Describe the motion of a person who is standing on the earths surface

marshall27 [118]

the definition of acceleration in kinematics, allows to find that the correct answer is:

2. speed up ( acceleration)

The kinematics study the movement of the body, the acceleration is defined as the change of the speed by the time in the interval

a = Δv / t

Where the bold letters indicate vectors, a is the acceleration, v the velocity and t the time

Analyzing this expression we see that for there to be a change in velocity, there must be an acceleration of the body.

Let's analyze the different claims

1. True. If it is stopped and you start to move there is an acceleration, therefore there is a change in speed, but after you are moving the acceleration becomes zero and there is no change in speed

2. True. Whenever there is acceleration there is a change in speed

3. False. Moving slowly does not change the acceleration, therefore there is no change in speed

4. True. If you are moving and you stop at this moment there is an acceleration, therefore there is a change in speed, but after being stopped the acceleration is zero and there is no change in speed

5. False. If you change the direction at the instant of change there is an acceleration but after you go in the opposite direction there is not acceleration therefore there is no change in speed.

In conclusion using the definition of acceleration in kinematics, we can find the answer the condition for a change in velocity, the correct statement is:

2. speed up ( acceleration)

Learn more here: brainly.com/question/5063616

3 0

2 years ago

A 1500 kg elevator, suspended by a single cable with tension 16.0 kN, is measured to be moving upward at 1.2 m/s. Air resistance

kirill115 [55]

Tension in the Cable is 0.87 m/s6^2, Elevator's speed after it has moved 10m is 1.6*10^5 N, Work done by gravity is 1.47 * 10^3 N, Elevator Kinetic Energy is 13897.5J and Elevator Speed after rising to 10m is 4.312 m/s.

Tension is a pulling force that operates in one dimension along the cables' axes in the opposite direction from the direction of the applied force. The combined weight of the elevator box and the passenger riding inside it, in the case of an elevator, provides the pulling force in the cables is called Tension.

A moving object or particle's kinetic energy, which depends on both mass and speed, is one of its characteristics. The type of motion can be vibration, translation, rotation around an axis, or any combination of these. Kinetic energy is a type of energy that an item or particle possesses as a result of motion.

We know that,

Tension in the Cable

T = m(g+a) = g+a = T/m = 16 * 103 / 1500 = 10.67 m/s2

a = 10.67 - 9.8 = 0.87 m/s6^2

Elevator's speed after it has moved 10m.

U^2 = u^2 +2as

= 1.22 +2*0.87*10

=1.6*10^5 N

Work done by gravity = mg * 10 = 14700 * 10 = 1.47 * 10^3 N

Elevator Kinetic Energy = 1/2 mv^2 = 1/2*1500*18.53 = 13897.5J

Elevator Speed after rising to 10m ,

U^2 = u^2 +2as = 1.2 +2*0.87*10 = 18.6

U =(18.6)^1/2=4.312 m/s

Learn more about Kinetic Energy here

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6 0

1 year ago

A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr

nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

a= μ g ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g =ω²X

We know for spring mass system natural frequency given as

$\omega=\sqrt{\dfrac{k}{M+m}}$

By putting the values

$\omega=\sqrt{\dfrac{150}{5.67+1}}$

ω = 4.47 rad/s

μ g =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

$\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2$

$kX^2=(m+M)v^2$

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0

3 years ago

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HELP ASAP Which statement describes the water particles when condensation takes place?

Vladimir [108]

The answer is going to be B

6 0

3 years ago

Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of

max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

$Q =\frac{KA(\delta T)}{L}$

$Q =\frac{25.87*1*20}{1}$

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0

3 years ago

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