A ship is launched toward the water on a slipway making an angle of 5 degree with horizontal. The coefficient of kinetic frictio
1 answer:
Newton's second law we can find the acceleration of the ship is
a = 0.074 m / s²
Given parameters
- The very coefficient of kinematic friction = 0.08
- The angle of the ramp tea = 5º
To find
- The acceleration of the ship
Newton's second law states that the force varies linearly with the mass and acceleration of the body
∑ F = m a
The bold indicate vectors, where F is the sum of the forces, m the masses and the acceleration
The force is a vector magnitude, therefore it must be solved for each axis of a coordinate system, in the exerted of inclined ramps the coordinate system used has the x-axis in the direction of the ramp and the y-axis perpendicular to it, see attached. In this cordinate system the weight of the body must be decomposed.
cos 5 = W_y / W
sin 5 = Wₓ / W
W_y = W cos 5
Wₓ = W sin 5
Let's solve Newton's second law for each axis
Y axis
N - W_y = 0
N = W_y = mg cos 5
X axis
Wₓ - fr = m a (1)
Friction force is the macroscopic result of contact interactions between two bodies and is described for solids by the expression
fr = μ N
fr = μ mg cos 5
we substitute in equation 1
mg sin 5 - μ m g cos 5 = m a
a = g (sin 5 - μ cos 5)
let's calculate
a = 9.8 (sin 5 - 0.08 cos 5)
a = 0.073 m / s²
In conclusion using Newton's second law we can find the acceleration of the ship is
a = 0.073 m / s²
Learn more about Newton's second law here:
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Answer:
The answer is
<h2>4 hrs</h2>Explanation:
To find the time taken we use the formula
where
d is the distance covered
v is the velocity
t is the time
From the question
d = 1200 miles
v = 300 mi/hr
We have
We have the final answer as
<h3>4 hrs</h3>Hope this helps you
Answer:
The tunnel probability for 0.5 nm and 1.00 nm are and
respectively.
Explanation:
Given that,
Energy E = 2 eV
Barrier V₀= 5.0 eV
Width = 1.00 nm
We need to calculate the value of
Using formula of
Put the value into the formula
(a). We need to calculate the tunnel probability for width 0.5 nm
Using formula of tunnel barrier
Put the value into the formula
(b). We need to calculate the tunnel probability for width 1.00 nm
Hence, The tunnel probability for 0.5 nm and 1.00 nm are and
respectively.
Answer:
The ball lands 154.3 ft from the origin at an angle of 13.6° from the eastern direction toward the south.
Explanation:
Hi there!
The position vector of the ball is described by the following equation:
r = (x0 + v0x · t + 1/2 · ax · t², y0 + v0y · t + 1/2 · ay · t², z0 + v0z · t + 1/2 · g · t²)
Where:
r = poisition vector of the ball at time t.
x0 = initial horizontal position.
v0x = initial horizontal velocity (eastward).
t = time.
ax = horizontal acceleration (eastward).
y0 = initial horizontal position.
v0y = initial horizontal velocity (southward).
ay = horizontal acceleration (southward)
z0 = initial vertical position.
v0z = initial vertical velocity.
g = acceleration due to gravity.
We have to find at which time the vertical component of the position vector is zero (the ball is on the ground) and then we can calculate the horizontal distance traveled by the ball at that time, using the equations of the horizontal components of the position vector.
Let´s place the origin of the system of reference at the throwing point so that x0 and y0 and z0 = 0.
y = z0 + v0z · t + 1/2 · g · t² (z0 = 0)
0 = 48 ft/s · t - 1/2 · 32 ft/s² · t²
0 = t (48 ft/s - 16 ft / s² · t) (t= 0, the origin point)
0 = 48 ft/s - 16 ft / s² · t
- 48 ft/s / -16 f/s² = t
t = 3.0 s
Now, we can calculate how much distance the ball traveled in that time.
First, let´s calculate the distance traveled in the eastward direction:
x = x0 + v0x · t + 1/2 · ax · t² (x0 = 0, ax = 0 there is no eastward acceleration)
x = 50 ft/s · 3 s
x = 150 ft
And now let´s calculate the distance traveled in southward direction:
y = y0 + v0y · t + 1/2 · ay · t² (y0 = 0 and v0y = 0, initially, the ball does not have a southward velocity).
y = 1/2 · ay · t²
y = 1/2 · (-8 ft/s²) · (3 s)²
y = -36 ft
Then, the final position vector will be:
r = (150 ft, -36 ft, 0)
The traveled distance is the magnitude of the position vector:
To calculate the angle, we have to use trigonometry (see attached figure):
cos angle = adjacent side / hypotenuse
cos α = x/r
cos α = 150 ft / 154.3 ft
α = 13.6°
The ball lands 154.3 ft from the origin at an angle of 13.5° from the eastern direction toward the south.
