8 * 0.3$

$P_w = 2940 \ Pa$

While $P_o$ is the pressure of oil which is mathematically represented as

$P_o = \rho_o * g * (h_t -h_w )$

Where $\rho _o$ is the density of oil with a constant value

$\rho _o = 900 \ kg / m^3$

substituting values

$P_o = 900 * 9.8 * (0.4 - 0.3)$

$P_o = 882 \ Pa$

Therefore

$P_g = 2940 - 882$

$P_g = 2058 \ P_a$

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2 years ago

Determine the inductive reactance for a 50 mH inductor that is across a 15 volt, 400 Hz source.

tresset_1 [31]

Answer:

Inductive reactance is 125.7 Ω

Explanation:

It is given that,

Inductance, $L=50\ mH=0.05\ H$

Voltage source, V = 15 volt

Frequency, f = 400 Hz

The inductive reactance of the circuit is equivalent to the impedance. It opposes the flow of electric current throughout the circuit. It is given by :

$X_L=2\pi fL$

$X_L=2\pi \times 400\ Hz\times 0.05\ H$

$X_L=125.66\ \Omega$

$X_L=125.7\ \Omega$

So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.

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Jlenok [28]

Oil is optically denser than water. When sound/light goes from optically denser medium to optically rarer medium, their velocity increase and they moves away for normal.

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$\Large{ \underline{ \boxed{ \pink{ \bf{Option \: (D)}}}}}$

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