Answer:
The acceleration increases.
Explanation:
From Newton's 2nd Law, we have 
. We can see that force is directly proportional to mass and acceleration. Therefore, as force increases, either mass or acceleration must increase as well, and vice versa. Since mass is maintained here, if you increase the force applied to the Frisbee, the acceleration will increase as well.
Explanation:
In total, the length is measured from the tip of the bow in a linear fashion to the stern of the formation of delight including any back-deck extensions. The measurement involves bow sprits; rudders; detachable engines and engine sections; handles; and various fittings and connections.
Importance in calculating a boat's length:
it affects the transportation costs (the longer the length, the higher the cost).
The pontoon's length counts as you find out how much rope you need to wrestle.
The cost of vessel settlement on marinas depends in part on the pontoon length. As more area is consumed by a more drawn pontoon, the docking charges are higher.
Transportation guidelines will probably not allow pontoons past a specific length on specific occasions of the day.
A scientific law is the simple mathematical expression of the relationship involved. A principle is the same relationship expressed in words. A theory is the explanation of the facts that make up the relationship.
Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr
<em>The answer is </em>Ninth <em>and </em>Tenth <em>grade so the answer would be</em> B
<em>I hope this helps you </em>
