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pickupchik [31]
1 year ago
6

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 73.5 m/s. Th

e driver of the Thunderbird realizes that she must make a pit stop, and she smoothly slows to a stop over a distance of 250 m. She spends 5.00 s in the pit and then accelerates out, reaching her previous speed of 73.5 m/s after a distance of 400 m. At this point how far has the Thunderbird fallen behind the Mercedes Benz, which has continued at a constant speed
Physics
1 answer:
tekilochka [14]1 year ago
6 0

Answer:

Thunderbird has fallen behind the Mercedes Benz by 1017.49 m

Explanation:

Given the data in question;

initial speed of the ford u1 = 73.5 m/s

distance d1 = 250 m

t1 = 5.00 s

d2 = 400 m

Now, let the time taken to stop be t2 and deceleration is a1

so,

a1 = u1² / (2 × d1)

a1 = (73.5)² / (2 × 250)

a1 = 10.8045 m/s²

Now , for acceleration is a2

a2 = v² / (2 × d2)

a2 = (73.5)² / (2 × 400)

a2 = 6.7528 m/s²

total time spend = 5 + u/a1 + u/a2

total time spend = 5 + (73.5/10.8045) + (73.5/6.7528)

total time spend = 22.687 sec

Now, distance Mercedes is ahead = 22.687 × 73.5 - 400 - 250

= 1667.4945 - 400 - 250

= 1017.49 m

Therefore, Thunderbird has fallen behind the Mercedes Benz by 1017.49 m

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Determine the length of a copper wire that has a resistance of 0.172 ? and cross-sectional area of 7.85 × 10-5 m2. The resistivi
KonstantinChe [14]

Answer:

Length of copper wire, l = 785 meters

Explanation:

Given that,

Resistance of the copper wire, R = 0.172 ohms

Area of cross section, A=7.85\times 10^{-5}\ m^2

Resistivity of copper, \rho=1.72\times 10^{-8}\ \Omega-m

The resistance of a wire is given by :

R=\rho\dfrac{l}{A}

l=\dfrac{RA}{\rho}

l=\dfrac{0.172\ \Omega\times 7.85\times 10^{-5}\ m^2}{1.72\times 10^{-8}\ \Omega-m}

l = 785 meters

So, the length of the copper wire is 785 meters. Hence, this is the required solution.

8 0
2 years ago
A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho
Gnoma [55]

Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)

80=5(50-x)

80=250-5x

5x=170

x=\frac{170}{5}

x=34 cm

4 0
2 years ago
a bicycle pump contains 20cm3 of air at a pressure of 100kpa the air is then pumped in a single stroke through a valve into a ty
riadik2000 [5.3K]
If we assume also that the temperature of the air does not change, we can use Boyle's Law:
p₁V₁ = p₂V₂

Now, we know: 
p₁ = 100kPa
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We can solve for p₂:
p₂ = (p₁V₁)/V₂
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Therefore your answer is: 120kPa
8 0
2 years ago
A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 t. what is the maximum hall voltage
Alex787 [66]
The equation for Hall voltage Vh is:

Vh=v*B*w, where v is the velocity of the strip, B is the magnitude of the magnetic field, and w is the width of the strip. 

v=25 cm/s = 0.25 m/s
B=5.6 T
w= 1.2 mm = 0.0012 m

We input the numbers into the equation and get:

Vh= 0.25*5.6*0.0012 = 0.00168 V

The maximum Hall voltage is Vh= 0.00168 V.
4 0
2 years ago
If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is u  =  7.73 \  m/s

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      s = h+  u t +  0.5 gt^2

Here s  is the distance covered by the bag at sea level which is zero

      0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>    0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>   u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}

=>   u  =  7.73 \  m/s

     

7 0
2 years ago
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