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Rasek [7]
3 years ago
14

Pls help i’ll give brainliest if you give a correct answer!!

Physics
2 answers:
barxatty [35]3 years ago
5 0

Answer:

third option is the most accurate

arsen [322]3 years ago
3 0

Answer:

C. The distance traveled by an object at a certain velocity.

Explanation:

YW!

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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
Particles in which state of matter are the most likely to interact with each other to cause a chemical reaction?
Mazyrski [523]
I think when particles are in the gas form they are most likely to cause a chemical reaction considering the fact that gas particles aren't very controllable and it would make sense that they would react unexpectedly
7 0
2 years ago
A Venturi tube may be used as a fluid flowmeter. Suppose the device is used at a service station to measure the flow rate of gas
leonid [27]

Answer

given,

flow rate = p = 660 kg/m³

outer radius = 2.8 cm

P₁ - P₂ = 1.20 k Pa

inlet radius = 1.40 cm

using continuity equation

 A₁ v₁ = A₂ v₂

 π r₁² v₁ = π r₁² v₂

 v_1= \dfrac{r_1^2}{r_2^2} v_2

 v_1= \dfrac{1.4^2}{2.8^2} v_2

 v_1= 0.25 v_2

Applying Bernoulli's equation

 \Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)

 \Delta P = \dfrac{1}{2}\rho (v_2^2-(0.25 v_2)^2)

 \Delta P = \dfrac{1}{2}\rho v_2^2 (1 - 0.0625)

 v_2=\sqrt{\dfrac{2\Delta P}{\rho(1 - 0.0625)}}

 v_2=\sqrt{\dfrac{2\times 1200}{660 \times(1 - 0.0625)}}

       v₂ = 1.97 m/s

b) fluid flow rate

Q = A₂ V₂

Q = π (0.014)²  x 1.97

Q = 1.21 x 10⁻³ m³/s

5 0
3 years ago
Hello, my younger sister is stuck on this and I just got on here to ask this question. I have my original account but I just mad
nignag [31]

Answer:

Hi there!

I'm sorry your sister is struggling!

I am an edgenuity student in grade 11, I could probably help!

A trend is a line on a graph that all points seem to follow. This is best explained when thinking about line of best fit. If all your points go upward each time and are closed together, we can find the line that gets closest to each point!

Think of a trend as a pattern. The line of best fit, created by analyzing the trends, helps us guess at what the data points beyond what we have will equal

Is there any clarification she needs?

4 0
3 years ago
An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?
NemiM [27]

1. Define Newtons second law of motion (this will help put things into perspective)

2.Get the mass of the object (in this case 75 kg)

3.The net force acting on the object...find it (in this case 500 N)

4.Change the equation to F=ma (500=75a)

5.Divide both sides by 75 and that is the acceleration.

7 0
3 years ago
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