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disa [49]
3 years ago
15

A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 1

20 m/s. He misses his target and the cannonball splashes into the briny deep. How far did the cannonball travel? (please show work)
Physics
1 answer:
Bumek [7]3 years ago
4 0

Answer:

202.8m

Explanation:

Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.

First calculate the total time travelled by using the second equation of motion

h = Ut + 1/2gt^2

Let assume that u = 0

And h = 3.5

Substitute all the parameters into the formula

3.5 = 1/2 × 9.8 × t^2

3.5 = 4.9t^2

t^2 = 3.5/4.9

t^2 = 0.7

t = 0.845s

To know how far the cannonball travel, let's use the equation

S = UT + 1/2at^2

But acceleration a = 0

T = 2t

T = 1.69s

S = 120 × 1.69

S = 202.834 m

Therefore, the distance travelled by the cannon ball is approximately 202.8m.

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The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a vel
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Answer:

Explanation:

We shall write the velocities given in vector form to make the solution easy.

The velocity of water with respect to earth that is waV(e) makes 30 degree with north or 60 degree with east so in vector form

waV(e) = 2.2 cos 60 i + 2.2 sin 60 j

waV(e) = 1.1 i + 1.9 j

Similarly , velocity of wind with respect to earth that is wiV(e) , is making 50 degree with west or - ve of x axes so we cal write it in vector form as follows

wiV(e) =  - 4.5 cos 50 i - 4.5 sin 50 j

wiV(e) = - 2.89 i - 3.45 j

Now we have to calculate velocity of wind with respect to water that is

wiVwa

wiV( wa) = wiV ( e)+ eV(wa)

=  wiV( e)- waV(e)

- 2.89 i - 3.45 j - 1.1 i -  1.9 j

= - 3.99 i - 5.35 j

Magnitude of this relative velocity

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6 0
3 years ago
You have arrived at the scene of a two car accident. You know the following pieces of information:
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The initial velocity of car 1 is 20 m/s to the right

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Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system must be conserved before and after the collision.

Therefore, we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 2000 kg is the mass of the first car

u_1 is the initial velocity of the first car

v_1 = 10 m/s is the final velocity of the first car (taking right as positive direction)

m_2 = 2000 kg is the mass of the second car

u_2 = 0 is the initial velocity of the second car

v_2 is the final velocity of the second car

We also know the initial momentum of car 1, which is

p_1 =40,000 kg m/s

And since momentum is the mass times the velocity, we find the initial velocity of car 1:

u_1 = \frac{p_1}{m_1}=\frac{40,000}{2,000}=20 m/s

with a positive sign, since the direction is to the right.

Now we can re-arrange the previous equation and solve for v2, the final velocity of car 2:

v_2 = \frac{m_1 u_1 -m_1 v_1}{m_2} = \frac{(2000)(20)-(2000)(10)}{2000}=10 m/s

And since the sign is positive, the direction is the same as the initial direction of car 1, so to the right.

Learn more about momentum here:

brainly.com/question/7973509  

brainly.com/question/6573742  

brainly.com/question/2370982  

brainly.com/question/9484203  

#LearnwithBrainly

8 0
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