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8_murik_8 [283]

2 years ago

14

A car is on a circular off ramp of an interstate and is traveling at exactly 25 mph around the curve. Does the car have velocity

? Yes No Does the car have acceleration? Yes No Is the car decelerating? Yes No

1 answer:

netineya [11]2 years ago

6 0

Answer:

The car has velocity and acceleration but is not decelerating

Explanation:

Since the car is traveling at 25 mph around the curve, it has a tangential velocity. This tangential velocity is constantly changing in direction (so the car could adapt to the curve and not moving forward in a straight line), there should be a centripetal acceleration in play here. This acceleration does not slow down the car so it's not decelerating.

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What material or device prevents the conducting of electricity?

PtichkaEL [24]

Answer:

Explanation:

An insulator. You can see ceramic insulators on telephone poles and power poles if you look carefully. If you live in a city, somewhere in that city is a power station. The insulators are huge. They have to be. The currents are very large in many cases.

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3 years ago

3. A car going 22m/s accelerates to pass a truck. Five seconds late the car is going 35m/s. Calculate the acceleration of the ca

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Answer:

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Explanation:

6 0

2 years ago

1. An express train, traveling at 36 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a loc

Colt1911 [192]

Answer:

(i) 12 seconds

(ii) 216 meters from the initial position

(iii) 132 meters from the initial position

(iv) No

Explanation:

Speed of express train =36 m/s

Speed of local train =11 m/s

The initial distance between the local train and passenger train =100 m.

Due to the application of breaks, the express train slows at the rare of $3.0 m/s^2.$

So, the acceleration of the express train, $a=-3 m/s^2$ .

(i) Let t be the time the express train takes to stop.

From the equation of motion,

v=u+at

where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.

In this case, v=0, u=36 m/s, $a=-3 m/s^2$

So, 0=36+(-3)t

$\Rightarrow t= 36/3=12$ seconds.

(ii) To compute the distance traveled, s, till the express train stops, using

$v^2=u^2+2as$

$\Rightarrow 0^2=36^2+2(-3)s$

$\Rightarrow s=\frac{36\times36}{6}$

$\Rightarrow s=216$ meters.

(iii) The local train is moving at a speed of 11 m/s

So, in 12 seconds, the distance, d, traveled by the local train

d= 11x12=132 meters [as distance= speed x time]

(iv) Let 0 be the reference position which is the initial position of the express train.

So, at the initial time, the position of the local train is at 100m.

After 12 seconds:

The position of the express train is at 216 m [using part (ii)]

and the position of the local train is at 100+132=232m [using part (iii)].

So, the local train is still ahead of the express train, hence the trains didn't collide.

6 0

2 years ago

Weight gain occurs when a person consumes too muchimany

MAVERICK [17]

The answer is calories !!

5 0

2 years ago

Read 2 more answers

A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i

AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition

- Kinetic and potential energy changes are negligible

Now from steam tables, we calculate specific volume $v$ and enthalpy $h$ as,

$h_1 = 95.96Btu/lb$ ( $h_1 = h_f$ at $2psia$ )

$v_1 = 0.016238ft^3/lb$ ( $v_1 = v_f$ at $2psia$ )

$w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb$

$w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb$

$h_3 = 1364.0Btu/lb$

$s_3 = 1.5073Btu/lb.R$

( at $P_3 = 1500psia$ & $T_3 = 800^0F$ )

$P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765$

( $S_f$ & $S_{fg}$ when pressure is 2psia)

$h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb$

$n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb$

Therefore,

$q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb$

To calculate the mass flow rate of steam in the cycle, we use the formula

$W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s$

where $1Kj = 0.947817 Btu$

The power output and the rate of heat addition are calculated thus,

$W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW$

$Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s$

The thermal efficiency of the cycle can be found thus;

$n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343$

= 34.3%

5 0

2 years ago