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3 years ago

Calculate the momentum of the titanic (m = 4.2 x 10^7 kg) moving at 14 knots (1 my = 1.852 km/h).

2 answers:

8 0

Momemtum is defined as

P = m*v

m is the mass
v is the velocity

14 knots = 14*(1.852) km/h = 25.928 km/h = 7.202 m/s

P = 4.2 x 10^7 kg * 14 knot = 4.2 x 10^7 kg * 25.928 <span> km/h
</span>
P = 4.2 x 10^7 kg *7.202 m/s = 302.493 kg*m/s

Liono4ka [1.6K]3 years ago

3 0

Answer:

$3.02\cdot 10^8 kg m/s$

Explanation:

- The mass of the Titanic is given by:

$m=4.2\cdot 10^7 kg$

- The velocity of the Titanic is given by:

$v=14 kn \cdot (1.852)=25.93 km/h \cdot ( \frac{1000 m/km}{3600 s/h})=7.2 m/s$

- The momentum of the Titanic is given by the product between mass and velocity:

$p=mv=(4.2\cdot 10^7 kg)(7.2 m/s)=3.02\cdot 10^8 kg m/s$

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At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counsel

sleet_krkn [62]

Answer:

47 m

Explanation:

Data obtained from the question include the following:

Length of dry leg 1 (L1) = 40 m

Length of dry leg 2 (L2) = 25 m

Length of swimming course (L) =..?

The length of the swimming course can be obtained by using pythagoras theory as shown below:

L² = L1² + L2²

L² = 40² + 25²

L² = 1600 + 625

L² = 2225

Take the square root of both side.

L = √2225

L = 47.1 ≈ 47 m

Therefore, the length of the swimming course is approximately 47 m.

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A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c

Nady [450]

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

$I= \Delta p = m \Delta v = m (v-u)$

where

m is the mass of the object

$\Delta v$ is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

$I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s$

(b) 155 N

The impulse can also be rewritten as

$I=F \Delta t$

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

$\Delta t$ is the duration of the collision