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1 year ago

9

a 160 kilogram space vehicle is traveling along a straight line at a constant speed of 800 meters per second. The magnitude of t

he net force on the space vehicle is?

1 answer:

Cloud [144]1 year ago

5 0

Anything that's moving in a straight line at a constant speed has zero acceleration, and that tells us that there is zero net force acting on it.

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A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the

lions [1.4K]

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

3 0

2 years ago

Noise-canceling headphones are an application of destructive interference. Each side of the headphones uses a microphone to pick

exis [7]

The question for this problem would be the minimum headphone delay, in ms, that will cancel this noise.
The 200 Hz. period = (1/200) = 0.005 sec. It will need to be delayed by 1/2, so 0.005/2, that is = 0.0025 sec. So converting sec to ms, will give us the delay of:Delay = 2.5 ms.

4 0

1 year ago

The position of a particle moving along the x axis may be determined from the expression x(t) = btu + ctv, where x will be in me

KIM [24]

As per given equation we have

$x = bt^u + ct^v$

now as per the dimensional analysis we can say that dimension of right side of equation must be equal to left side of the equation

now as per left side of equation its dimension is same as length or meter

now we can say it should be meter on right side also

$bt^u = M^0L^1T^0$

$b*T^8 = M^0L^1T^0$

$b = M^0L^1T^{-8}$

similarly for other term we have

$ct^v = M^0L^1T^0$

$c*T^7 = M^0L^1T^0$

$c = M^0L^1T^{-7}$

<em>so above are the dimensions of b and c</em>

8 0

1 year ago

Aaron Agin nodded off while driving home from play practice this past Sunday evening. His 1500-kg car hit a series of guardrails

Inessa [10]

Answer: 6.48m/s

Explanation:

First, we know that Impulse = change in momentum

Initial velocity, u = 19.8m/s

Let,

Velocity after first collision = x m/s

Velocity after second collision = y m/s

Also, we know that

Impulse = m(v - u). But then, the question said, the guard rail delivered a "resistive" impulse. Thus, our impulse would be m(u - v).

5700 = 1500(19.8 - x)

5700 = 29700 - 1500x

1500x = 29700 - 5700

1500x = 24000

x = 24000/1500

x = 16m/s

Also, at the second guard rail. impulse = ft, so that

Impulse = 79000 * 0.12

Impulse = 9480

This makes us have

Impulse = m(x - y)

9480 = 1500(16 -y)

9480 = 24000 - 1500y

1500y = 24000 - 9480

1500y = 14520

y = 14520 / 1500

y = 9.68

Then, the velocity decreases by 3.2, so that the final velocity of the car is

9.68 - 3.2 = 6.48m/s

5 0

1 year ago

Among the largest passenger ships currently in use, the Norway has been in service the longest. The Norway is more than 300 m lo

LenaWriter [7]

Answer:

$6.33\times 10^8\ kg\cdot m/s$

Explanation:

Mass of the ship (m) = 6.9 × 10⁷ kg

Speed of the ship (v) = 33 km/h

First, let us convert the speed from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

So, 33 km/h = $33\times \frac{5}{18}=9.17\ m/s$

Now, we know, the momentum of an object only depends on its mass and speed. Momentum is independent of the length of the object.

So, here, length of the ship doesn't play any role in the determination of the momentum.

Magnitude of momentum of the ship = Mass × Speed

= $(6.9\times 10^7\ kg)(9.17\ m/s)$

= $6.33\times 10^8\ kg\cdot m/s$

Therefore, the magnitude of ship's momentum is $6.33\times 10^8\ kg\cdot m/s$ .

6 0

2 years ago