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I am Lyosha [343]
2 years ago
5

A ship is launched toward the water on a slipway making an angle of 5 degree with horizontal. The coefficient of kinetic frictio

n between the bottom of the ship and slipway is 0.08. What is the acceleration of the ship along the slipway.?
Physics
1 answer:
kari74 [83]2 years ago
8 0

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:

brainly.com/question/13959891

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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22
alexgriva [62]
1) The law of motion of the projectile is
h(t) = 2+22.5 t-4.9 t^2
To find the velocity, we should compute the derivative of h(t):
v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t
So now we can calculate the speed at t=2 s and t=4 s:
v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s
v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
v(t)=0
So we have
22.5-9.8 t=0
And solving this we find
t=2.30 s

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
2+22.5t -4.9 t^2 = 0
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
t=4.68 s
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36  m/s
with negative sign, because it is directed downward.
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Answer: are u kidding

Explanation:

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what two objects have stored energy? A. ball rolling on the ground B. a small rock sitting on top of a big rock C. a stretched r
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The correct statements are:

B. a small rock sitting on top of a big rock

As the rock is at a height with respect to ground it has potential Energy

and

C. a stretched rubber band

A stretched rubber band has elastic potential energy


The others are actually moving and hence would consist of Kinetic energy. Potential energy is stored in objects that do not move and are stationary.

8 0
3 years ago
Read 2 more answers
6.7.1 Current and Current Density A sphere of radius 10 mm that carries a charge of 8 nC =8x 10°C is whirled in a circle at the
aev [14]

Answer:

Part a)

Rate of charge flow is known as electric current

Part b)

Average current flow is

i = 4\times 10^{-7} A

Part c)

As we know that current density is current per unit area

j = 1.27 \times 10^{-3} A/m^2

Explanation:

As we know that angular frequency of rotation is

\omega = 100\pi rad/s

now by basic definition of electric current

Part a)

Rate of charge flow is known as electric current

i = \frac{dq}{dt}

so here we have

i = \frac{Q}{T}

i = Qf

Part b)

here we know that

\omega = 2\pi f

100\pi = 2\pi f

f = 50 Hz

now we have

i = (8\times 10^{-9})(50)

i = 4\times 10^{-7} A

Part c)

As we know that current density is current per unit area

So we have

j = \frac{i}{A}

j = \frac{4\times 10^{-7}}{\pi(10\times 10^{-3})^2}

j = 1.27 \times 10^{-3} A/m^2

6 0
3 years ago
A student finds an unlabeled liquid container in his lab. He notices that the container has two liquids. Since the two liquids h
raketka [301]

Answer:

\rho = 1848.03 kg m^{-3}

Explanation:

given data:

density of water \rho = 1 gm/cm^3 = 1000 kg/m^3

height  of water  = 20 cm  =0.2 m

Pressure  p = 1.01300*10^5 Pa

pressure at bottom

P =  P_{fluid} + P_{h_2 o}

P   = P_{fluid}  + \rho g h

P_{fluid}  = P - \rho g h

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                 = 99340 Pa

p_{fluid}  = P_{atm} + \rho g h_{fluid}                       h_[fluid} = 0.307m

99340 = 104900 + \rho *9.8*0.307

\rho = 1848.03 kg m^{-3}

5 0
3 years ago
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