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3 years ago

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A child in a boat throws a 5.10-kg package out horizontally with a speed of 10.0 m/s. the mass of the child is 27.6 kg and the m

ass of the boat is 34.2 kg calculate the velocity of the boat immediately after, assuming it was initially at rest.

1 answer:

lapo4ka [179]3 years ago

7 0

I am totally lost with this sorry

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The velocity of a projectile at launch has a horizontal component vh and a vertical component vv. When the projectile is at the

Usimov [2.4K]

Answer:

- horizontal component of the velocity: $v_h$ (because it is constant)

- vertical component of the velocity: 0

- vertical component of the acceleration: $-g=-9.8 m/s^2$ (downward)

Explanation:

The motion of a projectile consists of two independent motions:

- Along the horizontal direction, there are no forces acting on the projectile (if we neglect air resistance), therefore the horizontal acceleration is zero and the horizontal component of the velocity, vh, is constant

- Along the vertical direction, there is only one force acting on the projectile: the force of gravity, downward, which produces a constant downward acceleration of $g=9.8 m/s^2$ . As a consequence, the vertical component of the velocity changes according to

$v_v(t) = v_v-gt$

where vv is the initial vertical velocity and t the time. According to this equation, the vertical component of the velocity decreases first, then becomes zero at the point of maximum height, then becomes negative (= changes direction and points downward)

So, in summary, at the highest point of the trajectory we have:

- horizontal component of the velocity: $v_h$ (because it is constant)

- vertical component of the velocity: 0

- vertical component of the acceleration: $-g=-9.8 m/s^2$ (downward)

3 0

3 years ago

PLS HELP BEING TIMED!!!

Elodia [21]

Answer:

$K_{i}+U_{g,i} = K_{f}+U_{g,f}$

Explanation:

A closed system is a system where exists energy interactions with surroundings, but not mass interactions. If we neglect any energy interactions from boundary work, heat, electricity, magnetism and nuclear phenomena and assume that process occurs at steady state and all effects from non-conservative forces can be neglected, then the equation of energy conservation is reduce to this form:

$\Delta K +\Delta U_{g} = 0$ (1)

Where:

$\Delta K$ - Change in kinetic energy of the system, measured in joules.

$\Delta U_{g}$ - Change in gravitational potential energy of the system, measured in joules.

If we know that $\Delta K=K_{i}-K_{f}$ and $\Delta U_{g} = U_{g,i}-U_{g,f}$ , then we get the following equation:

$K_{i}+U_{g,i} = K_{f}+U_{g,f}$ (2)

Where $i$ and $f$ stands for initial and final states of each energy component.

Hence, the right answer is $K_{i}+U_{g,i} = K_{f}+U_{g,f}$

7 0

3 years ago

Read 2 more answers

How would you find the speed of a person who walked

Alex Ar [27]

Answer:

v = 1.25 m/s

Explanation:

We have,

Distance covered by a person is 10 meters

Time taken by him to cover that distance is 8 seconds.

If we want to find the speed of a person, we must know distance covered by it and taken. In this case, we know both distance and time. His speed is given by :

$v=\dfrac{d}{t}\\\\v=\dfrac{10\ m}{8\ s}\\\\v=1.25\ m/s$

So, the speed of the person is 1.25 m/s.

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3 years ago

If the CD rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angu

gavmur [86]

Answer:

Explanation:

Given that,

Initial Angular velocity w=500rpm

Converting from rpm to rad/s

1rev =2πrad

1minutes =60secs

500rpm=500rev/mins

w = 500×2π/60

wi=52.36rad/s

The final angular velocity wf=0rad/s

Time to stop is t=2.6sec

We want to find angular acceleration α

Using the equation of angular motion

wf = wi + αt.

0 = 52.36 + 2.6α

-52.36=2.6α

α = -52.36/2.6

α = -20.14rad/s²

The angular acceleration is negative because it is decelerating.

Then, α=20.14rad/s²

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2 years ago

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What is the refractive index of air if light travels through it at 3.0 108 m/s?

olchik [2.2K]

Answer:

n = c/v = (3.00 x 108 m/s)/(2.76 x 108 m/s) = 1.09. This does not equal any of the indices of refraction listed in the table.

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2 years ago