Answer:
85 miles .
Explanation:
Displacement along the 110 South freeway = 260 - 150 = 110 miles
Displacement along the 110 North freeway = 150 - 175 = - 25 miles
Net displacement = 110 - 25 = 85 miles
So Joey's displacement from the 260 mile marker is 85 miles .
Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 =
/ T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.
Answer:
T = 4.905[N]
Explanation:
In order to solve this problem we must perform a sum of forces on the vertical axis.
∑Fy = 0
We have two forces acting only, the weight of the body down and the tension force T up, as the body does not move we can say that it is system is in static equilibrium, therefore the sum of forces is equal to zero.
![T-m*g=0\\T=0.5*9.81\\T=4.905[N]](https://tex.z-dn.net/?f=T-m%2Ag%3D0%5C%5CT%3D0.5%2A9.81%5C%5CT%3D4.905%5BN%5D)
W = force * displacement
W = 32 pounds * 10 feet
Now you need to convert it to newton and meters
W = 142 N * 3.048 m = 434 J
(I approximated the conversions- I hope it helps)
Answer:

Explanation:
Given that
Mass of rifle = M
Initial velocity ,u= 0
Mass of bullet = m
velocity of bullet = v
Lets take final speed of the rifle is V
There is no any external force ,that is why linear momentum of the system will be conserve.
Initial linear momentum = Final linear momentum
M x 0 + m x 0 = M x V + m v
0 = M x V + m v

Negative sign indicates that ,the recoil velocity will be opposite to the direction of bullet velocity.