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2 days ago

Two wires with equal lengths are made of pure copper. The diameter of wire A is three times the diameter of wire B. When 8 kg ma

sses are hung on the wires, wire B stretches more than wire A. You make careful measurements and compute Young's modulus for both wires. What do you find?a. YA > YBb. YA = YBc. YA < YB

Physics

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Two trains are headed towards each other on the same track unbeknownst to the engineers. One departs San Francisco. Its average

ValentinkaMS [3465]

Answer:

7.166 hours = 430 minutes.

Explanation:

As both trains are approaching each other on the same track, their relative speed is the sum of their individual speeds. Hence, the time until they intersect (and inevitably collide) is determined by how long it takes for speeds of 65 mph and 55 mph to cover the total distance of 860 miles. One train will cover part of the distance, while the other will cover the remainder. To calculate the required time, we can apply the formula:

1 hour ---> 120 miles

X ----> 860 miles; hence X = (860 miles * 1 hour)/120 miles = 43/6 hours = 7.16666 hours. To convert this into minutes, recall that 1 hour equals 60 minutes; therefore, 43/6 hours * 60 minutes/hour = 430 minutes.

7 0

3 months ago

A rocket takes off vertically from the launch pad with no initial velocity but a constant upward acceleration of 2.25 m/s^2. At

kicyunya [3294]

Answer:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

S = ut + $\frac{1}{2}$ a $t^{2}$

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + $\frac{1}{2}$ (2.25 x $15.4^{2}$ )

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s² (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

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

$0^{2}$ = $34.65^{2}$ + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m + 61.19 m = 328 m

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying $v^{2}$ = $u^{2}$ + 2gs

$v^{2}$ = $0^{2}$ + 2(9.81 x 328)

$v^{2}$ = 0 + 6435.36

v = $\sqrt{6435.36}$ = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec

7 0

3 months ago