<span><span>A dragster going at 5 m/s north increases
its velocity to 23 m/s north in 4 seconds. What is its acceleration during this
time interval? In order to calculate the acceleration of the boat we first have
to identify the suited formula for acceleration in this scenario. </span>
<span>Hence, a = vf – vi / t. Vi = 5 m/s Vf = 23 m/s T =
4 seconds </span>
<span>Solution: </span>
<span>A = 23 m/s – 5m/s / 4s </span>
A = 4.5m/s^2 </span>
OPTIONS :
A.) the force that the ball exerts on the wall
B.) the frictional force between the wall and the ball
C.) the acceleration of the ball as it approaches the wall
D.) the normal force that the wall exerts on the ball
Answer: D.) the normal force that the wall exerts on the ball
Explanation: The normal force acting on an object can be explained as a force experienced by an object when it comes in contact with a flat surface. The normal force acts perpendicular to the surface of contact.
In the scenario described above, Erica's tennis ball experiences an opposite reaction after hitting the wall.This is in relation to Newton's 3rd law of motion, which states that, For every action, there is an equal and opposite reaction.
The reaction force in this case is the normal force exerted on the ball by the wall perpendicular to the surface of contact.
Earth's rotation produces two high tides . This occurs because the moon revolves around the Earth in the same direction that the Earth rotates around its axis. ... Because the Earth rotates through two tidal “bulges” every lunar day, coastal areas experience two high and two low tides every 24 hours and 50 minutes. High tides occur 12 hours and 25 minutes apart.
1) The mass has the same value before and after the impact (We will neglect the microscopic effects in which there may be a minimum molecular loss of the material) so the equation will be
If we put the mass in one side of the expression (or just cancel directly) we have that
2) We must clear from the equation previously found the number two that is dividing the expression (It would go to the other side of the equation to multiply) and the exponent that would pass to the other of the expression as root therefore
3) We must simply replace the given values, the height is 3.67m and the gravity on the ground is therefore
Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s