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n200080 [17]
3 years ago
11

Problem I Marcella (see warmup problem, above) gets her car moving steadily at 4m/s but suddenly someone stops ahead to assist h

er and parks their car 14 meters from the front of her car. Marcella grabs the car bumper and pulls very hard, with 200 N of force. The work she does transfers energy out, it reduces the K of the car, as it gradually approaches the other car. a) What is the initial kinetic energy before she tries to stop the car? b) What is the final kinetic energy, when her car hits the other car? What is the speed? c) Suppose the other person also slowed her car, pushing it from the front. How much force would be needed to stop her car 1 meter from the other car? [1 m allows the person not to be crushed!]
Physics
1 answer:
Nady [450]3 years ago
3 0

Answer:

Explanation:

a) KE = (1/2) * m * (v^{2}) = F * d = 14m * 200N = 2800 m/N or 2.8 * 10^{3} m/N

b) 0J and 0m/s (if Marcella stopped after going 14 meters)

c)  Known from part (a) that KE = 2800 J = F1 * d1,

    2800J = F1 * (14m - 1m)  => F1 = 2800J/13m = 215.384 N

   

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Given that a fluid at 260°F has a kinematic viscosity of 145 mm^2/s, determine its kinematic viscosity in SUS at 260°F.
Oduvanchick [21]

Answer:

kinematic viscosity in SUS is = 671.64 SUS

Explanation:

given data

kinetic viscosity = 145 mm^2/s

we know

1 mm = 0.1 cm

so kinetic viscosity in cm is \nu =145 (0.1)^{2} =1.45 cm^{2}/s

other unit of kinetic viscosity is centistokes

1 cm^{2}/s = 100 cst

so 1.45 cm^2/s will be 145 cst

if the temperature is 260°f , then cst value should be multiplied by 4.632. therefore kinematic viscosity in SUS is = 4.362 *145 = 671.64 SUS

5 0
3 years ago
A car drives around a curve with a radius of 42 m at a velocity of 24m/s. What is the centripical acceleration of the car?
Vinvika [58]

Answer:

13.71

Explanation:

13.71 miles per hour

7 0
3 years ago
An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first
GarryVolchara [31]

Answer:

The right answer is "1.369 m/s²".

Explanation:

The given values are:

Distance (s)

= 260 m

Initial speed (u)

= 26 m/s

Reaction time (t')

= 0.51 s

During reaction time, the distance travelled by locomotive will be:

⇒  s'=ut'

        =26\times 0.51

        =13.26 \ m

Remained distance between locomotive and car:

⇒  x=s-s'

         =260-13.26

         =246.74 \ m

Now,

The final velocity to avoid collection is, V = 0 m/s

From third equation of motion:

⇒  V^2=u^2+2ax

On putting the estimated values, we get

⇒  0=(26)^2+2\times a\times 246.74

⇒  0=676+493.48a

⇒  493.48a=-676

⇒            a=-\frac{676}{493.48}

⇒            a=1.369 \ m/s^2

3 0
3 years ago
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Explanation:

upthrust is the same as Force

And the dimension of Force is MLT-2

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Hope it helps....

8 0
3 years ago
Need help pls and explain also
Alisiya [41]
This should be the answer

3 0
2 years ago
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