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ki77a [65]
3 years ago
12

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

sees the car, the locomotive is 300 m from the crossing and its speed is 18 m/s. If the engineer’s reaction time is 0.45 s, what should be the magnitude of the minimum deceleration to avoid an accident? Answer in units of m/s 2 .
Physics
2 answers:
nordsb [41]3 years ago
8 0

Answer:

a=0.555m/s^2

Explanation:

First we find the distance traveled from the moment the engineer reacts to the car, assuming uniform movement

X=VT

X=(18)(0.45)=8.1m

then we find the distance at which the deceleration begins, which is obtained by subtracting the total distance with the inner result

X=300-8.1=291.9

finally we use the equation for constant acceleration

Vf=0 final speed

Vo=18m/s= initial speed

X=291.9m

(Vf^2-Vo^2)/2X=a

(0-18^2)/(2*291.9)=a

a=0.555m/s^2

Leona [35]3 years ago
5 0

Answer:

- 0.56 m/s 2

Explanation:

To calculate the magnitude of deceleration to avoid the collision

distance = 300 m

initial speed = 18 m/s

reaction time ( t ) = 0.45 s

first calculate the distance traveled by the engineer/train during his reaction time

D₁ = initial speed * reaction time = 18 * 0.45 = 8.1 m

Then the remaining distance before impact between the car and Train would be

= distance - distance traveled during reaction time

= 300 m - 8.1 m = 291.9 m

from the equation of motion of objects

V² = U² + 2ax  (equation 1)

To avoid impact the final velocity( v )  would be ( 0 )

u = initial speed ( 18 m/s )

a = acceleration/deceleration ( ? )

x = remaining distance before collision ( 291.9 m )

hence equation 1 becomes

a = -\frac{18^{2} }{2*291.9}   =  - \frac{324}{583.8}  = - 0.56 m/s 2

note: the answer is in negative because the Locomotive is decelerating and not accelerating

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A cat walks along a plank with mass M= 7.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di
solong [7]

Answer:

Before the plank will tip cat will walk 1.652 m

Explanation:

Mass of the cat along with plank m_1=7kg

Center of mass of the plank d_1=0.850m

Mass of cat m_2=3.6kg

We have to find how far right of sawhorse B.

Plank will tip when weight of the cat about B is greater than the torque by the weight of the plank.

Balancing the the torque

m_1gd_1=m_1gd_2

7\times 9.8\times 0.850=3.6\times 9.8\times d_2

d_2=1.652m

5 0
3 years ago
(c) If η = 60% and TC = 40°F, what is TH, in °F?
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2b2t hope that helps
5 0
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In Ch. 1.6, the authors point out that interstellar space is not actually as empty as it seems. There is actually a lot of diffu
wel

Answer:

very small solid particles called interstellar dust.

Explanation:

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6 0
3 years ago
Are these two correct ?
gulaghasi [49]

Answer:

7. Your answer is correct dear, just add the unit

8. answer is 1.17m/s²

Explanation:

queation 7.

m = 3kg, F = 9N, a ?

F = ma

a = F/m = 9/3 = 3m/s²

Use the same approach for question 8

5 0
3 years ago
Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin
likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0
3 years ago
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