All categories

1 year ago

The positions of four objects as a function of time are shown

Physics

1 answer:

KengaRu [80]1 year ago

6 0

There is no movement in line C and the greatest velocity occurs at line D. The answers are:

1. 0.5 m/s

2. 0.25 m/s

3. 14m and -2m

4. -1 m/s

<h3>What is Position - time Graph ?</h3>

Position time graph is the graph of distance or displacement against time. The slope of the graph is velocity.

The given positions of four objects as a function of time are shown

on the graph to the right.

1.) The velocity of object A will be the slope m of the line A.

Slope m = Δx / Δt

m = (4 - 0) / (8 - 0)

m = 4 / 8

m = 0.5 m/s

Velocity at A = 0.5 m/s

2.) The average velocity of object B will be the slope m of the line B.

Slope m = Δx / Δt

m = (6 - 4) / (8 - 0)

m = 2 / 8

m = 0.25 m/s

The average velocity of object B is 0.25s

3.) The object moved a total distance during the first eight seconds will be 4m for A, 2m for B, and 8m for D

Total distance = 4 + 2 + 8 = 14m

It’s net displacement during the same time will be 2. That is,

Displacement = 8 - 6 = -2m

4.) The greatest speed occurred at line D. The velocity of the object moving at the greatest speed will be the slope of the line D

V = -Δx / Δt

V = -8/8

V = -1 m/s

Therefore, there is no movement in line C and the greatest velocity occurs at line D.

Learn more about velocity time graph here :brainly.com/question/769606

#SPJ1

You might be interested in

A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho

Aneli [31]

Answer:

h=2.86m

Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

$\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f$

There is no kinetic energy in the initial state, nor potential energy in the end,

$mgh+0=0+KE_f$

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

$mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)$

The inertia of the bodies is given by the equation,

$I=\frac{m(R_1^2+R^2_2)}{2}$

$I=\frac{2(0.2^2+0.1^2)}{2}$

$I=0.05Kgm^2$

On the other hand the angular velocity is given by

$\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s$

Replacing these values in the equation,

$(0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2$

Solving for h,

$h=2.86m$

5 0

3 years ago

Gravity on earth is 9.8 m/s squared, and gravity on the moon is 1.6 m/s squared. So if the mass of an object on earth is 40 kilo

garik1379 [7]

The mass of an object on Earth is the same as its mass on the Moon. The weight is different.

Weight = m * g

Weight ( Moon ) = 40 kg * 1.6 m/s² = 64 N

If the mass of an object on Earth is 40 kg, its mass on the Moon is 40 kg and its weight on the Moon is 64 N.

7 0

3 years ago

a circular cylinder and isused to maintain a water depth of 4 m. That is, when the water depth exceeds 4 m, thegate opens slight

stiv31 [10]

Answer:

W / A = 39200 kg / m²

Explanation:

For this problem let's use the equilibrium equation of / newton

F = W

Where F is the force of the door and W the weight of water

W = mg

We use the concept of density

ρ = m / V

m = ρ V

The volume of the water column is

V = A h

We replace

W = ρ A h g

On the other side the cylinder cover has a pressure

P = F / A

F = P A

We match the two equations

P A = ρ A h g

P = ρ g h

P = 39200 Pa

The weight of the water column is

W = 1000 9.8 4 A

W / A = 39200 kg / m²

3 0

3 years ago

Help with speed problems #1-3

-Dominant- [34]

1. Each plot represents the meters traveled by both the Hare and the Tortoise over a certain period of time (minutes).

2. The Tortoise lines show it lines is steadily increasing over a period of time. So as more time elapses the faster the tortoise becomes it travels more meters. The Tortoise line shows steady acceleration.

3. The Hare in the first 5 minutes had a rapid fast advancement up to 40 meters. But for the 5-20 mins. period the Hare did not move at all. Its speed stayed at the same place. But towards the end 20-25 mins. marks the Hare started moving again. At the end the Hare at first had a rapid acceleration but stopped for a long time then it sped up briefly.

8 0

3 years ago

A driver with a 0.80-s reaction time applies the brakes, causing the car to have acceleration opposite the direction of motion.

jeka94

Answer:

a) During the reaction time, the car travels 21 m

b) After applying the brake, the car travels 48 m before coming to stop

Explanation:

The equation for the position of a straight movement with variable speed is as follows:

x = x0 + v0 t + 1/2 a t²

where

x: position at time t

v0: initial speed

a: acceleration

t: time

When the speed is constant (as before applying the brake), the equation would be:

x = x0 + v t

a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:

x = 0m + 26 m/s * 0.80 s = <u>21 m </u>

b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):

v = v0 + a* t

0 = 26 m/s + (-7.0 m/s²) * t

-26 m/s / - 7.0 m/s² = t

t = 3.7 s

With this time, we can calculate how far the car traveled during the deacceleration.

x = x0 +v0 t + 1/2 a t²

x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>

4 0

3 years ago