Increase the force the engine provides and decrease the mass of the car
Answer:
Explanation:
From the position coordinates given , it appears that the ball moves simultaneously along x and y direction.
Displacement along x direction in one second = 4.4 - 1.8 = 2.6 m
So velocity along x direction V_x = 
Similarly velocity along y direction V_y(1) = 
In the next phase velocity changes both in x and y direction.
velocity in x - direction V_x(2) = [tex]\frac{2}{s}[/tex
Velocity in Y- direction V_y(2) = [tex]\frac{3.1}{s}[/tex
Acceleration in x direction = change of velocity in x direction
= ( 2 - 2.6 ) = -.6 m s⁻²
Acceleration in y direction = ( 3.1 - 2.6 ) = 0.5 m s⁻²
Total acceleration =\sqrt{( -.6 )² + ( .5 )²}
= .78 ms⁻²
By V=IR
A: 24=I*20
I = 1.2A
B: 220 = I*250
I = 0.88A
C: 6= I*3
I = 2 A
C,A,B
Answer:
Explanation:
Let the velocity of projectile be v and angle of throw be θ.
The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m
considering its vertical displacement
h = - ut +1/2 g t²
100 = - vsinθ x 5 + .5 x 9.8 x 5²
5vsinθ = 222.5
vsinθ = 44.5
It covers 160 horizontally in 5 s
vcosθ x 5 = 160
v cosθ = 32
squaring and adding
v²sin²θ +v² cos²θ = 44.4² + 32²
v² = 1971.36 + 1024
v = 54.73 m /s