Answer:
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The problem states that the distance travelled (d) is
directly proportional to the square of time (t^2), therefore we can write this in
the form of:
d = k t^2
where k is the constant of proportionality in furlongs /
s^2
<span>Using the 1st condition where d = 2 furlongs, t
= 2 s, we calculate for the value of k:</span>
2 = k (2)^2
k = 2 / 4
k = 0.5 furlongs / s^2
The equation becomes:
d = 0.5 t^2
Now solving for d when t = 4:
d = 0.5 (4)^2
d = 0.5 * 16
<span>d = 8 furlongs</span>
<span>
</span>
<span>It traveled 8 furlongs for the first 4.0 seconds.</span>
Answer:
a = 4.72 m/s²
Explanation:
given,
mass of the box (m)= 6 Kg
angle of inclination (θ) = 39°
coefficient of kinetic friction (μ) = 0.19
magnitude of acceleration = ?
box is sliding downward so,
F - f = m a
f is the friction force
m g sinθ - μ N = ma
m g sinθ - μ m g cos θ = ma
a = g sinθ - μ g cos θ
a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°
a = 4.72 m/s²
the magnitude of acceleration of the box down the slope is a = 4.72 m/s²
Answer:
t is appropriate to clarify that units such as time and angles the transformation is not in base ten, for example:
60 s = 1 min
60 min = 1 h
24 h = 1 day
Therefore, for this transformation, you must be more careful
the length transformation is base 10
Explanation:
In many exercises the units used are transformed by equations into other units called derivatives, in general the transformation of derived units is the product of the transformation of the constituent units.
In the example of velocity, the derivative unit is m / s, which is why it works in the same way that you transform length and time if in the equation it is multiplying it is multiplied and if it is dividing it is divided.
It is appropriate to clarify that units such as time and angles the transformation is not in base ten, for example:
60 s = 1 min
60 min = 1 h
24 h = 1 day
Therefore, for this transformation, you must be more careful
the length transformation is base 10
1000 m = 1 km
Answer:
Part A - 3N/m
Part B - see attachment
Part C - 4.9 × 10-³J
Part D - E = 1/2kd² + 1/2mv² + mgh
Explanation:
This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.
The free body diagram can be found in the attachment. Fp(force of pull), Ft(Force of tension) and W(weight).
The energy stored in the pring as a result of the stretching of d = 5.7cm is 1/2kd².
Part D
Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential
energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.