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Anni [7]
3 years ago
12

Car A is traveling north on a straight highway and car B is traveling west on a different straight highway. Each car is approach

ing the intersection of these highways. At a certain moment, car A is 0.4 km from the intersection and traveling at 75 km/h while car B is 0.3 km from the intersection and traveling at 70 km/h. How fast is the distance between the cars changing at that moment? km/h
Physics
1 answer:
Inessa05 [86]3 years ago
7 0

Answer:

102 km/h

Explanation:

x = distance of car A at any time from the intersection

x₀ = distance of car A at some time = 0.4 km

v_{A} = Speed of car A = 75 km/h

y = distance of car B at any time from the intersection

y₀ = distance of car B at some time = 0.3 km

v_{B} = Speed of car B = 70 km/h

d = distance between the two cars at any time

d₀ = distance between the two cars at some time

v = rate of change of distance between the cars

Using Pythagorean theorem

d²₀ = x₀² + y₀²

d²₀ = 0.4² + 0.3²

d₀ = 0.5 m

Distance between the two cars at any time is given using Pythagorean theorem as

d² = x² + y²

Taking derivative both side relative to "t"

2d \left ( \frac{dd}{dt} \right ) = 2x ( \frac{dx}{dt} \right ) + 2y ( \frac{dy}{dt} \right )

d_{o} v = x_{o} ( \frac{dx}{dt} \right ) + y_{o} ( \frac{dy}{dt} \right )

(0.5) v = (0.4) v_{A} + (0.3) v_{B}

(0.5) v = (0.4) (75) + (0.3) (70)

v = 102 km/h

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A 0.25 kg skeet (clay target) is fired at an angle of 30 degrees to the horizon with a speed of 25 m/s. When it reaches the maxi
kozerog [31]

Answer:

6.51 m and 37.13 m

Explanation:

from the question we were given

mass of skeet = 0.25 kg

speed of skeet = 25 m/s

angle = 30 degrees to the horizon

mass of pellet = 15 g

speed of pellet = 2000 m/s

without being hit by the pellet, the x and y components of the skeet velocity are  

Vx = 25 cos 30 = 21.65

Vy = 25 sin 30 = 12.5

now

V = U + (a x t)

where V = final velocity, U = initial velocity , a = acceleration, t = time and s = distance

-25 sin 30 = 25 sin 30 + (-9.8 x t)

-12.5 = 12.5 - 9.8 t

t = 2.55 s

also

V^2 = U^2 + 2as  ( s = vertical distance and V = 0 )

0 = (25 sin 30)^2 + 2 x (-9.8) x Y

19.6 Y = 156.25

Y =7.97 m

the distance traveled without the pellet hitting the skeet can be gotten using distance = speed x time

distance = 21.65 x 2.55 = 55.2 m

applying the conservation of linear momentum

on the x axis : (Ms x Us) + (Mp x Up) = (Ms x Vx) + (Mp x Vx)  ...equation 1

on the y axis :   (Ms x Us) + (Mp x Up) = (Ms x Vy) + (Mp x Vy) ...equation 2

(0.25 x 25 cos 30) + 0 = (0.25 +0.015) Vx

 Vx = 20.42m/s

0 + (0.015 x 200) = (0.25 + 0.015) Vy

 Vy = 11.32 m/s

now V^2 = U^2 + 2 as

 0 = 11.3^2 + (2 x (-9.8) x s)

s = 6.51 m                          

  • to find the extra distance moved after collision we apply

s = ut + 1/2 at^2

-7.98 = 11.32t + 1/2 (-9.8) t^2

4.9 t^2 - 11.32t + 7.98

t =  3.17 s

recall that distance = speed x time

distance = 20.42 x 3.17 = 64.73 m

the distance of the skeet before being hit would be half of the distance it travels without being hit, this is because the skeet was hit at its maximum height = 55.2 /2

= 27.6 m

therefore the extra distance traveled would be the change in distance = 64.73 -27.6 = 37.13 m

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Dafna11 [192]

Answer: -3x+13

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Scientists study how the continents move. Why might scientists use a model
tigry1 [53]

Answer:

B. It is too slow to observe directly

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They move too slow to be able to observe how they move.


I hope it helps! Have a great day!
bren~

3 0
2 years ago
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