Answer:
2.9 m
Explanation:
First find the time it takes to reach the floor.
y = y₀ + v₀ t + ½ at²
(0 m) = (1.6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 0.571 s
Next, find the distance it travels in that time.
x = x₀ + v₀ t + ½ at²
x = (0 m) + (5.0 m/s) (0.571 s) + ½ (0 m/s²) (0.571 s)²
x = 2.86 m
Rounded to two significant figures, the marble travels 2.9 meters in the x direction.
Answer: P= W/t so P=50/20 =2.5 W
Answer:
Maximum height, h = 11.32 meters
Explanation:
It is given that,
The baseball is thrown directly upward at time, t = 0
Initial speed of the baseball, u = 14.9 m/s
Ignoring the resistance in this case and using a = g = 9.8 m/s²
We have to find the maximum height the ball reaches above where it leaves your hand. Let the maximum height is h. Using third equation of motion as :

At maximum height, v = 0
and a = -g = -9.8 m/s²


h = 11.32 meters
Hence, the maximum height of the baseball is 11.32 meters.
Answer:
The chunk went as high as
2.32m above the valley floor
Explanation:
This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.
Applying the principle of energy conservation for the two ice we have based on the scenery
Momentum before impact = momentum after impact
M1U1+M2U2=(M1+M2)V
Given data
Mass of ice 1 M1= 5.20kg
Mass of ice 2 M2= 5.20kg
velocity of ice 1 before impact U1= 13.5 m/s
velocity of ice 2 before impact U2= 0m/s
Velocity of both ice after impact V=?
Inputting our data into the energy conservation formula to solve for V
5.2*13.5+5.2*0=(10.4)V
70.2+0=10.4V
V=70.2/10.4
V=6.75m/s
Therefore the common velocity of both ice is 6.75m/s
Now after impact the chunk slide up a hill to solve for the height it climbs
Let us use the equation of motion
v²=u²-2gh
The negative sign indicates that the chunk moved against gravity
And assuming g=9.81m/s
Initial velocity of the chunk u=0m/s
Substituting we have
6.75²= 0²-2*9.81*h
45.56=19.62h
h=45.56/19.62
h=2.32m
U use the train to play hope i helped please thank me