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3 years ago

7

How much must a woman weigh ( force) if the pressure she exerts while standing on one foot has an area of 0.6m squared exerts a

pressure of 16 Pa

1 answer:

lions [1.4K]3 years ago

4 0

Answer:

W = 9.6 N

Explanation:

Given that,

Area on 1 foot, A = 0.6 m²

Pressure, P = 16 Pa

The pressure is given by force acting per unit area. So,

$P=\dfrac{F}{A}\\\\P=\dfrac{W}{A}\\\\W=16\times 0.6\\\\W=9.6\ N$

So, the required weight is 9.6 N.

You might be interested in

Que fuerza será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s²

WARRIOR [948]

Answer:

La fuerza que será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s² es 80 N.

Explanation:

La segunda ley de Newton, llamada ley fundamental o principio fundamental de la dinámica, plantea que un cuerpo se acelera si se le aplica una fuerza.

De esta manera, esta ley establece que las aceleraciones que experimenta un cuerpo son proporcionales a las fuerzas que recibe. Dicho de otra forma, la aceleración de un cuerpo es proporcional a la fuerza neta que se le aplica. Cuanto mayor es la fuerza que se le aplica a un objeto con una masa dada, mayor será su aceleración.

La segunda Ley de Newton se expresa matemáticamente como:

F = m*a

Donde:

F es la fuerza neta. Se expresa en Newton (N)
m es la masa del cuerpo. Se expresa en kilogramos (Kg.).
a es la aceleración que adquiere el cuerpo. Se expresa en metros sobre segundo al cuadrado (m/s²).

En este caso:

m= 20 kg
a= 4 m/s²

Reemplazando:

F= 20 kg* 4 m/s²

Resolviendo:

F= 80 N

<u><em>La fuerza que será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s² es 80 N.</em></u>

4 0

3 years ago

A 5 kg mass is accelerated at a rate of 2.5 m/s/s. what is the force that is exerted on this object

Sav [38]

Answer:

<h2>12.5 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 5 × 2.5

We have the final answer as

<h3>12.5 N</h3>

Hope this helps you

3 0

3 years ago

Can someone please help me with these physics problems? I just don’t even know where to start.

KIM [24]

#1

for the block of mass 5 kg normal force is given as

$F_n = mg$

$F_n = 5*9.8 = 49 N$

friction force is given as

$F_f = \mu F_n$

$F_f = 0.1*49 = 4.9 N$

Net force is given as

$F_{net} = ma$

$F_{net} = 5*2 = 10 N$

now we know that

$F_{net} = F_{app} - F_f$

$10 = F_{app} - 4.9$

$F_{app} = 14.9 N$

#2

Normal force is given as

$F_n = mg$

$F_n = 6*9.8$

$F_n = 58.8 N$

now we know that

$F_{net} = F_{app} - F_f$

$F_{net} = 0$

as object moves with constant velocity

$F_{app} = F_f = 15 N$

now for coefficient of friction we can use

$F_f = \mu F_n$

$15 = \mu * 58.8$

$\mu = 0.255$

#3

net force upwards is given as

$F = 1.2 * 10^{-4} N$

mass is given as

$m = 7 * 10^{-5} kg$

now as per newton's law we can say

$F = ma$

$1.2 * 10^{-4} = 7 * 10^{-5} * a$

$a = 1.71 m/s^2$

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

$F_{net} = F_{app} - F_f$

$ma = F_{app} - F_f$

$5*6 = 40 - F_{f}$

$F_f = 40 - 30 = 10 N$

part b)

for coefficient of friction we can use

$F_f = \mu F_n$

$10 = \mu * F_n$

here normal force is given as

$F_n = mg = 5*9.8 = 49 N$

now we have

$\mu = \frac{10}{49} = 0.204$

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$20 = 0 + \frac{1}{2}a(5^2)$

$a = 1.6 m/s^2$

now net force is given as

$F_{net} = ma$

$F_{net} = 10*1.6 = 16 N$

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{0 - 25}{1.5} = -16.67 m/s^2$

now the force is gievn as

$F = ma$

$F = 5*16.67 = 83.3 N$

3 0

3 years ago

A small ball with mass 2.50 kg is mounted on one end of a rod 0.750 m long and of negligible mass. The system rotates in a horiz

alina1380 [7]

Answer:

1.40625 kg-m^2

Explanation:

Supposing we have to calculate rotational moment of inertia

Given:

Mass of the ball m= 2.50 kg

Length of the rod, L= 0.78 m

The system rotates in a horizontal circle about the other end of the rod

The constant angular velocity of the system, ω= 5010 rev/min

The rotational inertia of system is equal to rotational inertia of the the ball about other end of the rod because the rod is mass-less

$I_{sys}= mL^2= 2.50\times 0.75^2$

=1.40625 kg-m^2

m= mass of the ball and L= length of the ball

3 0

3 years ago

Define the term pressure

Setler [38]

Answer:

Pressure is the perpendicular force applied per unit area.

3 0

3 years ago

Read 2 more answers