Power (rate of using energy) = (voltage) x (current)
Power of this heating element = (120 V) x (3.2 A) = 384 watts
The easy way:
384 watts = 0.384 kilowatt
(0.384 kilowatt) x (5 hours) = 1.92 kilowatt-hour
Another way:
384 watts = 384 joules per second .
(384 joule/sec) x (3600 sec/hour) x (5 hours)
= (384 x 3600 x 5) (joules) = 6,912,000 joules.
<span> Let,
initial velocity = v m/sec,
Angle, x = 40 degrees,
horizontal-componant = v.cos(x) = 12 m/sec,
OR,
v = 12 / cos(40) = 12/0.766 = 15.67 meters/sec >================< ANSWER </span> Source(s): Fazaldin A <span> · 4 years ago </span>
First, the infinitive phrase must contain the word "to", so the only options are A and D.
In D "to" is a preposition: it specifies the spacial relation: to the grocery store.
The infinitive phrase is "to go" and the answer is A.
<span />
Explanation:
The liquid contains only one element. -The liquid is a pure substance. The number at the end of an isotope's name is the -mass number. While looking at xenon (Xe) on the periodic table, a student needs to find an element with a smaller atomic mass in the same group.
Answer:

Explanation:
As we know that ball travels horizontal distance of 24.7 m with uniform speed 49.4 m/s
so we will have



now in the same time ball is turned by angle

now we know that



now the tangential speed of a point at equator is given as


