Answer:
1) The velocity of the ball return to the thrower's hand is -15 meters per second.
2) The resulting velocity of the boat is [tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex].
Explanation:
1) Let suppose that ball experiments a free fall, that is an uniform accelerated motion, in which effects from gravity and Earth's rotation can be neglected. The velocity of the ball is represented by the following equations of motion:
Position
[tex]v_{o}\cdot t -\frac{1}{2}\cdot g\cdot t^{2} = 0[/tex]
[tex]t\cdot \left(v_{o}-\frac{1}{2}\cdot g\cdot t \right) = 0[/tex] (1)
Velocity
[tex]v = v_{o}-g\cdot t[/tex] (2)
Where:
[tex]t[/tex] - Time, measured in seconds.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v_{o}[/tex] - Initial velocity of the ball, measured in meters per second.
[tex]v[/tex] - Final velocity of the ball, measured in meters per second.
From (1), we get the time when the ball returns to the thrower's hand:
[tex]v_{o}-\frac{1}{2}\cdot g\cdot t = 0[/tex]
[tex]t = \frac{2\cdot v_{o}}{g}[/tex]
And then we apply this result in (2):
[tex]v = v_{o}-g\cdot \left(\frac{2\cdot v_{o}}{g} \right)[/tex]
[tex]v = -v_{o}[/tex] (3)
Then, the velocity of the ball return to the thrower's hand is -15 meters per second.
2) The resulting velocity of the boat ([tex]\vec v_{B}[/tex]) is represented by the vectorial sum of the velocity of the boat relative to the river ([tex]\vec v_{B/R}[/tex]) and the velocity of the river ([tex]\vec v_{R}[/tex]), both measured in meters per second, that is:
[tex]\vec v_{B} = \vec v_{R}+\vec {v}_{B/R}[/tex] (4)
If we know that [tex]\vec v_{R} = 6\,\hat{i}\,\left[\frac{m}{s} \right][/tex] and [tex]\vec v_{B/R} = 18\,\hat{j}\,\left[\frac{m}{s} \right][/tex], then the resulting velocity of the boat is:
[tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex]
The resulting velocity of the boat is [tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex].
A complex arrangement of pulleys forms what is called the block in a block and tackle. The rope used to lift the pulleys and the load is the tackle. A block and tackle is used to lift a truck engine, which has a weight and output force of nearly 8000 N. The input force required to lift this weight using the block and tackle is 400N. What is the mechanical advantage of the block and tackle?
Answer:
Mechanical advantage = 20
Explanation:
Given:
Output force = 8,000 N
Input force = 400 N
Find:
Mechanical advantage
Computation:
Mechanical advantage = Output force / Input force
Mechanical advantage = 8,000 / 400
Mechanical advantage = 20
Calculate Vector component in Y if the hypotenuse is 32 and angle is 45
Answer:
The correct option is;
c. 22.6
Explanation:
The given parameters are;
The hypotenuse of the vector = 32
The angle of the vector = 45°
Therefore, the vector component in the y-axis is given as follows;
[tex]v_y = v \times sin(\theta)[/tex]
Substituting the values from the question gives;
[tex]v_y = 32 \times sin(45^{\circ}) \approx 22.6[/tex]
The vector component in the y-axis, [tex]v_y[/tex], is approximately 22.6.
PLEASE HELP IF YOU KNOW THE ANSWER QUICK PLEASE!!!
Answer:
d
Explanation:
What are the Laws of conservation of atoms?
Answer:
Both the initial and final substances are composed of atoms because all matter is composed of atoms. According to the law of conservation of matter, matter is neither created nor destroyed, so we must have the same number and kind of atoms after the chemical change as were present before the chemical change.
Explanation:
Answer:
The law of conservation of atoms is simple but it has a great affect. An atom cannot be destroyed by such a force. Matter/atoms can't be created by any means. An atom exits only by nature and not by force.
Hope this helped! Please mark brainliest! Have a great day!
How do you solving kinematic equations for horizontal projectiles?
A cork floats on the surface of an incompressible liquid in a container exposed to atmospheric pressure. The container is then sealed and the air above the liquid is evacuated. The cork:
Question:
A cork floats on the surface of an incompressible liquid in a container exposed to atmospheric pressure. The container is then sealed and the air above the liquid is evacuated. The cork:
A. sinks slightly
B. rises slightly
C. floats at the same height
D. bobs up and down about its old position
Answer:
The correct answer is C) floats at the same height
Explanation:
The liquid is incompressible because its density very high and leaves no room for further compaction whether or not there is atmospheric pressure. So when you put a cork on the liquid, pressure or no pressure, there is no displacement hence it floats on the same height regardless of the absence of air.
Cheers!