Answer:
<u>FALSE.</u>
Explanation:
Newton's third law states that :
- <em>Every action has equal and opposite reaction</em>
- <em>That is , the magnitude is the same but the directions are opposite</em>
- <em>The action reaction forces DONOT operate on the same body.</em>
For example ,
If a block is kept on the ground , the action force is the normal force acting on it due to the ground. <em>BUT , NOTE THAT : the reaction force isn't the gravitational force on the body ! It is the normal force acting on the ground due to the block !</em>
Thus,
we conclude that action and reaction forces donot act on the same body and therefore , this case has the <u>answer : FALSE </u>
Answer:
A 30 lb weight is attached to the end of a spring. The spring is stretched 6 in. Find the equation of motion if the weight is released from rest a point 3 inches above equilibrium position 。x(,) =-2 sin(81) 32 x(t) =-32 cos(80 O x(r) =-icos(81)
Explanation:
I believe the answer is b) slowly heating the surface
Answer:
5.33*10^-3 seconds
Explanation:
c = d/t
c = speed of light constant (3.0*10^5 km/s)
d = distance (1600 km)
t = ?
3.0*10^5 = 1600/t
t = 1600/3.0*10^5
t = 5.33*10^-3 seconds
I hope this helped! :)
Answer:
Explanation:
a )
While breaking initial velocity u = 62.5 mph
= 62.5 x 1760 x 3 / (60 x 60 ) ft /s
= 91.66 ft / s
distance trvelled s = 150 ft
v² = u² - 2as
0 = 91.66² - 2 a x 150
a = - 28 ft / s²
b ) While accelerating initial velocity u = 0
distance travelled s = .24 mi
time = 19.3 s
s = ut + 1/2 at²
s is distance travelled in time t with acceleration a ,
.24 = 0 + 1/2 a x 19.3²
a = .001288 mi/s²
= 2.06 m /s²
c )
If distance travelled s = .25 mi
final velocity v = ? a = .001288 mi / s²
v² = u² + 2as
= 0 + 2 x .001288 x .25
= .000644
v = .025 mi / s
= .0025 x 60 x 60 mi / h
= 91.35 mph .
d ) initial velocity u = 59 mph
= 86.53 ft / s
final velocity = 0
acceleration = - 28 ft /s²
v = u - at
0 = 86.53 - 28 t
t = 3 sec approx .