Explanation:
It is given that,
Speed of the ball, v = 10 m/s
Initial position of ball above ground, h = 20 m
(a) Let H is the maximum height reached by the ball. It can be calculated using the conservation of energy as :
h' = 5.1 m
The maximum height above ground,
H = 5.1 + 20
H = 25.1 meters
So, the maximum height reached by the ball is 25.1 meters.
(b) The ball's speed as it passes the window on its way down is same as the initial speed i.e. 10 m/s.
Hence, this is the required solution.
The final velocity of the truck is found as 146.969 m/s.
Explanation:
As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.
Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.
Then, the final velocity will be
Thus, the final velocity of the truck is found as 146.969 m/s.
Calcium chloride contains ionic bonds.
Pennies contain metallic bonds.
Hydrochloric acid contains covalent bonds.
You're welcome.
To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description, which determine the velocity, such as the displacement of a particle as a function of time, that is to say
Where,
x = Displacement
v = Velocity
t = Time
Our values are given as,
Replacing we have that,
Therefore the distance from Earth to the Moon is 399.000 km
The distance between Mars and the Sun in the scale model would be 1140 m
Explanation:
In this scale model, we have:
represents an actual distance of
The actual distance between Mars and the Sun is 228 million km, therefore
On the scale model, this would corresponds to a distance of 
.
Therefore, we can write the following proportion:
And solving for , we find:
Learn more about distance:
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