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2 years ago

The drag force that resists the motion of a car traveling at 80 km h^- 1 is 300 N.

Physics

1 answer:

kobusy [5.1K]2 years ago

8 0

The power require to keep the car traveling is 6,666 W.

The power of the engine at the given efficiency is 3,999.6 W.

<h3>What is Instantaneous power?</h3>

This the product of force and velocity of the given object.

The power require to keep the car traveling is calculated as follows;

P = Fv

$P = 300\ N \ \times \ \frac{80 \ kmh^{-1}}{3.6 \ km h^{-1}/m/s} \\\\
P = 300 \ N \times 22.22 \ m/s\\\\
P = 6,666 \ W$

The power of the engine at the given efficiency is calculated as follows;

$E = \frac{P_{out}}{P _{in}} \times 100\%\\\\
60\% = \frac{P_{out}}{6,666} \times 100\%\\\\
0.6 = \frac{P_{out}}{6,666} \\\\
P_{out} = 3,999.6 \ W$

Learn more about efficiency here: brainly.com/question/15418098

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Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?

Alexandra [31]

11) $1.04\cdot 10^7 J$

12) $1.04\cdot 10^7 J$

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

$g=9.8 m/s^2$ (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

$PE=(8325)(9.8)(127)=1.04\cdot 10^7 J$

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

$KE_t + PE_t = KE_b + PE_b$

where

$KE_t$ is the kinetic energy at the top

$PE_t$ is the potential energy at the top

$KE_b$ is the kinetic energy at the bottom

$PE_b$ is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

$KE_t=0$

Also, at the bottom the height is zero, so the potential energy is zero

$PE_b=0$

Therefore, we find:

$KE_b=PE_t=1.04\cdot 10^7 J$

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

$KE=1.04\cdot 10^7 J$

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s$

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

$KE_1 + PE_1 = KE_2 + PE_2$

where

$KE_1$ is the kinetic energy at the top of the 1st hill

$PE_1$ is the potential energy at the top of the 1st hill

$KE_2$ is the kinetic energy at the top of the 2nd hill

$PE_2$ is the potential energy at the top of the 2nd hill

From the previous questions, we know that

$KE_1=0$

and

$PE_1=1.04\cdot 10^7 J$

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

$PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J$

So, the kinetic energy at the second hill is

$KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J$

And so, the speed is

$v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s$

4 0

3 years ago

Which red line shows the same action spectrum corrected for the unequal number of photons emitted across the visible spectrum?.

stealth61 [152]

The visible spectrum is composed of red, orange,yellow, green, blue, violet, indigo.

<h3>What is visible spectrum?</h3>

The visible spectrum refers to the portion of the electromagnetic spectrum that can be seen with the eyes. All other portions of the electromagnetic spectrum are invisible.

The question is incomplete as the details are missing. The visible spectrum is composed of red, orange,yellow, green, blue, violet, indigo.

Learn more about the visble spectrum: brainly.com/question/1596783

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2 years ago

Ivan walks 10 meters west to the water fountain, then runs 2 meters east to his class to avoid a lockout. His displacement would

otez555 [7]

Answer:

Explanation:

if he starts at ten and takes 10 steps left he'll be at -10... then if he takes 2 steps to the right , he's at -8 on the number-line

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3 years ago

Read 2 more answers

The substances referenced in the table are being considered for use in cooking materials. Since the materials must be melted dur

frutty [35]

Brass requires more energy than silver

Aluminum requires more energy than copper

Aluminum requires more energy than brass

Explanation:

The specific heat capacity of a substance indicates the amount of heat energy required to raise 1 kg of that substance by 1 degree in temperature.

Mathematically:

$C=\frac{Q}{m\Delta T}$

where

Q is the heat supplied to the substance

m is the mass of the substance

$\Delta T$ is the change in temperature

Therefore, the higher the specific heat capacity of a substance, the more energy is needed to increase its temperature.

Here we can compare the specific heat capacity of the materials mentioned:

Silver: $C=0.233 J/gK$

Brass: $C=0.380 J/gK$

Platinum: $C=0.130 J/gK$

Aluminium: $C=0.910 J/gK$

Copper: $C=0.390 J/gK$

Therefore, the correct statements are:

Brass requires more energy than silver

Aluminum requires more energy than copper

Aluminum requires more energy than brass

Learn more about specific heat:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0

3 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

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3 years ago