All categories

3 years ago

a 60 kg woman in a elevator is accelerating upward at a rate of 1.2 m/s2. What is the gravitational force acting upon the woman?

Physics

1 answer:

Flura [38]3 years ago

5 0

The gravitational force acting upon the woman is equal to <u>-588.6N</u>

Why?

To solve the problem, we need to consider that two forces are acting upon the woman, the first one is related to her weight and the second one is related to the acceleration of the elevator.

Gravitational force acting upon the woman:

$Force=m*gravityacceleration\\\\Force=60kg*-9.81\frac{m}{s^{2}}=-588.6N$

Hence, we have that the gravitational force acting upon the woman is equal to -588.6N.

Have a nice day!

You might be interested in

An airplane accelerates down a runway at 3 m/s2 for 32 s until is finally lifts off the ground. Determine the distance traveled

Alex Ar [27]

Answer:

12 or 24

Explanation:

i think it is i hope it is right

4 0

3 years ago

What is the maximum speed when the conditions are mass =450 kg, initial height= 30 m, and the roller coaster is initially at res

Zarrin [17]

Answer:

B. 24.2 m/s

Explanation:

Given;

mass of the roller coaster, m = 450 kg

height of the roller coaster, h = 30 m

The maximum potential energy of the roller coaster due to its height is given by;

$P.E_{max} = mgh\\\\PE_{max} = 450 *9.8*30\\\\PE_{max} = 132,300 \ J$

$P.E_{max} = K.E_{max} \ (law \ of \ conservation\ of \ energy)$

$K.E_{max} = \frac{1}{2}mv_{max}^2\\\\ v_{max}^2 = \frac{2K.E_{max}}{m}\\\\ v_{max}^2 = \frac{2*132300}{450}\\\\ v_{max}^2 =588\\\\v_{max} = \sqrt{588}\\\\ v_{max} = 24.2 \ m/s$

Therefore, the maximum speed of the roller coaster is 24.2 m/s.

3 0

3 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

3 years ago

A woman with a mass of 52.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of 0.470 rev/s abo

Charra [1.4K]

Explanation:

It is given that,

Mass of the woman, m₁ = 52 kg

Angular velocity, $\omega=0.47\ rev/s=2.95\ rad/s$

Mass of disk, m₂ = 118 kg

Radius of the disk, r = 3.9 m

The moment of inertia of woman which is standing at the rim of a large disk is :

$I={m_1r^2}$

$I={52\times 3.9^2}$

I₁ = 790.92 kg-m²

The moment of inertia of of the disk about an axis through its center is given by :

$I_2=\dfrac{m_2r^2}{2}$

$I_2=\dfrac{118\times (3.9)^2}{2}$

I₂ =897.39 kg-m²

Total moment of inertia of the system is given by :

$I=I_1+I_2$

$I=790.92+897.39$

I = 1688.31 kg-m²

The angular momentum of the system is :

$L=I\times \omega$

$L=1688.31 \times 2.95$

$L=4980.5\ kg-m^2/s$

So, the total angular momentum of the system is 4980.5 kg-m²/s. Hence, this is the required solution.

8 0

3 years ago

Technician A says crimping solderless connections is the best method of splicing wires. Technician B says crimping should not be

algol13

Answer:

Technician b is correct.

Explanation:

Crimping cable allows a firm connection in mechanical terms and allows a low resistance path for the signal or the current flow, solder although it is better in terms of electrical conduction, can be impractical if the cable is subjected to excessive movement.

A crimped cable with excessive movement can also be easily broken at the ends, where it joins the part of the cable that is crimped, for this reason, a cable that is in excessive motion is recomended to be spliced by joining cable with cable .

In order to decide which metod is better for splicing cables its necessary to evaluate each situation separatly.

7 0

3 years ago