1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kvasek [131]
4 years ago
11

Among the largest passenger ships currently in use, the Norway has been in service the longest. The Norway is more than 300 m lo

ng, has a mass of 6.9x107 kg, and can reach a top cruising speed of 33 km/h. Calculate the magnitude of the ship’s momentum.
Physics
1 answer:
LenaWriter [7]4 years ago
6 0

Answer:

6.33\times 10^8\ kg\cdot m/s

Explanation:

Mass of the ship (m) = 6.9 × 10⁷ kg

Speed of the ship (v) = 33 km/h

First, let us convert the speed from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

So, 33 km/h = 33\times \frac{5}{18}=9.17\ m/s

Now, we know, the momentum of an object only depends on its mass and speed. Momentum is independent of the length of the object.

So, here, length of the ship doesn't play any role in the determination of the momentum.

Magnitude of momentum of the ship = Mass × Speed

                                                             = (6.9\times 10^7\ kg)(9.17\ m/s)

                                                             = 6.33\times 10^8\ kg\cdot m/s

Therefore, the magnitude of ship's momentum is 6.33\times 10^8\ kg\cdot m/s.

You might be interested in
When thermal insulation, such as spun glass or rock wool, is placed beneath the roof of a house, then in cold weather the insula
siniylev [52]

Answer:

c. slow the flow of heat from inside the house to the outside.

Explanation:

Thermal insulation not completely stops the flow of heat but slow down the process of flowing of heat from inside the house to the outside.

7 0
3 years ago
Read 2 more answers
Oil having a specific gravity of 0.9 is pumped as illustrated with a water jet pump. The water volume flowrate is 1 m3 /s. The w
love history [14]

Answer:

The rate at which the pump moves oil is 1 m³/s

Explanation:

Assumptions:

  • there is steady-state flow
  • oil and water are incompressible
  • first fluid is water, second fluid is oil and third fluid is the mixture of oil and water.

\rho_1Q_1 + \rho_2Q_2 = \rho_3Q_3 -------equation (i)

where;

ρ is the fluid density

Q is the volumetric flow rate

Q_1 + Q_2 = Q_3--------equation (ii)

Substitute in Q₃ in equation i

\rho_1Q_1 + \rho_2Q_2 = \rho_3(Q_1 +Q_2)

divide through by ρ₁

\frac{\rho_1Q_1}{\rho_1}+ \frac{\rho_2Q_2}{\rho_1} =\frac{ \rho_3(Q_1 +Q_2)}{\rho_1}\\\\Note; \frac{\rho_2}{\rho_1} = \gamma_2 \ and \ \frac{\rho_3}{\rho_1} = \gamma_3\\\\Q_1 + \gamma_2Q_2 = \gamma_3(Q_1+Q_2)

Make Q₂ the subject of the formula

Q_2 = \frac{Q_1(1- \gamma_3)}{\gamma_3-\gamma_2} = \frac{1 (\frac{m^3}{s}) (1-0.95)}{0.95-0.9} = 1 \ \frac{m^3}{s}

Therefore, the rate at which the pump moves oil is 1 m³/s

6 0
3 years ago
Consider two soap bubbles with radius r1 and r2 (r1 <r2) connected via a valve. What happens if we open the valve​
Sedaia [141]

Complete Question

The  complete question is shown on the first uploaded image

Answer:

The pressure difference of the first bubble is   \Delta  P _1 =10  J/m^3

The pressure difference of the second bubble is  \Delta  P _2 =20  J/m^3

The pressure difference on the second bubble is higher than that of the first bubble so when the valve is opened pressure from second bubble will cause air to flow toward the first bubble making is bigger

Explanation:

From the question we are told that

    The  radius of the first bubble is  r_1 =  10 \ mm=0.01 \ m

      The radius of the second bubble is  r_2  =  5 \ mm  =  0.005 \ m

      The surface tension of the soap solution is  s =  25 \ mJ/m^2 = 25*10^{-3} J/m^2

Generally according to the Laplace's Law for a spherical membrane the pressure difference is mathematically represented as

         \Delta  P  =  \frac{4 s}{R}

Now the pressure difference for the first bubble is  mathematically evaluated as

        \Delta  P _1 =  \frac{4 s}{r_1}

substituting values  

       \Delta  P _1 =  \frac{4 *25 *10^{-3}}{0.01}

       \Delta  P _1 =10  J/m^3

Now the pressure difference for the second bubble is  mathematically evaluated as

        \Delta  P _2 =  \frac{4 s}{r_1}

       \Delta  P _2 =  \frac{4 *25 *10^{-3}}{0.005}

       \Delta  P _2 =20  J/m^3

3 0
3 years ago
All waves must travel up and down.<br><br><br> True<br><br><br> False
const2013 [10]

Answer:

True all waves that go up must come down

I feel like the follow the rule of gravity:everything that goes up must come d5

8 0
3 years ago
Consider a lawnmower of mass m which can slide across a horizontal surface with a coefficient of friction μ. In this problem the
inna [77]

Answer:

Fh = u*m*g / (cos(θ) - u*sin(θ))

Explanation:

Given:

- The mass of lawnmower = m

- The angle the handle makes with the horizontal = θ

- The force applied along the handle = Fh

- The coefficient of friction of the lawnmower with ground = u

Find:

Find the magnitude, Fh, of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

Solution:

- Construct a Free Body Diagram (FBD) for the lawnmower.

- Realize that there is horizontal force applied parallel to ground due to Fh that drives the lawnmower and a friction force that opposes this motion. We will use to Newton's law of motion to express these two forces in x-direction as follows:

                                     F_net,x = m*a

- Since, the lawnmower is to move with constant speed then we have a = 0.

                                     F_net,x = 0

- The forces as follows:

                                     Fh*cos(θ) - Ff = 0

Where, Ff is the frictional force:

                                     Fh = Ff /cos(θ)

Similarly, for vertical direction y the forces are in equilibrium. Using equilibrium equation in y direction we have:

                                    - W - Fh*sin(θ) + Fn = 0

Where, W is the weight of the lawnmower and Fn is the contact force exerted by the ground on the lawnmower. Then we have:

                                     Fn = W + Fh*sin(θ)

                                     Fn = m*g + Fh*sin(θ)

The Frictional force Ff is proportional to the contact force Fn by:

                                     Ff = u*Fn

                                     Ff = u*(m*g + Fh*sin(θ))

Substitute this expression in the form derived for Fh and Ff:

                                     Fh*cos(θ) = u*(m*g + Fh*sin(θ))

                                     Fh*(cos(θ) - u*sin(θ)) = u*m*g

                                     Fh = u*m*g / (cos(θ) - u*sin(θ))

5 0
3 years ago
Other questions:
  • How many steps in glycolysis have atp as a substrate or product?
    9·1 answer
  • Consider a cube with one corner at the origin and with sides of length 50 cm positioned along the x y z axes. There is an electr
    11·1 answer
  • A shopper does 157 J of work pushing a cart with 10.9 N force
    14·1 answer
  • Segment D - E : The bus is speed_____ It is gradually increasing it's
    15·1 answer
  • How does the number of batteries affect the strength of an electromagnet?
    13·1 answer
  • If each of the three rotor helicopter blades is 3.50 m long and has a mass of 120 kg , calculate the moment of inertia of the th
    10·1 answer
  • Definition of fluoresence
    15·2 answers
  • A 1.5 kilogram car is moving at 10 meters per second east. A braking force acts on the car for 5.0 seconds, reducing its velocit
    7·1 answer
  • How u do this atomic mass
    14·2 answers
  • For each statement, circle whether it is true or false.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!