I got you
Explanation:
normal force = 400 g cos 35
friction force up slope = .6 (400 g) cos 35
weight component down slope = 400 g sin 35
400 a = 400 g sin 35 - .6 (400 g cos 35)
a = g (sin 35 - .6 cos 35) = .082 g
I hope this helps you
Answer:
- 1.3 x 10⁻¹⁵ C/m
Explanation:
Q = Total charge on the circular arc = - 353 e = - 353 (1.6 x 10⁻¹⁹) C = - 564.8 x 10⁻¹⁹ C
r = Radius of the arc = 5.30 cm = 0.053 m
θ = Angle subtended by the arc = 48° deg = 48 x 0.0175 rad = 0.84 rad (Since 1 deg = 0.0175 rad)
L = length of the arc
length of the arc is given as
L = r θ
L = (0.053) (0.84)
L = 0.045 m
λ = Linear charge density
Linear charge density is given as

Inserting the values

λ = - 1.3 x 10⁻¹⁵ C/m
Answer:
The kinetic energy of the system after the collision is 9 J.
Explanation:
It is given that,
Mass of object 1, m₁ = 3 kg
Speed of object 1, v₁ = 2 m/s
Mass of object 2, m₂ = 6 kg
Speed of object 2, v₂ = -1 m/s (it is moving in left)
Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

E = 9 J
So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.
The force the escaping gas exerts of the rocket is 10.42 N.
<h3>
Force escaping gas exerts</h3>
The force the escaping gas exerts of the rocket is calculated as follows;
F = m(v - u)/t
where;
- m is mass of the rocket
- v is the final velocity of the rocket
- u is the initial velocity of the rocket
- t is time of motion
F = (0.25)(40 - 15)/0.6
F = 10.42 N
Thus, the force the escaping gas exerts of the rocket is 10.42 N.
Learn more about force here: brainly.com/question/12970081
#SPJ1
The mass m of the object = 5.25 kg
<h3>Further explanation</h3>
Given
k = spring constant = 3.5 N/cm
Δx= 30 cm - 15 cm = 15 cm
Required
the mass m
Solution
F=m.g
Hooke's Law
F = k.Δx
