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3 years ago

A 2,000kg car uses a braking force of 12,000 N to stop in 5 s

1 answer:

5 0

Knowing the amount of force F and the length of time t that force is applied to an object will tell you the resulting change in its momentum ΔP . This is:

F*t = ΔP

ΔP = F*t <em>but ΔP = mv</em>

mv = F*t

v = F*t/m

v = 12000 N × 5 s / 2000 kg

<h3>v = 30 m/s</h3><h2 /><h2>A/ The initial speed of the car was 30 m/s</h2>

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A truck driving east on a highway goes from the 42 mi marker to the 48 mi

Leona [35]

Average velocity is 1..2 mi/min east

Explanation:

Velocity = Displacement/Time

Here, displacement = 48 mi - 42 mi = 6 miles

Time = 5 minutes

⇒ Average Velocity = 6/5 = 1.2 mi/min east

3 0

3 years ago

Evaluate A car company wants to build a wind-powered car that converts 100 percent of the mechanical energy in the wind to the m

lorasvet [3.4K]

Answer:

There's one or two reasons, depending on what is meant by "wind-powered car".

The first reason is that it's impossible for any transfer of energy to be 100% efficient. There will always be frictional losses.

Secondly, if the company means that they want to attach a wind turbine to the car so that the car is powered by the same wind that it generates, that violates the conservation of energy.

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4 years ago

A piece of gym equipment states that the maximum load it can hold is 300 kg. Why do you think it is important not to go over thi

Crazy boy [7]

Answer:

The weight limit of 300kg is the maximum amount the machine can handle so it can be dangerous to exceed the maximum load.

5 0

3 years ago

Air enters a turbine operating at steady state with a pressure of 75 Ibf/in.^2, a temperature of 800º R and velocity of 400 ft/s

Arturiano [62]

Answer:

(a) W/m = 49.334 Btu/lb

(b) $\frac{E_{d} }{m}$ = 22.12 Btu/lb

Explanation:

For the given problem, it can be assumed that the system is operating at steady state and the effects of potential energy can be neglected.

(a) Using the thermodynamic table for air.

At the temperature ( $T_{1}$ )of 800 ºR and pressure ( $P_{1}$ ) of 75 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{1}$ ) = 191.81 BTu/lb

Specific entropy ( $s_{1}$ ) = 0.6956 Btu/(lb.ºR)

At the temperature ( $T_{2}$ )of 600 ºR and pressure ( $P_{2}$ ) of 15 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{2}$ ) = 143.47 BTu/lb

Specific entropy ( $s_{2}$ ) = 0.6261 Btu/(lb.ºR)

The work done can be calculated using energy rate equation:

$\frac{W}{m} = \frac{Q}{m} + (h_{1} - h_{2}) + \frac{V_{1}^{2} - V_{2}^{2}}{2}$

Q/m = heat transfer = -2 Btu/lb

$V_{1}$ = 400 ft/s

$V_{2}$ = 100 ft/s

$\frac{W}{m} = -2 + (191.81 - 143.47) + \frac{400^{2} - 100^{2}}{2}*[tex]\frac{1}{2*32.2*778}$ [/tex] = -2 + 48.34 + 29.938 = 49.334 Btu/lb

(b) To calculate the exergy destruction, we will use the equation for exergy rate:

$\frac{E_{d} }{m} = [1-\frac{T_{o} }{T_{b} }](\frac{Q}{m}) - \frac{W}{m} + [(h_{1} - h_{2}) -T_{o}(s_{1} - s_{2}) + \frac{V^{2} _{1} - V_{2} ^{2}}{2}]$

The equation above is further simplified to:

$\frac{Ed}{m} = T_{o}[(s_{2} -s_{1}) - Rln\frac{P_{2} }{P_{1} } - \frac{Q/m}{T_{b} }]$

Using a reference temperature (To) = 500 °R

Average surface temperature (Tb = 620°R

$\frac{Ed}{m} = 500*[(0.6261 -0.6956) - (1.986/28.97)ln\frac{15 }{75 } - \frac{-2}{620}}]$

$\frac{E_{d} }{m}$ = 500*[-0.0695 +0.068688*1.609 +0.003225] = 22.12 Btu/lb

5 0

4 years ago

A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a

EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$ , $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$ , $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$ , $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$ , then the final momentum of the big object is: