The lawnmower accelerates in the positive horizontal direction, so that the net horizontal force is, by Newton's second law,
(70 N) cos(-50°) = <em>m</em> (1.8 m/s²)
where <em>m</em> is the mass of the lawnmower. Solve for <em>m</em> :
<em>m</em> = ((70 N) cos(-50°)) / (1.8 m/s²)
<em>m</em> ≈ 25 kg
The lawnmower presumably doesn't get lifted off the ground, so that the net vertical force is 0. If <em>n</em> is the magnitude of the normal force, then by Newton's second law,
<em>n</em> - <em>m g</em> + (70 N) sin(-50°) = 0
<em>n</em> = <em>m g</em> + (70 N) sin(50°)
<em>n</em> = (25 kg) (9.8 m/s²) + (70 N) sin(50°)
<em>n</em> ≈ 300 N
Answer:
the acceleration 
Explanation:
Given that:
the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.
So; the acceleration for the first 6 miles can be calculated by using the formula:
v₂² = v₁² + 2a (Δx)
Making acceleration a the subject of the formula in the above expression ; we have:
v₂² - v₁² = 2a (Δx)
Thus;
Assume the car moves in the +x direction;
the acceleration
I'm sorry but I'm abouta fail a test if I dont d ig.i this for the points
Answer:
Gobernancia. ...
Compromiso con el mercado. ...
Seguridad. ...
Infraestructura. ...
Imperio de la ley. ...
Capital humano. ...
Gestión Financiera Pública. ...
Compromiso ciudadano.
Explanation:
Answer:
The universal law of gravitation.
PE = m * G M / R^2 potential energy of mass m due to attractive forces
If the kinetic energy of mass m is greater than the energy due to the attractive masses then then mass m can continue indefinitely away from the attracting masses.
