To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.
Newton's second law is defined as

Where,
m = mass
a = acceleration
From this equation we can figure the acceleration out, then



From the cinematic equations of motion we know that

Where,
Final velocity
Initial velocity
a = acceleration
x = displacement
There is not Final velocity and the acceleration is equal to the gravity, then





From the equation of motion where acceleration is equal to the velocity in function of time we have




Therefore the time required is 0.0705s
Answer:
Wet surfaces areaA=+25.3ft^2
Explanation:
Using F= K×A× S^2
Where F= drag force
A= surface area
S= speed
Given : F=996N S=20mph A= 83ft^2
K = F/AS^2=996/(83×20^2)
K= 996/33200 = 0.03
1215= (0.03)× A × 18^2
1215=9.7A
A=1215/9.7=125.3ft^2
Answer:
Total Work done =0.65 joule
Explanation:
Work done is given Mathematically as
W=F *d
Where w=work done in joules
F=applied force
d= distance moved
The work done to move the toy accros the first meter is
W1=0.5*1
W1=0.5joule
The work done to move the toy across the next 2m at an angle of 30° is
.W2=0.5*2cos30
W2=0.5*2*0.154
W2=0.154joule
Hence total work done is
W1+W2=0.5+0.154
Total Work done =0.65 joule
This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:
sin 65 = h / 150
Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters
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