Can someone answer pls??

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

What is the distance between a 900 kg compact car and a 1600 kg pickup truck if the gravitational force between them is about 0.

elena-14-01-66 [18.8K]

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

$r= \sqrt{\frac{Gm1m2}{F} }$

$r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m$

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required:

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

$r= \sqrt{\frac{Gm1m2}{F} }$

$r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m$

3 0

3 years ago

Read 2 more answers

What is the next step if the data from an investigation do not support the original hypothesis? A. The data are revised to suppo

Leno4ka [110]

I believe the answer is D. The hypothesis is revised and another experiment is conducted.

5 0

4 years ago

Read 2 more answers

Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a

Verizon [17]

Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.

Reasons:

The energy given to the block by the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = $0.5 \cdot k \cdot x^2$ = kinetic energy of

block = $0.5 \cdot m \cdot v^2$

Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

The force of the weight of the block on the string, $F = m \cdot g \cdot sin(\theta)$

The energy given to the block = $0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)$ = The kinetic energy of block as it leaves the spring = $\mathbf{0.5 \cdot m \cdot v^2}$

Which gives;

$0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2$

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and c are constants

The graph of the equation a·x² + c·v² = b is an ellipse

Therefore;

As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.

Please find attached a drawing related to the question obtained from a similar question online

The possible question options are;

As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x

The relationship is no longer linear and v will be more for the same value of x

The relationship is still linear, with lesser value of v

The relationship is still linear, with higher value of v

The relationship is still linear, but vary inversely, such that as x increases, v decreases

Learn more here:

brainly.com/question/9134528

6 0

2 years ago

In normal light conditions, how well do plants grow when there are 10 plants in the

Sati [7]

Answer

when there are ten they don't grow so well but when there is less than 10 they tend to grow

7 0

3 years ago

Can someone answer pls??​

Can someone answer pls??