Since the car is at rest, the force experienced by car will be normal face(exerted by surface to which car is in contact) and weight of the car.
As the car is at rest, net force on the car should be zero.
Answer:
F₁ = 499.61 N , this is the force that Bubba support
Explanation:
The trunk is in equilibrium with the two forces applied by man, let's use the equilibrium relation
let's set a reference frame at the extreme left and assume that the counterclockwise rotations are positive
Let's write the expression for the translational equilibrium
subscript 1 is for Bubba's mass and subscript 2 for his partner
F₁ + F₂ -W = 0
F₁ + F₂ = W
the expression for rotational equilibrium
∑ τ = 0
F₁ 2.2 + F₂ (6.2-0.9) - W 6.2/2 = 0
2.2 F1 + 5.3 F2 = 3.1 W
let's write our system of equations
F₁ + F₂ = W
2.2 F₁ + 5.3 F₂ = 3.1 W
we solve for F₁ in the first equation and substitute in the second
F₁ = W-F₂
2.2 (W- F₂) + 5.3 F₂ = 3.1 W
F₂ ( -2.2 +5.3) = W (3.1 - 2.2)
F₂ = 704 0.9 / 3.1
F₂ = 204.39 N
This is the force that the partner supports
we look for F1
F₁ = W-F₂
F₁ = 704 - 204.39
F₁ = 499.61 N
This is the force that Bubba support
Answer:
100 N
Explanation:
The following data were obtained from the question:
Mass (m) = 2 Kg
Velocity (v) = 5 m/s
Radius (r) = 50 cm
Force (F) =.?
Next, we shall convert 50 cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
50 cm = 50 cm × 1 m / 100 cm
50 cm = 0.5 m
Therefore, 50 cm is equivalent to 0.5 m.
Finally, we shall determine the magnitude of the net force on the ball by using the following formula:
F = mv²/r
Mass (m) = 2 Kg
Velocity (v) = 5 m/s
Radius (r) = 0.5 m
Force (F) =.?
F = mv²/r
F = 2 × 5²/ 0.5
F = 2 × 25/ 0.5
F = 50 / 0.5
F = 100 N
Therefore, the magnitude of the net force on the ball is 100 N.
Consider speed of bicyclist 1 as V1=+8m/s and bicyclist 2 as V2=-2m/s m=80kg
Since momentum has opposite signs moving to right or left
Therefore the combined momentum , P is given by :
P=m*V1+m*V2 =80*(8-2)kg*m/s=+480 kg m/s
Answer:
I agree with Katelyn's statement
Explanation:
Netwon's first law of motion states that an object will remain at rest or in uniform motion except acted upon by an external (unbalanced) force.
When balanced forces are acting on an object in motion, the object maintains its speed and direction.
This means that balanced forces do not change the speed and direction of an object in motion. We can say that the object is in equilibrium. It either remains at rest or in uniform motion.
Therefore, I agree with Katelyn's statement