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Vladimir [108]
3 years ago
7

On April 15, 1912, the luxury cruise liner Titanic sank after running into an iceberg. What was the cruise liner’s speed when it

collided with the iceberg if it had a mass of 4.23 x 108 kg ship and a momentum of 4.9 x 109 kg·m/s?
Physics
1 answer:
netineya [11]3 years ago
3 0

The cruise liner’s speed when it collided with the iceberg if it had a mass of 4.23 x 108 kg ship and a momentum of 4.9 x 109 kg·m/s is v= 11.584 m/s

Explanation:

Given :

The cruise liner’s speed when it collided with the iceberg has a

Mass (m) = 4.23 x 108 kg  and

Momentum (p) = 4.9 x 109 kg m/s

To calculate the momentum,

We need to use the formula of a momentum

Momentum = Mass x Velocity ( p = m v )

as we need to find the velocity the formula becomes

velocity = momentum / mass

v = p / m

= (4.9 x 109 kg m/s) / (4.23 x 108 kg)

v= 11.584 m/s

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A gas is contained in a vertical, frictionless piston-cylinder device. The piston has a mass of mpiston = 2 kg and a cross-secti
BlackZzzverrR [31]
<h2>Answer:</h2>

105146 Pa

<h2>Explanation:</h2>

1) We will make a Free-Body Diagram representing all the upward and downward pressures exerted on the piston.

  • Downward Pressures:
  1. Pressure exerted by the compressed spring (Pspring)
  2. Pressure due to weight of the piston (Pw)
  3. Atmospheric pressure (Patm)  
  • Upward Pressures:  
  1. Initial pressure inside the cylinder. (P1)              

2) We will formulate an equation balancing all upward and downward pressures.

     P1= Patm + Pw + Pspring

3) We will calculate each of the pressures separately.

  • Pspring

          P = F/A

          F= ks

          k= 38×1000 =38000 N m

          s= 2.5 /1000 = (2.5x10^-3) m

          F = 38000×(2.5x10^-3) = 95 N

          A = 30/10000 = (30x10^-4) m2

          P = 95 / (30x10^-4)

          Pspring ≅ 3167 Pa

  • Pw

          P = F/A

          F = W = mg

          W = 2×9.81 = 19.62 N

          A = 30/10000 = (30x10^-4) m2

          P = 19.62 / (30x10^-4)

          Pw = 654 Pa

  • Patm

        P = 1atm = 101325 Pa

        Patm = 101325 Pa

4) We will add all the downward pressures to reach the final answer (initial pressure inside the cylinder).

       P1= Patm + Pw + Pspring

       P1= 101325+654+3167

       P1= 105146 Pa

8 0
3 years ago
6)<br> Calculate the area of a body which experiences a pressure of 60000 Pa by a force of 120N
Vika [28.1K]

Answer:

area = force by pressure

or, a = 120N by 60000 Pa

or, a= 0.002 m²

4 0
4 years ago
A computer disk drive is turned on starting from rest and has a constant angular acceleration. If it took 0.750 seconds for the
Irina-Kira [14]

Answer:

a) Θ = ω₀*t + ½αt² To complete first revolution 2π rads = 0*t + ½αt² and to complete the first and second combined 4π rads = 0*t + ½α(t+0.810s)² Divide second by first: 2 = (t + 0.810s)² / t² This is quadratic in t and has roots at t = -0.336 s ← ignore and t = 1.96 s ◄ b) Use either equation from above: 2π rads = 0*t + ½α(1.96s)² α = 3.27 rad/s² ◄ Hope this helps!

Explanation:

6 0
4 years ago
A worker pushes a crate horizontally across a warehouse floor with a force of 245 N at an angle of 55 degrees below the horizont
aev [14]

Answer:

option A

Explanation:

given,

For exerted by the worker = 245 N

angle made with horizontal = 55°

we need to calculate Force which is not used to move the crate = ?

Movement of crate is due to the horizontal component of the force.

Crate will not move due to vertical force acting on the it.

F_y = F sin \theta

F_y = 245\times sin 55^0

F_y =200.69

hence, worker's force not used to move the crate is equal to 200.69

The correct answer is option A

6 0
3 years ago
Read 2 more answers
A horizontal disk with a radius of 23 m rotates about a vertical axis through its center. The disk starts from rest and has a co
lys-0071 [83]

Answer:

time is 0.42 sec

Explanation:

Given data

radius = 23 m

angular acceleration = 5.7 rad/s²

to find out

time

solution

we know that radius is constant so that

tangential acceleration At = angular acceleration × radius   ............. 1

tangential acceleration =  5.7 × 23 = 131.1 m/s²

and

radial acceleration Ar =  (angular velocity)² × radius    ........................2

we consider angular velocity = ω

this is acting toward center

so

compare 1 and 2

At = Ar

5.7 r =ω³ r

ω = √5.7 = 2.38746 rad/s

so

ω = 5.7 t

2.387 = 5.7 t

t =  2.387 / 5.7

t = 0.4187

time is 0.42 sec

8 0
3 years ago
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