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melomori [17]
3 years ago
12

a 59kg physics student jumps off the back of her laser sailboat (42kg). after she jumps the laser is found to be travelling at 1

.5m/s. what is the speed of the physics student? ​
Physics
1 answer:
evablogger [386]3 years ago
5 0

From the conservation of linear momentum of closed system,

Initial momentum = final momentum

Mass of the student, M = 59 kg

Mass of the laser boat, m = 42 kg

Initial speed of student + laser boat, u =0

Final speed of laser boat, v = 1.5 m/s

Final speed of the student = V

(M+m) u =M V +m v

0 = (59 kg) V + (42 kg) (1.5m/s)

V = - 1.06 m/s

Thus, the speed of the student is 1.06 m/s in the opposite direction of the motion of boat.

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A 1200-kg car moving at 15.6 m/s suddenly collides with a stationary car of mass 1500 kg. if the two vehicles lock together, wha
g100num [7]
Use conservation of momentum ;

m1u1 + m2u2 = m1v1 + m2v2

1200×15.6 + 0 = 2700v

v = 18720/2700

v = 6.933 or ~ 7 m/s
5 0
3 years ago
If gravity between the Sun and Earth suddenly vanished, Earth would continue moving in
Ksenya-84 [330]

Answer:

Earth would continue moving by uniform motion, with constant velocity, in a straight line

Explanation:

The question can be answered by using Newton's first law of motion, also known as law of inertia, which states that:

"an object keeps its state of rest or of uniform motion in a straight line unless acted upon by an external net force different from zero"

This means that if there are no forces acting on an object, the object stays at rest (if it was not moving previously) or it continues moving with same velocity (if it was already moving) in a straight line.

In this problem, the Earth is initially moving around the Sun, with a certain tangential velocity v. When the Sun disappears, the force of gravity that was keeping the Earth in circular motion disappears too: therefore, there are no more forces acting on the Earth, and so by the 1st law of Newton, the Earth will continue moving with same velocity v in a straight line.

6 0
3 years ago
The turntable in a microwave oven has a moment of inertia of 0.039 kg⋅m2 and is rotating once every 4.4 s . Part A What is its k
gayaneshka [121]

To solve this problem it is necessary to apply the concepts related to Kinetic Energy, specifically, since it is a body with angular movement, the kinetic rotational energy. Recall that kinetic energy is defined as the work necessary to accelerate a body of a given mass from rest to the indicated speed.

Mathematically it can be expressed as,

KE = \frac{1}{2} I\omega^2

Where

I = Moment of Inertia

\omega =Angular velocity

Our values are given as

I = 0.039kg\cdot m^2

A revolution is made every 4.4 seconds.

\theta = 1 rev \rightarrow 4.4s

\Rightarrow \omega = \frac{1rev}{4.4s}

If the angular velocity is equivalent to the displacement over the time it takes to perform it then

\omega = 0.2272rev/s(\frac{2\pi rad}{1rev})

\omega = 1.42rad/s

Replacing at our previous equation we have,

KE = \frac{1}{2} I\omega^2

KE = \frac{1}{2} (0.039)(1.42)^2

KE = 0.03993J

Therefore the kinetic energy is equal to 3.9*10^{-2}J

6 0
3 years ago
Imagine that you pushed a box, applying a force of 60 newtons, over a distance of 4 meters. How much would you have done?
grin007 [14]
You would have done 240 joules of work on the box.

Work = (force)  x  (distance)

         =  (60 newtons)  x  (4 meters)

         =         240 joules  . 
5 0
3 years ago
Show which of the following functional forms work or do not work as solutions to this differential equation (known as 'the wave
Akimi4 [234]

Answer:

E = A cos (Bx + Ct) and E = A cos (Bx + Ct + D)

Explanation:

The wave equation is

           d²y / dx² = 1 /v²  d²y / dt²

Where v is the speed of the wave

Let's review the solutions

In this case y = E

a) E = A sin Bt

 This cannot be a solution because the part in x is missing

      dE / dx = 0

b) E = A cos (Bt + c)

There is no solution missing the Part in x

         dE / dx = 0

c) E = A x² t²

The parts are fine, but this solution is not an oscillating wave, so it is not an acceptable solution of the wave equation

d) E = A cos (Bx + Ct) and E = A cos (Bx + Ct + D)

Both are very similar

Let's make the derivatives

       dE / dx = -A B sin (Bx + Ct + D)

      d²E / dx² = -A B² cos (Bx + Ct + D)

      dE / dt = - A C sin (Bx + Ct + D)

      d²E / dt² = -A C² cos (Bx + Ct + D)

We substitute in the wave equation

      -A B² cos (Bx + Ct + D) = 1 / v² (-A C² cos (Bx + Ct + D))

       B² = 1 / v² (C²)

       v² = (C / B)²

       v = C / B

We see that the two equations can be a solution to the wave equation, the last one is the slightly more general solution

8 0
3 years ago
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