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Aneli [31]
3 years ago
14

A car is behind a truck going 25 m/s on the highway. The car’s driver looks for an opportunity to pass, guessing that his car ca

n accelerate at 1.0 m/s^2. He gauges that he has to cover the 20 m length of truck, plus10 m clear room at the rear of the truck and 10 m more at the front of it. In the oncoming lane, he sees a car approaching, probably also travelling at 25 m/s. He estimates that the car is about 400 m away, should he attempt the pass? Give details
Physics
1 answer:
user100 [1]3 years ago
8 0

Answer:

No he should not attempt the pass

Explanation:

Let t be the time it takes for the car to pass the truck. The driver should ONLY attempt to pass when the distance covered by himself plus the distance covered by the oncoming car is less than or equal 400 m (a near miss)

At acceleration of 1m/s2 and a clear distance of 10 + 20 + 10 = 40 m, we can use the following equation of motion to estimate the time t in seconds

s = at^2/2

40 = 1t^2/2

t^2 = 80

t = \sqrt{80} = 8.94 s

Within this time frame, the first car would have traveled a total distance of the clear distance (40m) plus the distance run by the truck, which is

8.94 * 25 = 223.6m

So the total distance traveled by the first car is 223.6 + 40 = 263.6m

The distance traveled by the 2nd car within 8.94 s at rate of 25m/s is

8.94 * 25 = 223.6 m

So the total distance covered by both cars within this time frame

223.6 + 263.6 = 487.2m > 400 m

So no, he should not attempt the pass as we will not clear it in time.

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MA_775_DIABLO [31]

Answer:

A is a solid. C is a gas. In solid an liquid the particals are touching. In C, the particals have less affect on each other because they are so far apart.

5 0
3 years ago
Can we see a halo around the half moon?​
Hatshy [7]

Answer:

No we cannot

Explanation:

But what causes a ring to appear around the moon? This phenomenon is called a "moon halo." According to the National Weather Service, this ring of light, which is actually an optical illusion, forms around the moon when moonlight refracts off ice crystals in cirrus clouds, high up in the Earth's atmosphere.

8 0
2 years ago
Read 2 more answers
A capacitor is charged to 8.0×10−4 C , then discharged by connecting a wire between the two plates. 35us after the discharge beg
Anettt [7]

Answer:

The average current is 19.567 A

Solution:

As per the question:

Charge, Q = 8.0\times 10^{- 4}\ C

Time, t = 35\times 10^{- 6}\ s

Now,

We know that current is constituted by the rate of transfer of the charge per unit time. Thus we can write:

I = \frac{Q}{t}                 (1)

Now, the charge that was transferred is 86 % of the original value.

Therefore,

We replace Q by 0.86Q in eqn (1):

I = \frac{0.86\times 8.0\times 10^{- 4}}{35\times 10^{- 6}} = 19.657\ A

7 0
3 years ago
Calculate the maximum absolute uncertainty for R if:
Radda [10]

Answer:

ΔR = 9 s

Explanation:

To calculate the propagation of the uncertainty or absolute error, the variation with each parameter must be calculated and the but of the cases must be found, which is done by taking the absolute value

           

The given expression is      R = 2A / B

the uncertainty is                 ΔR = | \frac{dR}{dA} | ΔA + | \frac{ dR}{dB} | ΔB

we look for the derivatives

     \frac{dR}{dA} = 9 / B

     \frac{dR}{dB} = 9A ( - \frac{1}{B^2 } )

we substitute

     ΔR = \frac{9}{B}  ΔA + \frac{9A}{B^2}  ΔB

the values ​​are

     ΔA = 2 s

     ΔB = 3 s

 

     ΔR = \frac{9}{11}   2 + \frac{9 \ 32}{11^2 }  3

     ΔR = 1.636 + 7.14

     ΔR = 8,776 s

the absolute error must be given with a significant figure

     ΔR = 9 s

3 0
3 years ago
A 0.140 m high cylinder is filled with mercury(density=13600 kg/m^3). what is the pressure at the bottom of the cylinder?(unit=P
Arada [10]

Answer:18678.24pa or 0.184atm

Explanation: P= density *height *acceleration due to gravity

Density =13600kg/m^3

gravity =9.81m/s^2

H=0.140m

Pressure =13600*9.81*0.140

Pressure =18678.24pa or 0.184atm

8 0
3 years ago
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