Answer:
the work done by the 30N force is 4156.92 J.
For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:
W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J
Answer:
Given:
Initial velocity (u) = 30 m/s
Final speed (v) = 0 m/s
Acceleration (a) = - 1.5 m/,s²
To Find:
Time in which train will come to rest (t).
Explanation:
So,
Time in which train will come to rest = 20 seconds
Consider velocity to the right as positive.
First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right
Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left
Total momentum of the system is
P = m₁v₁ + m₂v₂
= 4*2 + 8*(-3)
= -16 (kg-m)/s
Let v (m/s) be the velocity of the center of mass of the 2-block system.
Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s
Answer:
The center of mass is moving at 1.33 m/s to the left.
Answer: higher and lower
Explanation:
charge in an electric field will experience a force in the direction of decreasing potential energy. Since the electric potential energy of a negative charge is equal to the charge times the electric potential the direction of decreasing electric potential energy is the direction of increasing electric potential.
Answer: Your answer is 1250J
Explanation:
K
E
=
1/2
m
v
2
The mass is
m
=
25
k
g
The velocity is v
=
10
m
s
−
1
So,
K
E
=
1
/2
x25
x
10
2^2=
1250
J
pls mark brainiest answer
