When an element tends to lose its valence electrons in chemical reactions, it behaves more like a

An automobile is traveling on a long, straight highway at a steady 80.0 mi/h when the driver sees a wreck 190 m ahead. At that i

kozerog [31]

Answer:

a) The car’s speed just after leaving the icy portion of the road is the first part

Explanation:

6 0

3 years ago

Name and explain 4 protein functions

Musya8 [376]

Answer:

Actin.

Arp2/3.

Collagen.

Coronin.

Dystrphin.

Elastin.

F-spondin.

Fibronectin.

Protein has many roles in your body. It helps repair and build your body's tissues, allows metabolic reactions to take place and coordinates bodily functions. In addition to providing your body with a structural framework, proteins also maintain proper pH and fluid balance.

Explanation:

5 0

3 years ago

A worker drives a 0.554 kg spike into a rail tie with a 2.30 kg sledgehammer. The hammer hits the spike with a speed of 62.7 m/s

Goryan [66]

Answer:

Increase in total energy will be equal to the increase in the internal energy i.e $1130.246$ Joules

Explanation:

Given

Weight of sledge hammer $= 2.30$ kilogram

Speed of sledge hammer $= 62.7$ meter per second

Kinetic energy is equal to half the product of mass and velocity square

$K E = \frac{1}{2} mv^2$

Substituting the value of mass and velocity, we get -

$KE = 0.5 * 2.30 * 62.7^2\\KE = 4520.9835$

It is given that one fourth of the energy is converted into internal energy

One fourth of kinetic energy is equal to

$\frac{1}{4} * 4520.9835\\= 1130.246$

Increase in total energy will be equal to the increase in the internal energy i.e $1130.246$ Joules

8 0

3 years ago

Read 2 more answers

an athlete runs 300 m up a hill at a steady speed of 3.0 m/s. She then immediately runs the same distance at 6.0 m/s . What is h

mina [271]

Answer:

4.0 m/s

Explanation:

In the first part of the run, the athlete runs a distance of

$d_1 = 300 m$

at a speed of

$v_1 = 3.0 m/s$

So, the time he/she takes is

$t_1 = \frac{d_1}{v_1}=\frac{300}{3.0}=100 s$

In the second part of the run, the athlete covers an additional distance of

$d_2 = 300 m$

with a speed

$v_2 = 6.0 m/s$

So, the time taken in this second part is

$t_2 = \frac{d_2}{v_2}=\frac{300}{6.0}=50 s$

So, the total distance covered is

d = 300 m + 300 m = 600 m

And the total time taken

t = 100 s + 50 s = 150 s

Therefore, the average speed for the entire trip is

$v=\frac{d}{t}=\frac{600}{150}=4.0 m/s$

4 0

4 years ago

A car traveling 75 km/h slows down at a constant 0.50 m/s2 just by "letting up on the gas." calculate (a) the distance the car c

Serjik [45]

Solution:

At 1st convert km/h to m/s. 1 km = 1000 m, 1 h = 3600 s, 1 km/m = 1000/3600 = 5/18 m/s

Initial velocity = 75 * 5/18 = 20.8 m/s

The car’s velocity decreases from 20.8 m/s to 0 m/s at the rate of 0.5 m/s each second. We have the final velocity, initial velocity, and the acceleration.

Now according to the equation determine the distance.

vf^2 = vi^2 + 2 * a * d

a = -0.5 m/s^2

0 = 20.8^2 + 2 * -0.5 * d

so d = 431.64 m

since we have the final velocity, initial velocity, and the acceleration. Use the following equation to determine time.

vf = vi + a * t

0 = 20.8 – 0.5 * t

Solve for t = 41 seconds

(c) the distance travels by it during the first and fifth second are.

d = vi * t + ½ * a * t^2

d1 = 20.8 * 1 – ½ * 0.5 * 1^2 = 20.55 m

The easiest way to the distance for the 5th second is:

d = vi * t + ½ * a * t^2, a = -0.5

d5 = 20.8 * 5 – ½ * 0.5 * 5^2 = 91.5 m

d6 = 20.8 * 6 – ½ * 0.5 * 6^2 = 106.8m

d6 – d5 = 15.3 m

this is the required solution.

3 0

3 years ago