Answer: B. The particles mover faster.
Explanation:
It is given that, two teams are playing tug of war.
Force applied by Team A, 
Force applied by Team B, 
We need to find the net force acting on the rope. It is equal to :
So, the net force acting on the rope is 35 N and it is acting toward right. Hence, this is the required solution.
Answer:
(C) 40m/s
Explanation:
Given;
spring constant of the catapult, k = 10,000 N/m
compression of the spring, x = 0.5 m
mass of the launched object, m = 1.56 kg
Apply the principle of conservation of energy;
Elastic potential energy of the catapult = kinetic energy of the target launched.
¹/₂kx² = ¹/₂mv²
where;
v is the target's velocity as it leaves the catapult
kx² = mv²
v² = kx² / m
v² = (10000 x 0.5²) / (1.56)
v² = 1602.56
v = √1602.56
v = 40.03 m/s
v ≅ 40 m/s
Therefore, the target's velocity as it leaves the spring is 40 m/s
Answer:
ball hit the ground from her feet is 1.83 m far away
Explanation:
given data
speed = 5.3 m/s
angle = 12°
height = 1 m
to find out
how far from her feet ball hit ground
solution
we consider here x is horizontal component and y is vertical component
so in vertical
velocity will be = v sin12
vertical speed u = 5.3 sin 12 = 1.1 m/s downward
and
in horizontal , velocity we know v = 5.3 m/s
so from motion of equation
s = ut + 0.5×a×t²
s is distance t is time a is 9.8
put all value
1 = 1.1 ( t) + 0.5×9.8×t²
solve it we get t
t = 0.353 s
and
horizontal distance is = vcos12 × t
so horizontal distance = 5.3×cos12 × ( 0.353)
horizontal distance = 1.83 m
so ball hit the ground from her feet is 1.83 m far away
