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Zina [86]
2 years ago
14

Please help me with this question ​

Physics
1 answer:
mariarad [96]2 years ago
5 0

Answer:

Please help me with this question Please help me with this questionPlease help me with this question Please help me with this question Please help me with this question Please help me with this question Please help me with this question Please help me with this questionPlease help me with this question

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Which best describes the purpose of a controlled experiment?
navik [9.2K]
A controlled experiment is best described as a safe, in depth, and insightful display that helps you understand the purpose of the experiment better
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3 years ago
A superball with a mass m = 61.6 g is dropped from a height h = [02]____________________ m. It hits the floor and then rebounds
aleksklad [387]

<em>There is not enough data to solve the problem, but I'm assuming the initial height as h = 10 m for the question to have a valid answer and the student can have a reference to solve their own problem</em>

Answer:

(a) \Delta P=1.67 \ kg.m/s

(b) \Delta P=0.86\ m/s

Explanation:

<u>Change of Momentum</u>

The momentum of a given particle of mass m traveling at a speed v is given by

P=m.v

When this particle changes its speed to a value v', the new momentum is

P'=m.v'

The change of momentum is

\Delta P=m.v'-m.v

\Delta P=m.(v'-v)

Defining upward as the positive direction, we'll compute the change of momentum in two separate cases.

(a) The initial height of the superball of m=61.6 gr = 0.0616 Kg is set to h= 10 m. This information leads us to have the initial potential energy of the ball just after it's dropped to the floor:

U=m.g.h=0.0616\cdot 9.8\cdot 10 =6.0368\ J

This potential energy is transformed into kinetic energy just before the collision occurs, thus

\displaystyle \frac{1}{2}mv^2=6.0368

Solving for v

\displaystyle v=\sqrt{\frac{6.0368\cdot 2}{0.0616}}

v=-14\ m/s

This is the speed of the ball just before the collision with the floor. It's negative because it goes downward. Now we'll compute the speed it has after the collision. We'll use the new height and proceed similarly as above. The new height is

h'=88.5\% (10)=8.85\ m

The potential energy reached by the ball at its rebound is

U'=m.g.h'=0.0616\cdot 9.8\cdot 8.85 =5.342568\ J

Thus the speed after the collision is

\displaystyle v'=\sqrt{\frac{5.342568\cdot 2}{0.0616}}

v'=13.17\ m/s

The change of momentum is

\Delta P=0.0616\cdot (13.17+14)

\Delta P=1.67 \ kg.m/s

(b) If the putty sticks to the floor, then v'=0

\Delta P=0.0616\cdot (0+14)

\Delta P=0.86\ m/s

3 0
4 years ago
A basketball player throws the ball at a 47 angle above the horizontal to a hoop which is located a horizontal distance L = 5.0
FromTheMoon [43]

Answer:

v_0 =1.71

Explanation:

the parabolic movment is described by the following equation:

y = tan(a)x-\frac{1}{2v_0^2(cos(a))^2}gx^2

where y is the height of the ball, a is the angle of launch, v_0 the initial velocity, g the gravity and x is the horizontal distance of the ball.

So, if we want that the ball reach the hood, we will replace values on the equation as:

0.8 = tan(47)(5)-\frac{1}{2v_0^2(cos(47))^2}(9.8)(5)^2

Finally, solving for v_0, we get:

v_0=\sqrt{\frac{-9.8(5)^2}{(0.8-tan(47)(5))2cos^2(47)}}

v_0 =1.71

4 0
4 years ago
The element in an incandescent light bulb that releases light energy is _____. a phosphor mercury vapor a thin tungsten filament
Dimas [21]
The answer to this physics question would be 'a thin tungsten filament'. The element in an incandescent light bulb that releases light energy is a thin tungsten filament. All the other choices are not applicable as answers to this question.
6 0
4 years ago
Read 2 more answers
A student initially 10.0 m East of his school walks 17.5 m West. The magnitude of the student's displacement, relative to the sc
Fantom [35]

Answer:

1. 7.5 m

2. towards west side

explanation:

I hope it will help you

3 0
3 years ago
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