First establish the summation of the forces acting int the
ladder
Forces in the x direction Fx = 0 = force of friction (Ff) –
normal force in the wall(n2)
Forces in the y direction Fy =0 = normal force in floor (n1)
– (12*9.81) –( 60*9.81)
So n1 = 706.32 N
Since Ff = un1 = 0.28*706.32 = 197,77 N = n2
Torque balance along the bottom of the ladder = 0 = n2(4 m) –
(12*9.81*2.5 m) – (60*9.81 *x m)
X = 0.844 m
5/ 3 = h/ 0.844
H = 1.4 m can the 60 kg person climb berfore the ladder will
slip
The answer is 9.8 ms^-2, because there is only one force acting on the object so the acceleration will be numerically equal to the gravitational field strength.
If you increase the mass m of the car, the force F will increase, while acceleration a is kept constant. Because F and m are directly proportional.
If you increase the acceleration a of the car, the force F will increase, while mass m is kept constant. Because F and a are directly proportional.
How can Newton's laws be verified experimentally; is by setting this experiment, and changing one variable while keeping the other constant, then observe the change in F.
Hope this helps.
Answer:
y = 128.0 km
Explanation:
The minimum separation of two objects is determined by Rayleygh's diffraction criterion, which establishes that two bodies are solved if the first minino of diffraction of one coincides with the central maximum of the second, with this criterion the diffraction equation remains
the diffraction equation for the first minimum is
a sin θ = λ
In the case of circular openings, the equation must be solved in polar coordinates, leaving the expression, we use the approximation that the sine of tea is very small.
θ = 1.22 λ / d
d = 15 cm
to find the distance we can use trigonometry
tan θ = y / L
tan θ = sin θ / cos θ = θ
substituting
y / L = λ / d
y = L λ /d
let's calculate
y = 384 10⁸ 500 10⁻⁹ / 0.15
y = 1.28 10⁵ m
Let's reduce to km
y = 1.28 10⁵ m (1km / 10³ m)
y = 128.0 km
the correct answer is 120 km away
Answer:
Stretch in the spring = 0.1643 (Approx)
Explanation:
Given:
Mass of the sled (m) = 9 kg
Acceleration of the sled (a) = 2.10 m/s
²
Spring constant (k) = 115 N/m
Computation:
Tension force in the spring (T) = ma
Tension force in the spring (T) = 9 × 2.10
Tension force in the spring (T) = 18.9 N
Tension force in the spring = Spring constant (k) × Stretch in the spring
18.9 N = 115 N × Stretch in the spring
Stretch in the spring = 18.9 / 115
Stretch in the spring = 0.1643 (Approx)
