<h2>
Component of the velocity of the ball in the horizontal direction just before the ball hits the ground = 7.31 m/s</h2>
Explanation:
In horizontal direction there is acceleration or deceleration for a ball tossed upward at an initial angle of 43° off the ground.
So the horizontal component of velocity always remains the same.
Horizontal component of velocity is the cosine component of velocity.
Initial velocity, u = 10 m/s
Angle, θ = 43°
Horizontal component of velocity = u cosθ
Horizontal component of velocity = 10 cos43
Horizontal component of velocity = 7.31 m/s
Since the horizontal velocity is unaffected, we have
Component of the velocity of the ball in the horizontal direction just before the ball hits the ground = 7.31 m/s
A) -3.75 meters/second
A=(20^2-80^2)/(2x800)
=(400-6400)/1600
=-6000/1600
=-3.75
B) 16 seconds
t=(20-80)/-3.75
=-60/-3.75
=16
Answer:
number line
Explanation: hope this helps lol
It would be D) the rope is pulled to the right. This is because their is a greater force in that direction.
Answer:
If you hold the temperature of an ideal gas constant, what happens to its volume when you triple the pressure? For T fixed, P is proportional to 1/V or V is proportional to 1/P. Tripling P reduces V to 1/3. ... If T is constant, the speeds of the average speeds and kinetic energy of the atomic particles remain constant.
I hope this helps!
