It is 29 and a half days long
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 .
The mode in this case would be 125 because it occurs the most in the sequence of numbers.
Answer:
The magnitude will be "353.5 N". A further solution is given below.
Explanation:
The given values is:
F = 500 N
According to the question,
In ΔABC,
⇒ 
⇒ 
then,
⇒ 
⇒ 
Now,
The corresponding angle will be:
⇒ 
⇒ 
⇒ 
Aspect of F across the AC arm will be:
= 
On putting the values of F, we get
= 
= 
Component F along the AC (in magnitude) will be:
= 
= 
= 
