While a block slides forward 1.35 m, a force pulls back at a 135 direction, doing -17.8 J of work. what is the magnitude of the
1 answer:
<h2>18.6467 N</h2>
Explanation:
The Work done by any force is defined as the force applied times the displacement of point of application of force in the direction of force.
This is better represented as a scalar product of Force vector and displacement vector.
Here, the angle between force and displacement is .
∴ Magnitude of force =
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Answer:
a)he angle is 90 with respect to the horizontal (x axis)
b) its speed is zero both vertically and horizontally
c) vertical path
d) a parabolic movement
e) v₀ = (9.5i ^ + 8.232 j ^) m / s
Explanation:
This is a relative problem and movement.
.v ’= v + u
Where v ’is the speed with respect to the mobile system, v the speed with respect to the fixed system and the speed between the two reference systems
a) The child and the can is in the truck, so they go at the speed of the truck, when he throws the can he continues at this speed on the x-axis and therefore as the two advance the same distance the more hands of the child, consequently the can is thrown vertically
The angle is 90 with respect to the horizontal (x axis)
b) with respect to the truck, the can is still, so its speed is zero both vertically and horizontally
c) The child sees that the can follows a vertical path
d) A stationary observer on the ground, sees that the can has a constant speed in the same direction of the truck and when they throw it vertical goal has a vertical movement, the sum of these two movements gives a parabolic movement of the same uncle as a projectile launch
e) the initial speed has two components
X Axis v_lata = v_camion = 9.5 m / s
Y Axis speed given by the child
Let's look for the travel time
x = v₀ₓ t
t = x / v₀ₓ
t = 16 / 9.5
t = 1.68 s
y = t - ½ g t²
When he returns to the child's hand and = 0
0 = v_{oy} t - ½ g t²
v_{oy} = ½ g t
v_{oy} = ½ 9.8 1.68
v_{oy} = 8,232 m / s
Speed is
v₀ = (9.5i ^ + 8.232 j ^) m / s