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3 years ago

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Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.40 m/s

. Sue applies her brakes but can accelerate only at −1.90 m/s2 because the road is wet. Will there be a collision? Yes No

2 answers:

schepotkina [342]3 years ago

8 0

Answer:

Yes there will be a collision.

Explanation:

We shall use the relative motion between the 2 cars to solve

Relative speed of Sue with respect to slow moving van is

$V_{r}=V_{sue}-V_{van}\\\\V_{r}=34-5.40=28.6m/s$

Now in order to stop this vehicle the distance it requires to stop shall be calculated by third equation of kinematics as

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

since the finally the car shall stop thus v = 0 m/s

Applying values in the above equation and solving for 's' we get

$0=28.6^{2}+2\times -1.90\times s\\\\\therefore s=\frac{28.6^{2}}{2\times 1.90}\\\\s=215.25m$

Since distance required to stop is greater than available distance of 160 m thus the two cars shall collide.

emmasim [6.3K]3 years ago

5 0

Answer:

Yes, there will be collision.

Explanation

Initial relative velocity u = 34-5.4 = 28.6 m/s

required final relative velocity v = 0

Acceleration , suppose be a .

Distance traveled =d = 160 m

Using the formula for relative motion,

v² - u² = 2as

0 - 28.6²=2as

a = - 2.55 m s⁻².

Since this required value is more than maximum braking deceleration so , there will be collision.

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Explanation:

Formula for maximum efficiency of a Carnot refrigerator is as follows.

$\frac{W}{Q_{H_{1}}} = \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}$ ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

$\frac{W}{Q_{C_{2}}} = \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}}$ ...... (2)

Now, equating both equations (1) and (2) as follows.

$Q_{C_{2}} \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}}$ = $Q_{H_{1}} \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}$

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= $\frac{250}{600} (\frac{(600 - 300)K}{300 K - 250 K})$

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3 years ago

It is known that heat is added to a gas in sealed container. The container is fitted with a moveable piston.

jasenka [17]

Answer:

d. Not enough information is given to answer this question.

Explanation:

From first law of thermodynamics

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Q- W = ΔU

Ii Q > W ,then temperature of the gas will increase.

If Q< W ,Then temperature of the gas will decreases.

If work done on the gas:

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There are three cases because they did not give any information about the work.That is why option d is correct.

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3 years ago

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tresset_1 [31]

Answer:

a) Acceleration of the car is given as

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As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

$m_1v_1 + m_2v_2 = (m_1 + m_2)v$

now we have

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a) For acceleration of car we know that it is rate of change in velocity of car

so we have

$a_{car} = \frac{v_f - v_i}{t}$

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b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

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3 years ago

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Answer:

v= 4.0 m/s

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When standing on the bathroom scale within the moving elevator, there are two forces acting on Henry's mass: Normal force and gravity.
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Normal force, can adopt any value needed to match the acceleration of the mass, according to Newton's 2nd Law.
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$F_{g} = m*g = 95 kg * 9.8 m/s2 = 930 N (1)$

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As the gravity is larger than normal force, this means that the acceleration is downward, so, we choose this direction as the positive.

Solving for a, we get:

$a =\frac{F_{net} }{m} =\frac{100 N}{95 kg} = 1.05 m/s2$

We can find the speed after the first 3.8 s (assuming a is constant), applying the definition of acceleration as the rate of change of velocity:

$v_{f} = a* t = 1.05 m/s * 3.8 m/s = 4.0 m/s$

Now, if during the next 3.8 s, normal force is 930 N (same as the weight), this means that both forces are equal each other, so net force is 0.
According to Newton's 2nd Law, if net force is 0, the object is either or at rest, or moving at a constant speed.
As the elevator was moving, the only choice is that it is moving at a constant speed, the same that it had when the scale was read for the first time, i.e., 4 m/s downward.

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3 years ago