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disa [49]
2 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]2 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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Answer:

1- t^3

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3- t1

Explanation:

The acceleration produced in a body, while travelling in a circular motion, due to change in direction of motion is called centripetal acceleration. The formula of the centripetal acceleration is as follows:

ac = v²/r

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for a constant radius the centripetal acceleration will be directly proportional to the speed of object. The speed of pendulum will be lowest at t1 due to zero speed initially. Then the speed will increase gradually having greater speed at t^2 and the highest speed and centripetal acceleration at t^3. Therefore, the three instants in tie can be written in following order from greatest centripetal acceleration to lowest:

<u>1- t^3</u>

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Which type of compound is salt
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Read 2 more answers
A 110-kg object and a 410-kg object are separated by 3.80 m.
aniked [119]

Answer:

a)   Fₙ = 2,273 10⁻⁷ N   and   b)    x₃ = 1,297 m

Explanation:

This problem can be solved using the law of universal gravitation and Newton's second law for the equilibrium case. The Universal Gravitation Equation is

    F = G m₁ m₂ / r₁₂²

a) we write Newton's second law

       Σ F = F₁₃ - F₃₂

Body 1 has mass of m₁ = 110 kg and we will place our reference system, body 2 has a mass of m₂ = 410 kg and is in the position x₂ = 3.80 m

Body 3 has a mass of m₃ = 41.0 kg and is in the middle of the other two bodies

      x₃ = (x₂-x₁) / 2

     x₃ = 3.80 / 2 = 1.9 m

     Fₙ = -G m₁ m₃ / x₃² + G m₃ m₂ / x₃²

     Fₙ = G m₃ / x₃² (-m₁ + m₂)

Calculate

     Fₙ = 6.67 10⁻¹¹ 41.0 / 1.9² (- 110 + 410)

     Fₙ = 2,273 10⁻⁷ N

Directed to the right

b) find the point where the force is zero

The distance is

     x₁₃ = x₃ - 0

    x₃₂ = x₂ -x₃= 3.8 -x₃

We write the park equation net force be zero

     0 = - F₁₃ + F₃₂

     F₁₃ = F₃₂

     G m₁ m₃ / x₁₃² = G m₃ m₂ / x₃₂²

     m₁ / x₁₃² = m₂ / x₃₂²

Let's look for the relationship between distances, substituting

     m₁ / x₃² = m₂ / (3.8 - x₃)²

     (3.8 - x₃) = x₃ √ (m₂ / m₁)

     x₃ + x₃ √ (m₂ / m₁) = 3.8

     x₃ (1 + √ m₂ / m₁) = 3.8

     x₃ = 3.8 / (1 + √ (m₂ / m₁))

     x₃ = 3.8 / (1 + √ (410/110))

     x₃ = 1,297 m

When body 3 is in this position the net force on it is zero

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Explain the roles of piston valve of a water pump.​
Shtirlitz [24]

Answer:

The upstroke of the piston draws water, through a valve, into the lower part of the cylinder. On the downstroke, water passes through valves set in the piston into the upper part of the cylinder. On the next upstroke, water is discharged from the upper part of the cylinder via a spout.

Explanation:

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2 years ago
A car is traveling along a road, and its engine is turning over with an angular velocity of +188 rad/s. The driver steps on the
andreev551 [17]

Answer:

3751.80514 radians

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

\theta = Angle of rotation

t = Time taken

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=188\times 15.6+\dfrac{1}{2}\times 0\times 15.6^2\\\Rightarrow \theta=2932.8\ rad

The angular displacement would be 2932.8 rad

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=293\times 15.6+\dfrac{1}{2}\times 0\times 15.6^2\\\Rightarrow \theta=4570.8\ rad

The angular displacement would be 4570.8 rad

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{293-188}{15.6}\\\Rightarrow \alpha=6.73076\ rad/s^2

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\dfrac{293^2-188^2}{2\times 6.73076}\\\Rightarrow \theta=3751.80514\ rad

Actual value of the angular displacement is 3751.80514 radians

5 0
3 years ago
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