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zvonat [6]
3 years ago
11

Just think about this.

Physics
2 answers:
diamong [38]3 years ago
8 0
This ain’t the place, bud. If you have a QUESTION, then you can post it here.
Aliun [14]3 years ago
8 0
Hi don’t listen to the other person who commented you can post whatever you want even if it isn’t a question
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A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav
gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2        (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

v^2=v_o^2+2gy

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}

With this value you can find the spring constant k from the equation (1):

mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

W=F_e\\\\mg=kx

you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

The net fire is stretched 2.72 m

5 0
3 years ago
A source for conducting formal research would include_
Wittaler [7]
C. written reports !
5 0
3 years ago
Read 2 more answers
Question 30
stellarik [79]

Answer: 0.69\°

Explanation:

The angular diameter \delta of a spherical object is given by the following formula:

\delta=2 sin^{-1}(\frac{d}{2D})

Where:

d=16 m is the actual diameter

D=1338 m is the distance to the spherical object

Hence:

\delta=2 sin^{-1}(\frac{16 m}{2(1338 m)})

\delta=0.685\° \approx 0.69\° This is the angular diameter

3 0
3 years ago
What is the speed of a car that traveled 500 meter in 30 seconds?
GenaCL600 [577]
<h2>Greetings!</h2>

To find speed, you need to remember the formula:

Speed = distance ÷ time

So plug the given values in:

500 ÷ 30 = 16.66

<h3>So the speed is 16.66m/s (metres per second)</h3>
<h2>Hope this helps!</h2>
3 0
3 years ago
A satellite of mass m is moving in a circular orbit around the earth at a constant speed v and at an altitude h above the earth'
qwelly [4]

The satellite executes a rotation motion around the earth, because Earth's force of attraction plays the role of centripetal force:

Fa=Fcp=>k*Mp*m/(Rp+r)²=mv²/(Rp+r)=>v=√(k*Mp/(Rp+r))=√(6.67*10⁻¹¹*5.98*10²⁴/(6371*10³+1000*10³))=√(39.88*10¹³/(7371*10³))=√(5.41*10⁷)=7355.53 m/s


Check the calculations again !


7 0
3 years ago
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