Answer:
The power exerted by the student is 51.2 W
Explanation:
Given;
extension of the elastic band, x = 0.8 m
time taken to stretch this distance, t = 0.5 seconds
the spring constant, k = 40 N/m
Apply Hook's law;
F = kx
where;
F is the force applied to the elastic band
k is the spring constant
x is the extension of the elastic band
F = 40 x 0.8
F = 32 N
The power exerted by the student is calculated as;
P = Fv
where;
F is the applied force
v is velocity = d/t
P = F x (d/t)
P = 32 x (0.8 /0.5)
P = 32 x 1.6
P = 51.2 W
Therefore, the power exerted by the student is 51.2 W
Answer:
Explanation:
<em>The top 5 winners of NASA's design contest winner received amount of </em><em> </em>(Dollar One hundred thousand)<em>. </em>The respected amount was divided into 5 parts as there were 5 winners as well.
So, the amount was divided among the winners are as follows:
<em>Winner 1 ⇒ </em>
<em>Winner 2 ⇒ </em>
<em>Winner 3 ⇒ </em>
<em>Winner 4 ⇒ </em>
<em>Winner 5 ⇒</em> 
Hence by adding all of the 5, we will get the answer, which is .
Answer:
Spring C
Explanation:
According to Hooke's law, a force require to compress or extend a spring is directly proportional to the distance from its mean position.
The force of compression or expansion of the spring is given by the formula,
F = k x
Where,
k - spring constant. It is the constant value of the particular spring
x - distance of expansion or compression from mean position.
The x value of different spring,
A = 25 cm
B = 10 cm
C = 100 cm
D = 1 cm
The highest value of x for the same mass has the smallest spring constant according to the equation.
Hence, C has the smallest spring constant.
Answer:
the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
Explanation:
Given that;
speed of car V = 120 km/h = 33.3333 m/s
Reaction time of an alert driver = 0.8 sec
Reaction time of an alert driver = 3 sec
extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec
now, extra distance that car will travel in case of sleepy driver will be'
S_d = V × 2.2 sec
S_d = 33.3333 m/s × 2.2 sec
S_d = 73.3333 m
hence, number of car of additional car length n will be;
n = S_n / car length
n = 73.3333 m / 5m
n = 14.666 ≈ 15
Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
Less than. A machine should have equal or more output than input to have a high efficiency. :)
