All categories

3 years ago

A girl has a weight of 450 N and her feet have a total area of 0.0300 m2. Calculate the pressure her feet put on the ground.

Physics

1 answer:

GenaCL600 [577]3 years ago

4 0

Answer:

15000N/m²

Explanation:

Given parameters:

Weight = 450N

Total area = 0.03m²

Unknown:

Pressure under her feet = ?

Solution:

Pressure is the force per unit area on a body.

Pressure = $\frac{Force }{Area}$

Weight is a type of force;

Pressure = $\frac{450}{0.03}$ = 15000N/m²

You might be interested in

What is impulse? How does this relate to momentum?

lawyer [7]

Impulse is a force acting briefly on a body and producing a finite change of momentum.
This relates to momentum because impulse is a change in momentum. Impulse = momentum. Since force is a vector quantity, impulse is also a vector in the same direction. Impulse applied to an object produces equivalent vector change in its linear momentum, also in the same direction. m•(triangle)v

4 0

3 years ago

Two metal spheres are suspended from strings. The

nikdorinn [45]

Answer:

B. A repulsive force of 8.0*10^3 N.

Explanation:

As we know by Coulomb's law that the electrostatic force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1 = -2 \times 10^2 C$

$q_2 = -4 \times 10^{-8} C$

r = 3.0 m

now we have

$F = \frac{(9 \times 10^9)(2 \times 10^2)(4\times 10^{-8})}{3^2}$

$F = 8000 N$

since both charges are similar charges so they will repel each other by the force we calculated above so correct answer will be

B. A repulsive force of 8.0*10^3 N.

3 0

3 years ago

Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,

klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble= $2\hat{i}-3\hat{k}$ N

Position vector= $r=0.5\hat{j}-2\hat{k}$ m

a.We have to find the resulting torque on the pebble about origin.

Torque= $r\times F$

Substitute the values then we get

$Torque= (0.5j-2k)\times (2i-3k)$

Torque= $-k-1.5i-4j$ N-m

By using $i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0$

b. $r=2i-3k$

$r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k$

Torque about point (2,0,-3)

$\tau=(-2i+0.5j+k)\times (2i-3k)$

$\tau=-6j-k-1.5i+2j=-1.5i-4j-k$ N-m

6 0

3 years ago

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ $\frac{du}{dy}$

= µ $\frac{v}{h}$ ............1

put here value

= 1.75× $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

and

area between air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

and now apply newton second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0

3 years ago

A 31.0 kg crate, initially at rest, slides down a ramp 2.6 m long and inclined at an angle of14.0° with the horizontal. Using th

mart [117]

The answer is 3.5m/s

8 0

3 years ago