According to Newton’s first law of motion, if a person is swinging a rock st the end of a string and the string breaks, what wil
1 answer:
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We can solve the problem by using Newton's second law of motion:
where
F is the net force applied to the object
m is the object's mass
a is the acceleration of the object
In this problem, the force applied to the car is F=1050 N, while the mass of the car is m=760 kg. Therefore, we can rearrange the equation and put these numbers in, in order to find the acceleration of the car:
The equation also tells us that the acceleration and the force have same directions: therefore, since the force exerted on the car is horizontal, the correct answer is
<span>B) 1.4 m/s2 horizontally.</span>
where
F is the net force applied to the object
m is the object's mass
a is the acceleration of the object
In this problem, the force applied to the car is F=1050 N, while the mass of the car is m=760 kg. Therefore, we can rearrange the equation and put these numbers in, in order to find the acceleration of the car:
The equation also tells us that the acceleration and the force have same directions: therefore, since the force exerted on the car is horizontal, the correct answer is
<span>B) 1.4 m/s2 horizontally.</span>
Answer:
Explanation:
I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.
The relation between the linear acceleration and the angular acceleration is
a = R α .... (1)
According to the diagram,
T x R = I x α
T x R = I x a / R from equation (1)
T = I x a / R² .... (2)
mg - T = ma .... (3)
Substitute the value of T from equation (2) in equation (3)
T is the acceleration in the system
Substitute the value of a in equation (2)
This is the tension in the string.
Answer:
a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
v =
let's perform the derivative
v = 15 t² - 18t - 24
0 = 15 t² - 18t - 24
let's solve the quadratic equation
t =
t1 = -0.8 s
t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
a = dv / dt
a = 30 t - 18
a = 0
30 t = 18
t = 18/30
t = 0.6 s
we substitute this time in the expression of the position
x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
x = 1.08 - 3.24 - 14.4 - 8
x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
x₀ = -8 m
the position for t = 0.6 s
x_f = - 24.56 m
the distance
ΔX = x_f - x₀
Δx = | -24.56 -(-8) |
Δx = 16.56 m