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pshichka [43]
3 years ago
10

A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force)

acts on her. The student has a mass of 71.0 kg, and the height of the water slide is 12.7 m. If the kinetic frictional force does -5.10 103 J of work, how fast is the student going at the bottom of the slide
Physics
1 answer:
zloy xaker [14]3 years ago
7 0

Answer:

10.27m/s

Explanation:

Given data

work W= -5.10 10^3 J

mass m= 71kg

final height of slide h2= 12.7m

initial height of slide h1=0m

initial velocity v1= 0m/s

final velocity v2=?

Step two:

required

Final velocity

The work-energy theorem is expressed as'

W=1/mv_2^2 +mgh_2-(1/mv_1^2+mgh_1)

make V2 subject of formula we have final speed

v_2=\sqrt{\frac{2W}{m}+v_1^2-2g(h_1-h_2) } \\\\

substitute our given data we have

v_2=\sqrt{\frac{2*(-5.1*10^3)}{71}+0^2-2*9.81(12.7) } \\\\v_2=\sqrt{143.66-249.174 } \\\\v_2=\sqrt{105.514 } \\\\v_2=10.27m/s

The student going at  10.27m/s

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A rocket on Earth experiences an upward applied force from its thrusters. As a result of this force, the rocket accelerates upwa
gayaneshka [121]

Answer:

F=m(11.8m/s²)

For example, if m=10,000kg, F=118,000N.

Explanation:

There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:

F-mg=ma

Solving for the force F, we obtain that:

F=ma+mg=m(a+g)

Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:

F= m(2m/s^{2}+9.8m/s^{2})\\\\F=m(11.8m/s^{2})

From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:

F=(10,000kg)(11.8m/s^{2})\\\\F=118,000N

In this case, the force from the rocket's thrusters is equal to 118,000N.

5 0
4 years ago
A helicopter is hovering above the ground. Jim reaches out of the copter (with a safety harness on) at 180 m above the ground. A
vesna_86 [32]

Answer:

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

Explanation:

Hi there!

The equation of height and velocity of the package are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the package at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² because we consider the upward direction as positive).

v = velocity of the package at a time t.

First, let´s find the time it takes the package to reach the maximum height. For this, we will use the equation of velocity because we know that at the maximum height, the velocity of the package is zero. So, we have to find the time at which v = 0:

v = v0 + g · t

0 = 50.5 m/s - 9.8 m/s² · t

Solving for t:

-50.5 m/s / -9.81 m/s² = t

t = 5.15 s

Now, let´s find the height that the package reaches in that time using the equation of height. Let´s place the origin of the frame of reference on the ground so that the initial position of the package is 10 m above the ground:

h = h0 + v0 · t + 1/2 · g · t²

h = 10 m + 50.5 m/s · 5.15 s - 1/2 · 9.81 m/s² · (5.15 s)²

h = 140 m

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

6 0
3 years ago
3 simple machines you find in a pinball machine
cestrela7 [59]

lever (The things that flick the ball around)

Inclined plane (The hill the ball rolls down)

Wedge (The bumps that stop the ball from rolling certain places)

6 0
2 years ago
Imagine that someone pushes one marble toward a motionless marble. Would there still be action-reaction forces involved in the c
Finger [1]
All the energy that was in the first marble was transfers into the second marble
8 0
4 years ago
A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point
Alexus [3.1K]

Answer:

(2.5,0)

Explanation:

The particle can be described by the following equations:

x=Rsin(-\omega t)+2.5\\y=Rcos(-\omega t)\\\frac{dx}{dt}=-\omega Rcos(-\omega t)\\\frac{dy}{dt}=\omega Rsin(-\omega t)\\\frac{d^2x}{dt^2}=-\omega^2Rsin(-\omega t)\\\frac{d^2y}{dt^2}=-\omega^2Rcos(-\omega t)

For R = 2.5, ω = 2 and t = 0:

x=2.5\\y=2.5\\\\\frac{dx}{dt}=-5\\ \frac{d^2y}{dt^2}=-10

The center of the circle would be at point (2.5,0)

5 0
4 years ago
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