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ella [17]
3 years ago
12

What is the mass of a large ship that has a momentum of 1.60×109kg·m/s, when the ship is moving at a speed of 48.0 km/h? (b) Com

pare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s.
Physics
1 answer:
erastova [34]3 years ago
6 0

a) The mass of the ship is 1.2\cdot 10^8 kg

b) The ship has a larger momentum than the shell

Explanation:

a)

The momentum of an object is given by:

p=mv

where

m is the mass of the object

v is its velocity

For the ship in this problem, we have

p=1.60\cdot 10^9 kg m/s is the momentum

v=48.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=13.3 m/s is the velocity

Solving for m, we find the mass of the ship:

m=\frac{p}{v}=\frac{1.60\cdot 10^9}{13.3}=1.2\cdot 10^8 kg

b)

The momentum of the artillery shell is given by

p=mv

where

m is its mass

v is its velocity

For the shell in this problem,

m = 1100 kg

v = 1200 m/s

Substituting,

p=(1100)(1200)=1.32\cdot 10^6 kg m/s

So, we see that the ship has a larger momentum.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

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With a frequency of 500 hz, what is the period of a wave
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A hot-air balloon is ascending at the rate of 10 m/s and is 74 m above the ground when a package is dropped over the side. (a) H
Reika [66]

Answer:

The answer to your question is:

a)  t1 = 2.99 s ≈ 3 s

b)  vf = 39.43 m/s

Explanation:

Data

vo = 10 m/s

h = 74 m

g = 9.81 m/s

t = ?   time to reach the ground

vf = ?   final speed

a)    h = vot + (1/2)gt²

     74 = 10t + (1/2)9.81t²

     4.9t² + 10t -74 = 0                  solve by using quadratic formula

   

   t = (-b ± √ (b² -4ac) / 2a

   t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)

   t = (-10 ± √ 1550.4 ) / 9.81

  t1 = (-10 + √ 1550.4 ) / 9.81               t2 = (-10 - √ 1550.4 ) / 9.81

  t1 = (-10 ± 39.38 ) / 9.81                    t2 = (-10 - 39.38) / 9.81

   t1 = 2.99 s ≈ 3 s                               t2 = is negative then is wrong there are

                                                                   no negative times.

b) Formula vf = vo + gt

                  vf = 10 + (9.81)(3)

                  vf = 10 + 29.43

                  vf = 39.43 m/s

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A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the
Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

E=\dfrac{Q}{4\pi\epsilon_0 r^2}

Substitute numerical values:

E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

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