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galina1969 [7]
4 years ago
12

A Boeing 747 ""Jumbo Jet"" has a length of 59.7 m. The runway on which the plane lands intersects another runway. The width of t

he intersection is 25.0 m. The plane decelerates through the intersection at a rate of 5.70 m/s2 and clears it with a fi nal speed of 45.0 m/s. How much time is needed for the plane to clear the intersection?
Physics
1 answer:
Reika [66]4 years ago
8 0

Answer:

1.7 seconds

Explanation:

To clear the intersection, the total distance to be covered = 59.7 + 25 =84.7m

first we need to find the initial speed to just enter the intersection by using the third equation of motion

v^2 - u^2 = 2*a*s

45^2 - u^2 = 2 * -5.7 * 84.7

u^2 = 45^2 +965.58

u^2 = 2990.58

u = 54.7 m/s

Now for time we apply the first equation of motion

v-u =a * t

t = (v-u)/a = (45 - 54.7)/-5.7 = 1.7seconds

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Answer:

c. prone; supine

Explanation:

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2 years ago
Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.18 m away from a waterfall 0.294 m in heigh
Karolina [17]

Answer:

v = 7.65 m/s

t = 0.5882 s

Explanation:

We are told that the salmon started downstream, 3.18 m away from a waterfall.

Thus, range = 3.18 m

Since the horizontal velocity component is constant, then;

Range = vcosθ × t

Thus,

vcosθ × t = 3.18 - - - (eq 1)

We are told the salmon reached a height of 0.294 m

Thus, using distance equation;

s = v_y•t + ½gt²

g will be negative since motion is against gravity.

s = v_y•t - ½gt²

Thus;

0.294 = v_y•t - ½gt²

v_y = vsinθ

Thus;

0.294 = vtsinθ - ½gt² - - - (eq 2)

From eq(1), making v the subject, we have;

v = 3.18/tcosθ

Plugging into eq 2,we have;

0.294 = (3.18/tcosθ)tsinθ - ½gt²

0.295 = 3.18tanθ - ½gt²

We are given g = 9.81 m/s² and θ = 45°

0.295 = (3.18 × tan 45) - ½(9.81) × t²

0.295 = 3.18 - 4.905t²

3.18 - 0.295 = 4.905t²

4.905t² = 2.885

t = √2.885/4.905

t = 0.5882 s

Thus;

v = 3.18/(0.5882 × cos45)

v = 7.65 m/s

8 0
3 years ago
True or False: Speed is a vector quantity.
Travka [436]

Answer:

it's false coz

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3 0
3 years ago
Read 2 more answers
A playground slide is in the form of an arc of a circle with a maximum height of 3.0 m, with a radius of 8.5 m, and with the gro
julia-pushkina [17]

Answer:

a) s \approx 6.676\,m

Explanation:

a) Let assume that the ground is not inclined, since the bottom of the playground slide is tangent to ground. Then, the length of given by the definition of a circular arc:

s = \frac{\pi}{4}\cdot R

s=\frac{\pi}{4}\cdot (8.5\,m)

s \approx 6.676\,m

The bottom of the slide has a height of zero. The physical phenomenon around Dr. Ritchey's daughter is modelled after Principle of Energy Conservation. The child begins at rest:

U_{g,A} = K_{B} + W_{fr}

m\cdot g \cdot h_{A} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + f\cdot s

The average frictional force is cleared within the expression:

f = \frac{m\cdot (g\cdot h_{A}-\frac{1}{2}\cdot v_{B}^{2})}{s}

f = \frac{(12\,kg)\cdot [(9.807\,\frac{m}{s^{2}} )\cdot (3\,m)-\frac{1}{2}\cdot (4.5\,\frac{m}{s} )^{2} ]}{6.676\,m}

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4 0
3 years ago
What would happen if a solid temperature increases
Ber [7]
It will increase in size
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3 years ago
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