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3 years ago

Which explanation describes the forces involved for a person to walk down the sidewalk?

Physics

1 answer:

Svet_ta [14]3 years ago

4 0

The person walking down the sidewalk follows the newton's third law of motion.

Explanation:

A person is able to walk down the sidewalk by using the reaction forces from the ground.
In simple term, feet pushes the ground and the reaction forces makes the feet able to walk.
Another important force included in the walking mechanism is friction. With out friction one cannot walk down the sidewalks.
Hence the forces involved in the walking of a person down the sidewalk are:

Friction force
Action and reaction force between ground and person's feet.

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Geologists apply various methods to study the layers of the Earth. Which of the following is NOT a method used to study the Eart

77julia77 [94]

Answer:

A. Scientists use seismic computer models to measure the atmospheric conditions above the Earth's crust

Explanation:

why would use atmosphere to study the layers of earth? dont think thats possible

8 0

3 years ago

Compute the density in g/cm^3 of a piece of metal that had mass of 0.485 kg and a volume of 52cm^3

steposvetlana [31]

Answer:

9.3 g/cm³

Explanation:

First, convert kg to g:

0.485 kg × (1000 g / kg) = 485 g

Density is mass divided by volume:

D = (485 g) / (52 cm³)

D = 9.33 g/cm³

Rounding to two significant figures, the density is 9.3 g/cm³.

8 0

3 years ago

A diverging lens with a focal length of 14 cm is placed 12 cm to the right of a converging lens with a focal length of 21 cm. An

IRINA_888 [86]

Answer:

-0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing

and

0.023 m right of diverging lens

Explanation:

given data

focal length f2 = 14 cm = -0.14 m

Separation s = 12 cm = 0.12 m

focal length f1 = 21 cm = 0.21 m

distance u1 = 38 cm

to find out

final image be located and Where will the image

solution

we find find image location i.e v2

so by lens formula v1 is

1/f = 1/u + 1/v ...............1

v1 = 1/(1/f1 - 1/u1)

v1 = 1/( 1/0.21 - 1/0.38)

v1 = 0.47 m

and

u2 = s - v1

u2 = 0.12 - 0.47

u2 = -0.35

so from equation 1

v2 = 1/(1/f2 - 1/u2)

v2 = 1/(-1/0.14 + 1/0.35)

v2 = -0.233 m

so -0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing

and

for Separation s = 45 cm = 0.45 m

v1 = 1/(1/f1 - 1/u1)

v1 =0.47 m

and

u2 = s - v1

u2 = 0.45 - 0.47 =- 0.02 m

v2 = 1/(1/f2 - 1/u2)

v2 = 1/(-1/0.14 + 1/0.02)

v2 = 0.023

so here 0.023 m right of diverging lens

6 0

3 years ago

an object is dropped from the top of the building through the air friction is present to the ground below how does this kinetic

olga55 [171]

They are equal. Due to the Law of Conservation of Energy

4 0

3 years ago

A 0.550 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.500 N to

Inessa [10]

(1) The effective spring constant of the system is 7.14 N/m.

(2) The maximum x-acceleration of the glider is 0.9 m/s².

(3) The x-coordinate of the glider at time t= 0.650T is 0.28 m.

(4) The kinetic energy of the glider at x=0.00 m is 0.0175 J.

<h3>The effective spring constant of the system</h3>

The effective spring constant of the system is calculated as follows;

F = kx

where;

k is spring constant

k = F/x

k = 0.5/0.07

k = 7.14 N/m

<h3>Maximum acceleration of the glider</h3>

a = ω²x

where;

ω is angular speed

ω = √k/m

ω = √(7.14/0.55)

ω = 3.6 rad/s

a = (3.6)² x 0.07

a = 0.9 m/s²

<h3>Period of the oscillation</h3>

T = 2πx/v

T = 2πx/(ωx)

T = 2π/ω

T = 2π/(3.6)

T = 1.75 seconds

t = 0.65T

t = 0.65 x 1.75

t = 1.14 seconds

x = vt

x = (ωx)t

x = (3.6 x 0.07) x 1.14

x = 0.28 m

<h3>kinetic energy of the glider</h3>

apply the principle of conservation of energy

Kinetic energy = work done by the spring

Kinetic energy = average force x distance

Kinetic energy = ¹/₂(0.5 N) x 0.07 m

Kinetic energy = 0.0175 J

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

The complete question is below:

A 0.550 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.500 N to displace the glider to a new equilibrium position, x= 0.070 m.

1. Find the effective spring constant of the system.

2. The glider is now released from rest at x= 0.070 m. Find the maximum x-acceleration of the glider.

3. Find the x-coordinate of the glider at time t= 0.650T, where T is the period of the oscillation.

4. Find the kinetic energy of the glider at x=0.00 m.

3 0

1 year ago