Correct answer choice for question 1 is:
A) Figurative language
Explanation:
Figurative language uses terms of speech to be more powerful, effective, and impactful. Characters of speech such as metaphors, similes, and implications go away the usual meanings of the words to give users new acumens. On the other hand, alliterations, imageries, or echoes are symbolic tools that review the minds of the users. Figurative language can perform in various forms with the use of various literary and eloquent articles.
Correct answer choice for question 2 is :
D) Words that trigger the imagination to remember by appealing to the five senses
Explanation:
Imagery, in a literary text, is an author's use of rich and detailed language to add intensity to their work. It advances to human senses to increase the reader's knowledge of the work. Powerful kinds of comparison join all of the senses. There are seven main kinds of imagery, each resembling a sense, feeling, action, or reaction. Olfactory imagery concerns, scents, or the sense of smell. Gustatory imagery concerns tastes or the sense of taste. Tactile imagery concerns physical forms or the feeling of touch.
Answer:
48.22 kg
Explanation:
Applying the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
Note: Since both skaters were initially at rest, then their total momentum before collision is equal to zero.
And the velocity of the second skater will be in opposite direction to the first.
0 = mv+m'v'.................... Equation 1
Where m = mass of the first skater, m' = mass of the second skater, v = final velocity of the first skater, v' = final velocity of the second skater.
make v' the subject of the equation
m' = -mv/v'................. Equation 2
Given: m = 62 kg, v = 0.7 m/s, v' = -0.9 m/s (opposite direction to the first)
Substitute into equation 1
m' = -62(0.7)/-0.9
m' = 48.22 kg
Answer:
1.6s
Explanation:
Given that A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction,
To know how fast the ball will roll when it reaches the bottom of the incline, we need to calculate the acceleration at which it is rolling.
Since the frictional force is negligible, at the top of the incline plane, the potential energy = mgh
Where h = 5.2sin25
h = 2.2 m
P.E = 1.2 × 9.8 × 2.2
P.E = 25.84 j
At the bottom, K.E = P.E
1/2mv^2 = 25.84
Substitutes mass into the formula
1.2 × V^2 = 51.69
V^2 = 51.69/1.2
V^2 = 43.07
V = 6.56 m/s
Using the third equation of motion
V^2 = U^2 + 2as
Since the object started from rest,
U = 0
6.56^2 = 2 × a × 5.2
43.07 = 10.4a
a = 43.07/10.4
a = 4.14 m/s^2
Using the first equation of motion,
V = U + at
Where U = 0
6.56 = 4.14t
t = 6.56/4.14
t = 1.58s
Therefore, the time the ball rolls when it reaches the bottom of the incline is approximately 1.6s
Answer:
moving charges b)
I think this is the answer
Answer:
Explanation:
40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.
