All categories

3 years ago

Consider the following equations of motion.

1 answer:

6 0

A) No, the equations presented above are the product of the derivation of position and velocity when the acceleration is constant.

The equations change to polynomial function of the second degree for the description of the acceleration when described as a function of time.

B) Yes, when the acceleration is zero it is concluded that the velocity is constant, therefore they could be used to describe the position as a function of the change in velocity.

You might be interested in

WILLL GIVE 5 STARS BRAINIEST AND THANKS AND 20 POINTS EACH ANSWER!!!!

vekshin1

Answer:

The time she takes to reach the water from when she jumps off the platform is 1.71 s

Explanation:

According to the equations of motion, we have;

v = u - g·t

v² = u² - 2·g·s

s₁ = u₁·t₁ + 1/2·g₁·t₁²

The given parameters are;

The height of the platform (assumption: above the water) = 10 m

The velocity with which she jumps, u = 3 m/s

The acceleration due to gravity, g = 9.81 m/s²

The height to which she jumps, s, is found as follows;

v² = u² - 2·g·s

At maximum height, v = 0, which gives;

0 = 3² - 2×9.81×s

2×9.81×s = 3² = 9

s = 9/(2×9.81) = 0.4587 m

s = 0.4587 m

The time to maximum height, t, is found as follows;

v = u - g·t

0 = 3 - 9.81×t

9.81×t = 3

t = 3/9.81 = 0.3058 s

The total distance, s₁ from maximum height to the water surface = s + 10 = 0.4587 + 10 = 10.4587 m = 10.46 m

The time to reach the water from maximum height, t₁, is found as follows;

s₁ = u₁·t₁ + 1/2·g₁·t₁²

Where;

s₁ = The total distance from maximum height to the water surface = 10.46 m

u₁ = The initial velocity, this time from the maximum height = 0 m/s

g₁ = The acceleration due to gravity, g (positive this time as the diver is accelerating down) = 9.81 m/s²

t₁ = The time to reach the water from maximum height

Substituting the values gives;

s₁ = u₁·t₁ + 1/2·g₁·t₁²

10.46 = 0·t₁ + 1/2·9.81·t₁²

t₁²= 10.46/(1/2×9.81) = 2.13 s²

t₁ = √2.13 = 1.46 s

Total time = t₁ + t = 1.46 + 0.3058 = 1.7066 ≈ 1.71 s.

Therefore, the time she takes to reach the water from when she jumps off the platform = 1.71 s.

8 0

3 years ago

What is 902 in proper scientific notation?

Julli [10]

Hope it help you 9.02x10^2

3 0

3 years ago

A rotating object has an angular acceleration of α = 0 rad/s2. Which one or more of the following three statements is consistent

Murrr4er [49]

Answer:

A,B and C

Explanation:

Statement A

At all times, angular velocity is $\omega = 0\,{\rm{rad/s}$

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 0\,{\rm{rad/s}}$

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero hence statement A is valid.

Statement B

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 10\,{\rm{rad/s}}$ .The final and initial velocities remain the same.

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{10\,{\rm{rad/s}} - 10\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement B is valid

Statement C

Angular velocity is defined as the change in the angular position with respect to time.

Angular velocity and angular displacement are related by

$\theta = \omega t$

Which can also be modified as:

${\theta _{\rm{f}}} - {\theta _{\rm{i}}}$

Note that the final position is ${\theta _{\rm{f}}}$ and initial position is ${\theta _{\rm{i}}}$

Modifying the equation to find the angular velocity we obtain

$\omega = \frac{{{\theta _{\rm{f}}} - {\theta _{\rm{i}}}}}{t}$

When the angular displacement has the same value at all times, the equation becomes

$\begin{array}{c}\\\omega = \frac{{{\theta _{\rm{i}}} - {\theta _{\rm{i}}}}}{t}\\\\ = 0\\\end{array}$

The angular velocity becomes zero.

Angular acceleration and angular velocity are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

The expression above can be rearranged as follows:

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

At all times, the angular velocity is $\omega = 0\,{\rm{rad/s}}$ hence initial and final velocities remain the same

We obtain

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement C is valid.

Therefore, statements A,B and C are consistent .

4 0

3 years ago

A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N

Sergio [31]

Answer:

200cm

Explanation:

Answer:

100cm

Explanation:

Using

F= ( N/2L)(√T/u)

F1 will now be (0.5*2)( √600/0.015)

=> L( wavelength)= 200/2cm = 100cm

7 0

3 years ago

A 7.00-g bullet, when fired from a gun into a 1.00-kg block of wood held in a vise, penetrates the block to a depth of 8.00 cm.

gulaghasi [49]

Answer:

In the second case there's no way to know what depth the bullet will penetrate into the block.

Explanation:

Since the block is on a <em>frictionless </em>surface, when hitted by the bullet, this last one could barely penetrate the surface of the block and, both, start moving as one (<em>perfectly inellastic collition</em>) since here, there's no vise to hold the block into place.

4 0

3 years ago