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Ilia_Sergeevich [38]
3 years ago
11

Three kids are riding on a snow sled traveling horizontally without friction at 19.8 m/s. The masses of Kid A, B, and C are 42.8

kg, 31.5 kg, and 25.9 kg, respectively.
What is the new velocity of the sled if Kid A jumps off vertically from the sled? (Kid A velocity = 0)Three kids are riding on a snow sled traveling horizontally without friction at 19.8 m/s. The masses of Kid A, B, and C are 42.8 kg, 31.5 kg, and 25.9 kg, respectively.

What is the new velocity of the sled if Kid A jumps off vertically from the sled? (Kid A velocity = 0)
Physics
1 answer:
Kamila [148]3 years ago
4 0

Answer:

34.6 m/s

Explanation:

From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg

Final mass will be 31.5+25.9=57.4 kg

From formula of momentum

M1v1=m2v2

Making v2 the subject of the formula then

V2=\frac {M1v1}{m2}

Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then

V2=\frac {100.2 kg\times 19.8 m/s}{57.4 kg}=34.56376 m/s\approx 34.6 m/s

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How does friction affect the kinetic energy of an object?
Law Incorporation [45]

Answer: Friction also prevents an object from starting to move, such as a shoe placed on a ramp. When friction acts between two surfaces that are moving over each other, some kinetic energy is transformed into heat energy. Friction can sometimes be useful.

Explanation:

8 0
3 years ago
An observer sees a full moon in the night sky. What will the Moon's phase be 10 days later?
algol [13]
I'm pretty sure the moon would be a crescent within the 10 day period because it takes about 28 days for the moon to go through the different stages.
5 0
3 years ago
An object starts at rest. Its acceleration over 30 seconds is shown in the graph below:
ddd [48]

Answer:

The instantaneous speed of the object after the first five seconds is 12.5 m/s.

(C) is correct option.

Explanation:

Given that,

An object starts at rest. Its acceleration over 30 seconds.

We need to calculate the instantaneous speed of the object after the first five seconds

We know that,

Area under the acceleration -time graph gives speed.

According to figure,

speed = area\ of\ tringle

speed=\dfrac{1}{2}\times b\times h

speed =\dfrac{1}{2}\times5\times5

speed-12.5\ m/s

Hence, The instantaneous speed of the object after the first five seconds is 12.5 m/s.

6 0
3 years ago
Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for sl
ikadub [295]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

  • We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
  • Both movements are independent each other, due to they are perpendicular.
  • In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
  • Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

       v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)

  • Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

       t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s  (2)

  • In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
  • Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

       \Delta y = v_{oy}  * t - \frac{1}{2} *g*t^{2} (3)

  • In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

       v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)

  • Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

       \Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)

  • Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.
4 0
3 years ago
suppose you increase your walking speed from 4 m/s to 12 m/s in a period of 1 second what is your acceleration
KiRa [710]

a   = 12 ms⁻¹ - 4ms⁻¹ / 1 s

a  =    8 ms⁻²

7 0
3 years ago
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