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nikdorinn [45]
2 years ago
7

How many lines does each stanza contain? 2 3 4 6

Physics
2 answers:
Musya8 [376]2 years ago
7 0
6 lines the stanza contains
slamgirl [31]2 years ago
4 0

Explanation:

a stanza can contain 4-6 lines or more than that, that's all upon the writer.

hope this helps you

have a great day :)

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17 points
const2013 [10]

The velocity of the cannonball is 150 m/s,  the right option is B. 150 m/s.

The question can be solved, using Newton's second law of motion.

Note: Momentum of the cannon = momentum of the cannonball.

<h3>Formula:</h3>
  • MV = mv................. Equation 1

<h3>Where:</h3>
  • M = mass of the cannon
  • m = mass of the cannonball
  • V = velocity of the cannon
  • v = velocity of the cannonball

Make v the subject of the equation.

  • v = MV/m................ Equation 2

From the question,

<h3>Given: </h3>
  • M = 500 kg
  • V = 3 m/s
  • m = 10 kg.

Substitute these values into equation 2.

  • v = (500×3)/10
  • v = 150 m/s.

 

Hence, The velocity of the cannonball is 150 m/s,  the right option is B. 150 m/s.

Learn more about Newton's second law here: brainly.com/question/25545050

3 0
2 years ago
A ball on a cart is moving at a rate of 2 m/s. The cart suddenly stops and the ball continues to travel in the same direction at
sukhopar [10]

Answer:

The first

Explanation:

Cause it will continue in motion till another force is applied

7 0
3 years ago
Read 2 more answers
a 10.0 kg sphere is released from rest in an ocean. as it falls, the water applies a resistive force r
dimaraw [331]

The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

spring constant k= 2250 Nm

now according to principal of conservation of energy we observe,

the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.

mgh= μ (mgl) +1/2 kx²

10 x 10 x 3= μ(600) +(1125) (0.09)

μ(600) =300 - 101.25

μ = 198.75÷600

μ =0.33125

The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

Learn more about kinetic friction here-

brainly.com/question/13754413

#SPJ4

4 0
1 year ago
What is the relationship between these animals (Predator-Prey) (Parisite-Host) (mutualism) or (commensalism)
ohaa [14]
1.commensalism
2. pred-prey
3. parasite-host
4.commensalism
8 0
3 years ago
A hiker climbs a mountain. Starting at the base of the mountain, he first moved up 520m at a 32.0 degree angle. What is the fina
balu736 [363]

Answer:

\displaystyle \vec{d}=

Explanation:

<u>Displacement Vector</u>

Suppose an object is located at a position  

\displaystyle P_1(x_1,y_1)

and then moves at another position at

\displaystyle P_2(x_2,y_2)

The displacement vector is directed from the first to the second position and can be found as

\displaystyle \vec{d}=

If the position is given as magnitude-angle data ( z , α), we can compute its rectangular components as

\displaystyle x=z\ cos\alpha

\displaystyle y=z\ sin\alpha

The question describes the situation where the initial point is the base of the mountain, where both components are zero

\displaystyle P_1(0,0)

The final point is given as a 520 m distance and a 32-degree angle, so  

\displaystyle x_2=520\ cos32^o= 440.99\ m

\displaystyle y_2=520\ sin32^o=275.6\ m

The displacement is

\displaystyle \vec{d}=

5 0
2 years ago
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