How fast will a bug that is going 640cm in 320 seconds travel across the

Are particularly useful in explaining deformation and strengthening in metallic materials.

vampirchik [111]

Answer: value

Explanation:

8 0

4 years ago

The number of ______ always differs in atoms of different elements

zhenek [66]

The number of protons always differs in atoms of different elements.

8 0

3 years ago

In 1780, in what is now referred to as "Brady's Leap," Captain Sam Brady of the U.S. Continental Army escaped certain death from

zysi [14]

The minimum speed with which Captain Brady had to run off the edge of the cliff to make it safely to the far side of the river is around 6 meters per second.

<h3>Further explanation</h3>

This is a free fall 2-dimensional type of problem, therefor we can write equations for both dimensions which model the fall of captain Brady. Let's call x the distance travelled by the captain on the horizontal direction and y the distance travelled on the vertical direction.

Lets suppose that Brady jumped with a complete horizontal velocity from a point which we will call the origin (meaning zero horizontal and vertical displacement), and let's call ta the time it took for captain Brady to reach the river (meaning the time he spent on the air). The equations of motion for the captain will be:

$x= V \cdot t$

$y= - \frac{g \cdot t^2}{2}$

We know that at time ta the captain would have traveled 6.7 m on the horizontal direction, and 6.1 m in the vertical direction. Therefor we can write that:

$6.7= V \cdot ta$

$-6.1= - \frac{g \cdot {ta}^2}{2}$

Which gives us a system of 2 equations and 2 unknowns (V and ta). From the second equation we can solve for ta as:

$ta = \sqrt{\frac{2 \cdot 6.1}{g}} =1.12 s$

And solving for V on the first equation, we find that:

$V= \frac{6.7}{1.12} = 5.98 \frac{m}{s}$

Which is almost 6 meters per second.

<h3>Learn more</h3>

Free fall of an arrow: brainly.com/question/1597396
Concept of free fall: brainly.com/question/1708231

<h3>Keywords</h3>

Free fall, projectile, gravity

7 0

3 years ago

Read 2 more answers

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.