What force in Newton is required to accelerate a car starting from rest to 20 m/s in 15 seconds if the mass of the car is 2500 k
2 answers:
We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.
Where,
- F is force
- m is mass
- a is acceleration
In our case,
- F = ?
- m = 2500 kg
- a = 20m/s
<em>Thus, The force of 50000 Newton is required to accelerate a car of 2500 kg...~</em>
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Answer:
THE RUBBER BALL
Explanation:
From the question we are told that
The mass of the rubber ball is
The initial speed of the rubber ball is
The final speed at which it bounces bank
The mass of the clay ball is
The initial speed of the clay ball is
The final speed of the clay ball is
Generally Impulse is mathematically represented as
where is the change in the linear momentum so
For the rubber is
=>
For the clay ball
=>
So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball
Answer:
F = 233.52 N, θ' = 351.41º
Explanation:
In this exercise we must find the net force applied on the donkey.
For this we use Newton's second law, where we create a reference frame with the horizontal x axis
let's decompose the forces
Jack
= 80.5 N
Jill
cos 45 = F_{2x} / F₂2
sin 45 = F_{2y} / F₂2
F_{2x} = F₂ cos 45
F_{2y} = F₂ sin 45
F_{2x} = 81.7 cos 45 = 57.77 N
F_{2y} = 81.7 sin 45 = 57.77 N
Jane
cos (270 + 45) = F_{3x} / F₃3
sin 315 = F_{3y} / F₃
F_{3x} = 131 cos 315 = 92.63 N
F_{3y} = 131 sin 315 = -92.63 N
the force can be found in each axis
X axis
F_{x} = F_{1x} + F_{2x} + F_{3x}
F_{x} = 80.5 +57.77 + 92.63
F_{x} = 230.9 N
Axis y
F_{y} = F_{1y} + F_{2y} + F_{3y}
F_{y} = 0 + 57.77 -92.63
F_{y} = -34.86 N
we can give the result in two ways
a) F = (230.9 i ^ - 34.86 j ^) N
b) in the form of module and angle
we use the Pythagorean theorem
F = √(Fₓ² + F_{y}²
F = √(230.9² + 34.86²)
F = 233.52 N
let's use trigonometry for the angle
tan θ =
θ = tan⁻¹ (\frac{F_y}{F_x} })
θ = tan⁻¹ (-34.86 / 230.9)
θ = -8.59º
if we measure this angle from the positive side of the x-axis counterclockwise
θ' = 360 -θ
θ‘= 360- 8.59
θ' = 351.41º