It's usually taken from <span>400 to 700 nm.</span>
Answer:
Polar covalent bond
Explanation:
When two atoms of unequal electro- negativity comes together to form bonds , there is unequal sharing of electrons between them , one of them acquaring partial positive and partial negative charge . Then they make bond due to mutual attraction.
Bond formation in HCl in gaseous state is an example .
From the average velocity calculated, if the car should travel 363.0 miles, the time elapsed will be 7.6 hours.
<h3>How long will it take to travel 363.0 miles at the given average velocity? </h3>
Note that; Velocity is the speed at which an object moves in a particular direction. It is expressed as;
v = distance / time
Given that;
- Distance covered by the car d = 263.0 miles
- Time elapsed t = 5.5 hours
First we determine the velocity of the car.
v = distance / time
v = 263.0 miles / 5.5 hours
v = 47.8181 miles per hour.
Now, if the car travels 363.0 miles, time elapsed will be;
velocity = distance / time
time = distance / velocity
time = 363.0 miles / 47.8181 miles per hour
time = 7.6 hours
Therefore, from the average velocity calculated, if the car should travel 363.0 miles, the time elapsed will be 7.6 hours.
Learn more about speed here: brainly.com/question/7359669
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Answer:
2.726472 s more or 1.5874 times more time is taken than 10-lb roast.
Explanation:
Given:
- The cooking time t is related the mass of food m by:
t = m^(2/3)
- Mass of roast 1 m_1 = 20 lb
- Mass of roast 2 m_2 = 10 lb
Find:
how much longer does a 20-lb roast take than a 10-lb roast?
Solution:
- Compute the times for individual roasts using the given relation:
t_1 = (20)^(2/3) = 7.36806 s
t_2 = (10)^(2/3) = 4.641588 s
- Now take a ration of t_1 to t_2, to see how many times more time is taken by massive roast:
t_1 / t_2 = (20 / 10)^(2/3)
- Compute: t_1 / t_2 = 2^(2/3) = 1.5874 s
- Hence, a 20-lb roast takes 1.5874 times more seconds than 10- lb roast.
t_2 - t_1 = 2.726472 s more
