All categories

3 years ago

Objects fall at constant velocity

2 answers:

5 0

No they don't. Without air resistance, objects fall with constant
acceleration.

On Earth, the acceleration is 9.8 m/s² ... the speed of a falling object
at any time is 9.8 m/s faster than it was 1 second earlier.

ch4aika [34]3 years ago

5 0

False,
Object do not intend to fall at a constant velocity.

You might be interested in

What is the concentration of H^ + at apH = 2 ? Mol / L What is the concentration of H^ + ions at apH = 6 ? Mol/L How many more H

Nonamiya [84]

Answer:

The concentration of hydrogen ion at pH is equal to 2 : $= [H^+]=0.01 mol/L$

The concentration of hydrogen ion at pH is equal to 6 : $[H^+]'=0.000001 mol/L$

There are 0.009999 more moles of $H^+$ ions in a solution at a pH = 2 than in a solution at a pH = 6.

Explanation:

The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.

$pH=-\log [H^+]$

The hydrogen ion concentration at pH is equal to 2 = [H^+]

$2=-\log [H^+]\\$

$[H^+]=10^{-2}M= 0.01 M=0.01 mol/L$

The hydrogen ion concentration at pH is equal to 6 = [H^+]

$6=-\log [H^+]\\\\$

$[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L$

Concentration of hydrogen ion at pH is equal to 2 = $[H^+]=0.01 mol/L$

Concentration of hydrogen ion at pH is equal to 6 = $[H^+]'=0.000001 mol/L$

The difference between hydrogen ion concentration at pH 2 and pH 6 :

$= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L$

Moles of hydrogen ion in 0.009999 mol/L solution :

$=0.009999 mol/L\times 1 L=0.009999 mol$

There are 0.009999 more moles of $H^+$ ions in a solution at a pH = 2 than in a solution at a pH = 6.

8 0

3 years ago

Jack has two boxes. One is 148g and one is 78g. If jack pushes both boxes with the same amount of force, which will accelerate f

Step2247 [10]

The 78g box, since it has less weight, would accelerate faster. If you had a frictionless surface, and you conducted this experiment, both boxes, without any outside forces, would accelerate at the same rate forever. However, in this problem we must assume the surface is not frictionless. Friction is determined by weight; the more weight, the more friction. Since the 78g box has less weight, it has less friction, making it easier to push with less force.

8 0

4 years ago

A drowsy cat spots a flowerpot that sails first up and then down past an open window. the pot was in view for a total of 0.49 s,

Alika [10]

For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. 

An initial condition has been supplied by the problem:

s=1.80m when t=0.245s

This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2)

to solve for v, the initial velocity of the pot as it enters the cat's view through the window. Substituting and solving (note that gravitational acceleration is negative since this is opposite our coordinate orientation):

(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2

v=8.549m/s

Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

(v final)=(v initial)+(a)(t)

to solve for the time it will take for the pot to reach the apex of its flight. Because (v final)=0, this equation will look like

0=(v)+(a)(t)

Substituting and solving for t:

0=(8.549m/s)+(-9.81m/s^2)(t)

t=0.8714s

Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2)

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

s=3.725m

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window:

3.725m – 1.8m=1.925m

Answer:

1.925m

3 0

3 years ago

Identify the forces acting on the cars and explain why the cars do not slide down the hill

Dvinal [7]

Examples of outside forces acting on a car is gravity, wind, and other cars. Cars do not slide down hills because their weight, combined with the friction of their tires against the road, hold them in place.

5 0

4 years ago

What best describes the motion of the medium particles in a longitudinal wave? The medium does not move. The medium moves in all

Citrus2011 [14]

In the motion of the medium particles in a longitudinal wave, the medium vibrates parallel to the direction of the wave.

<h3>What is a longitudinal wave?</h3>

A longitudinal wave is a wave that is transversing along the length. When the displacement of medium and travel of wave is the same in that condition wave is known as the longitudinal wave.

It requires some medium to travel. A mechanical and sound wave is an example of a longitudinal wave.

Hence in the motion of the medium particles in a longitudinal wave, the medium vibrates parallel to the direction of the wave.

To learn more about the longitudinal wave refer to the link;

brainly.com/question/8497711

3 0

2 years ago