Answer:
K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
Em₀ = K = ½ mv²
final point. When it is at tea = 50º
Em_f = K + U
Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
h = L - L cos 50
Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
Em₀ = Em_f
½ mv² = ½ m v_b² + mg L (1 - cos50)
K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
K_b = 36 +42.0
K_b = 78 J
Answer:
the distance between two cars becomes 
when blue car travels at legal speed
Explanation:
Initially the relative speed of the two cars is given as
now the relative acceleration of blue car with respect to white car
now the distance between two cars till the relative speed of two cars comes to zero
so the distance between two cars becomes when blue car travels at legal speed
Given: Mass m = 18 Kg; Height h = 3.6 m; Time t = 60 s
Required: Power output in unit of Watts
Formula: P = mgh/t
P = (18 Kg)(9.8 m/s²)(3.6 m)/60 s
P = 10.58 J/s or Watts
Answer:
Horizontal component of the initial velocity=v ×cos20°
horizontal displacement= 35m
time taken. (t)=35÷(v×cos20°)
vertical component of the initial velocity= v×sin20°
vertical displacement= -5m(since it is opposite to the direction of the initial velocity. )
application of s=ut+1/2at×t. vertically,
-5=vsin20°×t -1/2 gt×t
-5= vsin20×35/vcos20-1/2×10×35×35/(vcos20×vcos20)
simplify further to obtain v and hence find the time taken
Answer:
distance = 32 kilometers
displacement = 0 kilometers
Explanation:
they traveled 15 km to and back which would mean the distance they traveled is 32 km. but the displacement would still be 0 because their starting and ending points are the same.
