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Ghella [55]
3 years ago
13

A motorboat accelerates uniformly from a ve- locity of 7.0 m/s to the west to a velocity of 2.4 m/s to the west. If its accelera

tion was 2.7 m/s2 to the east, how far did it travel during the acceleration? Answer in units of m.
Physics
1 answer:
Lena [83]3 years ago
8 0

Answer:

Distance, d = 8 meters

Explanation:

Given that,

Initial velocity of the motorboat, u = 7 m/s (west)

Final velocity of the motorboat, v = 2.4 m/s (west)

Acceleration of the motorboat, a=-2.7\ m/s^2 (east)

We need to find the distance covered by it. It can be calculated as :

v^2-u^2=2as

(2.4)^2-(7)^2=2\times (-2.7)s

s = 8 meters

So, the distance covered by the motor boat is 8 meters. Hence, this is the required solution.

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Explanation:

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2 years ago
During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of
olasank [31]

Answer:

3.416 m/s

Explanation:

Given that:

mass of cannonball m_A = 72.0 kg

mass of performer m_B = 65.0 kg

The horizontal component of the ball initially \mu_{xA} = 6.50 m/s

the final velocity of the combined system v = ????

By applying the linear momentum of conservation:

m_A \mu_{xA}+m_B \mu_{xB} = (m_A+m_B) v

72.0 \ kg \times 6.50 \ m/s+65.0 \ kg \times 0 = (72.0 \ kg+65.0 \ kg) v

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v = \dfrac{468\  kg m/s }{137 \ kg}

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8 0
3 years ago
What do you mean by velocity ratio of a wheel and axle​
IgorC [24]

Answer:

Explanation:

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a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f
34kurt

Answer:

    \frac{h_{liquid} }{ h_{body} } = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

         B = ρ_liquid  g V_liquid

let's write the translational equilibrium condition

         B - W = 0

let's use the definition of density

        ρ_body = m / V_body

        m = ρ_body  V_body

        W = ρ_body  V_body  g

we substitute

          ρ_liquid  g  V_liquid = ρ_body  g  V_body

          \frac{\rho_{body}   }{\rho_{liquid} } } =  \frac{V_{liquid}   }{V_{body} } }

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

          V = A h_bogy

Thus

          \frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }

we substitute

           5/9 = \frac{h_{liquid} }{ h_{body} }

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