A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800.0 N on him. Th
1 answer:
Answer:
a) = 692 N
b) = 932 N
Explanation:
a)
According to newton's second law of motion, acceleration of an object is directly proportional to the net force acting on it. When there is no net force force acting on the body, there is no acceleration. A force is a push or a pull, and the net force ΣF is the total force, or sum of the forces exerted on an object in all directions.
∝ a
= ma
= ma
Given data:
= 800 N
Mass = m = 90 kg
acceleration = a = 1.2 m/s²
= ?
800 - = (90)(1.2)
= 692 N
b)
According to newton's second law of motion,
∝ a
= ma
= ma
Given data:
If we assume the same friction and acceleration between player's feet and ground as calculated in part a
= 692 N
acceleration = a = 1.2 m/s²
We take the equal mass to the total mass of both the players because when the winning player push losing player backward, he exert force on the ground not only due to his mass but also due to the mass of losing player.
Mass = M = m₁ + m₂ = 110 kg + 90 kg
= 200 kg
= ?
- 692 N = (200)(1.2)
= 692 + 240
= 932 N
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Answer:
Δt =
Explanation:
Hi there!
Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.
The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):
v = d/t
Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:
Vr = D/ (2 · Δt1)
For the other half of the trip the expression of velocity will be:
Vw = D/(2 · Δt2)
The total time traveled is the sum of both Δt:
Δt(total) = Δt1 + Δt2
Then, solving the first equation for Δt1:
Vr = D/ (2 · Δt1)
Δt1 = D/(2 · Vr)
In the same way for the second equation:
Δt2 = D/(2 · Vw)
Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)
Δt(total) = D/2 · (1/Vr + 1/Vw)
The time needed by Rick to complete the trip was:
Δt(total) = D/2 · (1/Vr + 1/Vw)
Now let´s calculate the time it took Tim to do the trip:
Tim walks half of the time, then his speed could be expressed as follows:
Vw = 2d1/Δt Where d1 is the traveled distance.
Solving for d1:
Vw · Δt/2 = d1
He then ran half of the time:
Vr = 2d2/Δt
Solving for d2:
Vr · Δt/2 = d2
Since d1 + d2 = D, then:
Vw · Δt/2 + Vr · Δt/2 = D
Solving for Δt:
Δt (Vw/2 + Vr/2) = D
Δt = D / (Vw/2 + Vr/2)
Δt = D/ ((Vw + Vr)/2)
Δt = 2D / (Vw + Vr)
The time needed by Tim to complete the trip was:
Δt = 2D / (Vw + Vr)
Let´s find the diference between the time done by Tim and the one done by Rick:
Δt(tim) - Δt(rick)
2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))
= Δt
Let´s check the result. If Vr = Vw:
Δt = 2D/2Vr - D/2Vr - D/2Vr
Δt = D/Vr - D/Vr = 0
This makes sense because if both move with the same velocity all the time both will do the trip in the same time.
To solve this problem we will apply the definitions given in Newtonian theory about the Force of gravity, and the Force caused by weight. Both will be defined below, and in equal equilibrium condition to clear the variable concerning acceleration due to gravity. Finally, with the values provided in the statement, it will be replaced.
The equation for the gravitational force between the Earth and the object on the surface of the Earth is
Where,
G = Universal gravitational constant
= Mass of Earth
= Distance between object and center of earth
= Mass of Object
The equation for the gravitational pulling force on the object due to gravitational acceleration is
Equation the two expression we have
This the acceleration due to gravity which is composite constant.
Replacing with our values we have then
The value of composite constant is . Here, the composite constant is nothing but the acceleration due to gravity which is constant always.