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

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

8 0

4 years ago

A swimmer does 3560 j of work in 55 s what is the swimmers power output?

ankoles [38]

Explanation:

Here it is given 

Work (W) = 3560 J

Time (t) = 55 sec

power (P) = ?

We know we have the formula 

 $p = \frac{w}{t}$ 

P = 3560/55

P = 64.73 watt

8 0

3 years ago

Imagine you and a friend are trying to rearrange the furniture in your classroom. You push on a desk with a force of 50 N to the

Nastasia [14]

Answer:

The net force on the desk is zero.

4 0

3 years ago

For small angles, does the pendulum's period of oscillation depend on initial angular displacement from equilibrium? Explain.

aksik [14]

Answer:

No, the pendulum's period of oscillation does not depend on initial angular displacement.

Explanation:

Given that,

For small angle, the pendulum's period of oscillation depend on initial angular displacement from equilibrium.

We know that,

The time period of pendulum is defined as

$T=2\pi\sqrt{\dfrac{l}{g}}$

Where, l = length of pendulum

g = acceleration due to gravity

So, The time period of pendulum depends on the length of pendulum and acceleration due to gravity.

It does not depend on the initial angular displacement.

Hence, No, the pendulum's period of oscillation does not depend on initial angular displacement.

6 0

4 years ago

A rocket takes off against the force of gravity. Consider this a non-isolated system. Derive the rocket equation formula relatin

Zielflug [23.3K]

Answer:

Explanation:

The solution of the question has been put in attachment.

8 0

3 years ago