Answer
given,
mass of the ball = 6.3 kg
speed of the ball = 10.4 m/s
angle made with horizontal = 43°
m_a = 1.8 kg v_a = 2.2 m/s
m_b = 1.6 kg v_b = 1.8 m/s
mass of third particle = 6.3 - 1.8 - 1.6
= 2.9 kg
u cos θ = 10.4 x cos 43° = 7.61 m/s
by using conservation momentum along x-axis
6.3 x 7.61 = 1.8 × (-2.2) + 0 + 1.6 × V₃ₓ
V₃ₓ = 32.44 m/s (toward right)
by using conservation momentum along y-axis
0 = 0 + 1.6 x 1.8 + 1.6 × V₃y
V₃y = -1.8 m/s (indicate downward)
velocity of the third particle

v = 32.49 m/s


θ = 3.176° (downward with horizontal)
Answer:
25 m/s
Explanation:
This question can be solved using equation of motion

where
v is the final velocity
u is the initial velocity
s is the distance covered while moving from initial to final velocity
a is the acceleration
_____________________________________________
Given
box moved for distance of 62.5 m
Friction slows the box at –5.0 m/s2----> this statement means that there is deceleration , speed of truck decreases by 5 m/s in every second until the box comes to rest. Friction causes this deceleration.
thus in this problem
a = -5.0 m/s2
V = 0 as body came to rest due to friction deceleration
u the initial velocity we have to find
the initial velocity of box will be the same as speed of truck, as the box was in the truck and hence box will pick the speed of truck.
so if we find speed of box, we will be able get sped of truck as well.
using equation of motion

Thus, initial speed with the truck was travelling was 25 m/s.
Complete Question:
In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.
a) Draw a diagram in a (x,y) plane of the four wires with wire 4 perpendicular to the origin. Indicate the current's directions.
b) Draw a diagram of all magnetic fields produced at the position of wire 3 by the other three currents.
c) Draw a diagram of all magnetic forces produced at the position of wire 3 by the other three currents.
d) What are magnitude and direction of the net magnetic force per meter of wire length on wire 3?
Answer:
force, 1.318 ₓ 10⁻⁴
direction, 18.435°
Explanation:
The attached file gives a breakdown step by step solution to the questions
Answer:
"2Ω" is the net resistance in the circuit.
Explanation:
The given resistors are:
R1 = 3Ω
R2 = 6Ω
The net resistance will be:
⇒ 
On substituting the values, we get
⇒ 
On taking L.C.M, we get
⇒ 
⇒ 
⇒ 
On applying cross-multiplication, we get
⇒ 
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