7 I'LL GIVE BRAINLIST!!!
2 answers:
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Answer:
a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.
b) The weight of the sled is 490.35 newtons.
c) A force of 147.105 newtons is needed to start the sled moving.
d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.
Explanation:
a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:
- External force exerted on the sled, measured in newtons.
- Friction force, measured in newtons.
- Normal force from the ground on the mass, measured in newtons.
- Weight, measured in newtons.
b) The weight of the sled is determined by the following formula:
(1)
Where:
- Mass, measured in kilograms.
- Gravitational acceleration, measured in meters per square second.
If we know that and
, the weight of the sled is:
The weight of the sled is 490.35 newtons.
c) The minimum force needed to start the sled moving on the horizontal ground is:
(2)
Where:
- Static coefficient of friction, dimensionless.
- Weight of the sled, measured in newtons.
If we know that and
, then the force needed to start the sled moving is:
A force of 147.105 newtons is needed to start the sled moving.
d) The minimum force needed to keep the sled moving at constant velocity is:
(3)
Where is the kinetic coefficient of friction, dimensionless.
If we know that and
, then the force needed to keep the sled moving at a constant velocity is:
A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.
Answer:
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m
Explanation:
For this exercise we must use conservation of energy
the electric potential energy is
U =
for the proton at x = -1 m
U₁ =
for the electron at x = 1 m
U₂ =
starting point.
Em₀ = K + U₁ + U₂
Em₀ =
final point
Em_f =
energy is conserved
Em₀ = Em_f
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²( )
we substitute the values
½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ ) = 9 109 (1.6 10-19) ²(
)
2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( )
2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷
r₂² -1 = (4.443 10⁸)⁻¹
r2 =
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m