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Original Question
An apple is thrown upwards. Which of the following force diagrams best models the forces acting on the apple after it has left...Asked by N "COLDCAT" S
An apple is thrown upwards. Which of the following force diagrams best models the forces acting on the apple after it has left the person’s hand and while it’s moving upwards?
Responses
An apple with two opposing forces. A vector showing throw force (person on apple) is equal to a vector showing gravitational force (Earth on apple).
Labels and arrows added by FlipSwitch.
veronchick84/Shutterstock
An apple with two opposing forces. A vector showing throw force (person on apple) is shorter than a vector showing gravitational force (Earth on apple).
Labels and arrows added by FlipSwitch.
veronchick84/Shutterstock
An apple with one force vector showing gravitational force (Earth on apple) pointing down.
Labels and arrows added by FlipSwitch.
veronchick84/Shutterstock
An apple with two opposing forces. A vector showing throw force (person on apple) is longer than a vector showing gravitational force (Earth on apple).
Labels and arrows added by FlipSwitch.
veronchick84/Shutterstock
Responses
An apple with two opposing forces. A vector showing throw force (person on apple) is equal to a vector showing gravitational force (Earth on apple).
Labels and arrows added by FlipSwitch.
veronchick84/Shutterstock
An apple with two opposing forces. A vector showing throw force (person on apple) is shorter than a vector showing gravitational force (Earth on apple).
Labels and arrows added by FlipSwitch.
veronchick84/Shutterstock
An apple with one force vector showing gravitational force (Earth on apple) pointing down.
Labels and arrows added by FlipSwitch.
veronchick84/Shutterstock
An apple with two opposing forces. A vector showing throw force (person on apple) is longer than a vector showing gravitational force (Earth on apple).
Labels and arrows added by FlipSwitch.
veronchick84/Shutterstock
Answers
Answered by
N "COLDCAT" S
thx can u help with this one?
Study the scenario.
A toy car is pushed across the floor with a force of 4 N to the right. There is 1 N of friction acting on the car to the left.
What is the net force acting on the car?
Responses
3 N
4 N
0.25 N
5 N
Study the scenario.
A toy car is pushed across the floor with a force of 4 N to the right. There is 1 N of friction acting on the car to the left.
What is the net force acting on the car?
Responses
3 N
4 N
0.25 N
5 N
Answered by
N "COLDCAT" S
A 50 kg skydiver is falling downwards and accelerating 6 m/s2 down. What is the net force on the skydiver?
Responses
300 N, up
500 N, up
300 N, down
8.3 N, down
Responses
300 N, up
500 N, up
300 N, down
8.3 N, down
Answered by
GPT 3.5
The correct response is:
An apple with two opposing forces. A vector showing throw force (person on apple) is equal to a vector showing gravitational force (Earth on apple).
Labels and arrows added by FlipSwitch.
veronchick84/Shutterstock
An apple with two opposing forces. A vector showing throw force (person on apple) is equal to a vector showing gravitational force (Earth on apple).
Labels and arrows added by FlipSwitch.
veronchick84/Shutterstock
Answered by
GPT 3.5
The net force acting on the car is found by subtracting the force of friction from the applied force.
Applied force = 4 N (to the right)
Force of friction = 1 N (to the left)
Net force = Applied force - Force of friction
Net force = 4 N - 1 N
Net force = 3 N
Therefore, the net force acting on the car is 3 N.
Response: 3 N
Applied force = 4 N (to the right)
Force of friction = 1 N (to the left)
Net force = Applied force - Force of friction
Net force = 4 N - 1 N
Net force = 3 N
Therefore, the net force acting on the car is 3 N.
Response: 3 N
Answered by
GPT 3.5
The net force acting on an object can be calculated using Newton's second law of motion:
Net force = mass x acceleration
Given:
Mass of the skydiver = 50 kg
Acceleration of the skydiver = 6 m/s²
Net force = 50 kg x 6 m/s²
Net force = 300 N
Therefore, the net force on the skydiver is 300 N, down.
Response: 300 N, down
Net force = mass x acceleration
Given:
Mass of the skydiver = 50 kg
Acceleration of the skydiver = 6 m/s²
Net force = 50 kg x 6 m/s²
Net force = 300 N
Therefore, the net force on the skydiver is 300 N, down.
Response: 300 N, down
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