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il63 [147K]
2 years ago
8

A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the officials’ boat pulls

up next to the spy’s boat, both boats reach the edge of a 4.6 m waterfall. The spy’s speed is 17 m/s and the officials’ speed is 28 m/s. How far apart will the two vessels be when they land below the waterfall? The acceleration of gravity is 9.81 m/s 2 . Answer in units of m.
Physics
1 answer:
babunello [35]2 years ago
7 0

Answer:

They are 7.4m apart.

Explanation:

Here we have a parabolic motion problem. we need the time taken to land so:

Y=Yo+Vo*t+\frac{1}{2}*a*t^2

considerating only the movement on Y axis:

0=4.6-(9.81)*t^2\\t=0.68s

Because we have a contant velocity motion on X axis:

xs=vs*t\\xs=17m/s*(0.68s)\\xs=11.6m

and

xg=vg*t\\xg=28m/s*(0.68s)\\xg=19m

the distance between them is given by:

d=|xg-xs|\\d=7.4m

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djyliett [7]

Answer:

32 seconds

Explanation:

m1 = 80 kg

m2 = 10 kg

v2 = 5m/s

According to the property of conservation of momentum, assuming that both you and the bag are stationary before the safety rope comes lose:

m_{1} v_{1} =m_{2} v_{2} \\80v_{1} =10*5 \\v_{1} = 0.625\ m/s

Since the space station is 20 meters away, the time taken to reach it is given by:

t = \frac{20}{0.625}\\t=32\ s

It takes you 32 seconds to reach the station.

7 0
2 years ago
What qualities did Galilea, Sir Isaac Newton, and Albert Einstein have?
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3 years ago
Explain why the several different types of microscopes are all neccassary?
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Answer:

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7 0
2 years ago
In 0.601 s, a 13.1-kg block is pulled through a distance of 4.19 m on a frictionless horizontal surface, starting from rest. The
Naya [18.7K]

Answer:

0.615 m

Explanation:

We need to determine the force on the spring first. By Newton's second law of motion, force is the product of the mass and acceleration. The mass is given.

The acceleration is determined using the equation of motion.

Given parameters:

Initial velocity, <em>u</em> = 0.00 m/s

Distance, <em>s</em> = 4.19 m

Time, <em>t</em> = 0.601 s

We use the equation

s = ut+\frac{1}{2}at^2

With <em>u</em> = 0.00 m/s,

s = \frac{1}{2}at^2

a = \dfrac{2s}{t^2}

a = \dfrac{2\times4.19\text{ m}}{(0.601\text{ s})^2} = 23.2\text{ m/s}^2

The force is

F = (13.1\text{ kg})(23.2\text{ m/s}^2) = 303.92 \text{ N}

From Hooke's law, the extension, <em>e</em>, of a string is given by

e = \dfrac{F}{k}

where <em>k</em> is the spring constant.

Hence,

e = \dfrac{303.92\text{ N}}{494\text{ N/m}} = 0.615\text{ m}

4 0
3 years ago
A 300-W blender did 10,500J of work. how long was the blender running?
forsale [732]

Answer:

The blender was running for 35 seconds.

Explanation:

Given:

Power of the blender is, P=300\ W

Work done by the blender is, W=10500\ J

We know that, power is defined as the rate at which work is done by the body.

Here, we are given the power and work done and we are asked to find time taken.

Let the time of running of the blender be 't' seconds.

So, the power of the blender is given as:

P=\frac{W}{t}

Rewriting the above in terms of 't', we have

t=\frac{W}{P}

Plug in the given values and solve for 't'. This gives,

t=\frac{10500}{300}\\t=35\ s

Therefore, the blender was running for 35 seconds.

6 0
3 years ago
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