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lukranit [14]
2 years ago
11

The figure below shows a man in a boat on a lake. The man's mass is 74 kg, and the boat's is 135 kg. The man and boat are initia

lly at rest when the man throws a package of mass m = 15 kg horizontally to the right with a speed of vi = 5.0 m/s.
What is the velocity of the boat after the package is thrown? Neglect any resistance force from the water. (Give the magnitude in m/s, and select the correct direction from the options given.)

Physics
1 answer:
vazorg [7]2 years ago
7 0

The velocity of the boat after the package is thrown is 0.36 m/s.

<h3>Final velocity of the boat</h3>

Apply the principle of conservation of linear momentum;

Pi = Pf

where;

  • Pi is initial momentum
  • Pf is final momentum

v(74 + 135) = 15 x 5

v(209) = 75

v = 75/209

v = 0.36 m/s

Thus, the velocity of the boat after the package is thrown is 0.36 m/s.

Learn more about velocity here: brainly.com/question/6504879

#SPJ1

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8 0
3 years ago
An archer wants to hit a target that is dropped from a tower. At the sound of a horn, the archer is to shoot an arrow; at the sa
bija089 [108]

It has to be D because the arrow will drop as it moves, if it were a gun, you'd lead the target so fire below it, but due to it being an arrow, you aim high not low. Also, they didnt specify how fast anything is, so you'd probably miss if you actually did it.


6 0
3 years ago
Read 2 more answers
Difference between kilogram and kilometre in points​
skelet666 [1.2K]

Answer:

Kilogram(kg) is the SI unit for mass while kilometre(km) is a unit for length. They are both similar in that they are 10^3 of a unit, thus kilo. As kilogram represents mass, it is a measure of how much matter is present in an object. While kilometre is a measure of distance/how long or short an object is.

7 0
3 years ago
A sled starts off with an initial velocity of 8 m/s and slows down to 2 m/s after 3 seconds. What was its acceleration?
melomori [17]

Answer:

-2m/s

Explanation:

3 0
2 years ago
A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio
natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}

For the maximum velocity of the ball at the top of the vertical circular motion,

v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}

where g is the acceleration due to gravity,

Substituting the known values,

\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0
1 year ago
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