Giúp em toàn bộ nhé ............

As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o

Soloha48 [4]

Answer:

4.15 x 10^6 N

Explanation:

Area, A = 1.43 cm^2 = 1.43 x 10^-4 m^2

mass, m = 60.5 kg

Weight, F = m g = 60.5 x 9.8 = 592.9 N

Pressure = Force / Area

P = Weight / Area

P = 592.9 / (1.43 x 10^-4)

P = 4.15 x 10^6 N

3 0

3 years ago

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

During take-off a 5kg model rocket, initially at rest, burns fuel for 4.6s causing its speed to increase from rest to 39m/s duri

MrRa [10]

Answer:

The height is "89.61 m". A further explanation is given below.

Explanation:

According to the question,

Mass of rocket,

m = 5 kg

Time,

t = 4.6 s

Initial speed of rod

u = o m/s

Final speed,

v = 39 m/s

drag,

= 60 N

So,

⇒ acceleration, $a=\frac{v-u}{t}$

$=\frac{39-0}{4.6}$

$= 8.478 \ m/s^2$

⇒ $F_{net}= Net \ force$

$=ma$

$=5\times 8.478$

$=42.89 \ N$

Now,

The thrust will be:

⇒ $Thrust-weight-drag=Net \ force$

⇒ $Thrust-(5\times 9.8)-60=42.89$

⇒ $Thrust-49-60=42.89$

⇒ $Thrust-109=42.89$

⇒ $Thrust=42.89+109$

⇒ $Thrust=151.89 \ N$

The height will be:

⇒ $h=\frac{1}{2} at^2$

$=\frac{1}{2}\times 8.47\times (4.6)^2$

$=\frac{1}{2}\times 8.47\times 21.16$

$=89.61 \ m$

5 0

3 years ago

Need answer fast plz

vazorg [7]

Answer:

a = 1.20m\ $s^{2}$

Explanation:

225 x 9.8

= 441N

-

710 - 441

= 269N

-

$\frac{269}{225}$ = 1.19

a = 1.20

8 0

3 years ago

A 150kg person stands on a compression spring with spring constant 10000n/m and nominal length of 0.50.what is the total length

Ivahew [28]

Answer:

<em>The total length of the spring would be 0.65 m</em>

Explanation:

The Concept

Hooke's law evaluates the increment of spring in relation to the force acting on the body. Hooke's law states that for a spring undergoing deformation, the force applied is directly proportional to the deformation experienced by the spring. Hooke's law is represented thus;

F = k x ..................1

where F is the force applied to the spring

k is the spring constant

x is the spring stretch or extension

Step by Step Calculations

We have to obtain x before adding it to the nominal length, We make x the subject formula in equation 1;

x = F/k

but F = m x g

so, x = (m x g)/k

given that, the mass of the person m =150 kg

g is the acceleration due to gravity = 9.81 m/ $s^{2}$

k is the spring constant = 10000 N/m

then x = (9.81 m/ $s^{2}$ x 150 kg)/10000 N/m

x = 0.14715 m

the extension experienced by the spring after the compression is 0.14715 m

The total length of the spring would be;

L = 0.14715 m + 0.5 m = 0.64715

L ≈ 0.65 m

Therefore the total length of the spring would be 0.65 m

4 0

4 years ago