(a) At a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).
(b) If the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.
<h3>
Conservation of mechanical energy</h3>
The effect of height and gravity on speed on the given planet Epislon is determined by applying the principle of conservation of mechanical energy as shown below;
ΔK.E = ΔP.E
¹/₂m(v²- u²) = mg(hi - hf)
¹/₂(v²- u²) = g(0 - hf)
v² - u² = -2ghf
v² = u² - 2ghf
where;
- v is the final velocity at upper level
- u is the initial velocity
- hf is final height
- g is acceleration due to gravity
when u² = 2gh, then v² = 0,
when gravity reduces, u² > 2gh, and v² > 0
Thus, at a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).
<h3>Final speed</h3>
v² = u² - 2ghf
where;
- u is the initial speed = 5 m/s
- g is acceleration due to gravity and its less than 9.8 m/s²
- v is final speed
- hf is equal height
Since g on Epislon is less than 9.8 m/s² of Earth;
5² - 2ghf > 3 m/s
Thus, if the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.
Learn more about conservation of mechanical energy here: brainly.com/question/6852965
Answer
given,
diameter of the pipe is = (14 ft)4.27 m
minimum speed of the skater must have at very top = ?
At the topmost point of the pipe the normal force will be equal to zero.
F = mg
centripetal force acting on the skateboard
equating both the force equation
r = d/2 = 14/ 2 = 7 ft
or
r = 4.27/2 = 2.135 m
g = 32 ft/s² or g = 9.8 m/s²
v = 14.96 ft/s
or
v = 4.57 m/s
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
Answer:No, it doesn't move easily downward because it will try to resist the movement ,due to a resistance force of inertia that it possess at rest.
Explanation:when an object has higher or larger mass it tends to resist any motion given to it unlike the one with lower mass.
The larger the mass the more resistance force an object has.
