Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
Answer:
# of Snickers bars 2
Explanation:
Power output= 0.30 HP
=0.3*746
= 0.30 HP (746 W=1.00 HP)
= 224 W
time required 2 h 49 m = 10140 seconds
Since power is work divided by time, then work is:
Work done by the jet = P*t
= 224 *(10140)
= 2.3 MJ (2.3 x 
J)
Converting MJ to Cal
2.3 MJ=549 Cal
# of Snickers bars = 549 Cal / 280 Cal
= 2.0 bars (rounded from 1.96)
Unlike acceleration and velocity, speed does not need to specify the direction of motion. Speed is a scalar quality.
Yolanda might put more items to the desk to make it heavier, requiring more force.
We need to learn more about the force acting on an object in order to locate the solution.
<h3>How can the force that is required to modify the motion be increased?</h3>
- We are aware that the word for force is,
F=ma
where m denotes the object's mass and an its acceleration
- There are two ways to increase the force required to alter the motion of the table.
- One is to increase the mass, and the other is to accelerate it more quickly.
- Otherwise, there will be a lot of friction between the surfaces, making it difficult to move without exerting a lot of force.
We can infer from this that Yolanda could add items to the desk to increase its mass, necessitating the use of additional force.
Learn more about the force here:
brainly.com/question/4075805
#SPJ1
Answer:
Main Asteroid Belt: The majority of known asteroids orbit within the asteroid belt between Mars and Jupiter, generally with not very elongated orbits. The belt is estimated to contain between 1.1 and 1.9 million asteroids larger than 1 kilometer (0.6 mile) in diameter, and millions of smaller ones.
Explanation:
