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3 years ago

My questions are in the picture: Please help.

Physics

1 answer:

frutty [35]3 years ago

8 0

Answer:

a.)

I.) Open system

II.) Closed system

b.)

The total momentum of the object and earth system stay the same as the object fall toward the earth

The kinetic energy of the object increases as the object falls toward the earth.

Explanation:

a.)

I.) The system containing only the object is an open system because of the influence of external forces and presence of matters. External forces in this scenario means gravitational force acting on the object. And air is the matter that has influence on the object.

II.) The system containing only the object and earth is a closed system. Because in a closed system, there are no external dissipative forces acting on it. Universe is a closed system. And the mechanical energy of a close system is conserved. The mechanical energy will remain constant. In other words, it will not change (become more or less). This is called the Law of Conservation of Mechanical Energy.

b.) The total momentum of the object and earth system stay the same as the object fall toward the earth. Because momentum will always be conserved.

The kinetic energy of the object increases as the object falls toward the earth. When the object was initially at rest, kinetic energy equals zero

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Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat

kvasek [131]

Answer:

a) $a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

b) $k = \frac{9.8}{9.74}=1.006$

Explanation:

Part a

For this case we can begin finding the period like this:

$T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s$

Then we know that the centripetal acceleration is given by:

$a= \frac{v^2}{r}$

And the velocity is given by:

$v=\frac{2\pi r}{T}$

If we replace this into the acceleration we got:

$a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}$

And we can replace the values and we got:

$a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

Part b

For this case we want to find a value of k such that:

$a= k 9.8$

Where a = 9.74, so then we can solve for k like this:

$k = \frac{9.8}{9.74}=1.006$

8 0

3 years ago

Help me out with this pe question plz :c

blagie [28]

Answer:

the first one is a group 1 and the second one is d all of the above

Explanation:

6 0

3 years ago

Read 2 more answers

What is the difference between speed and velocity?

mixer [17]

Speed is the rate of change of distance with time while velocity is the rate of change of displacement with time.
Speed is a scalar quantity while velocity is a vector quantity.
Speed cannot be negative but velocity can be negative.

Hope you could get an idea from here.

Doubt clarification - use comment section.

8 0

2 years ago

When you whirl a can at the end of a string in a circular path, what is the direction of the force that acts on the can

andreyandreev [35.5K]

Answer:

Toward the centre of the circular path

Explanation:

The can is moved in a circular path: this means that it is moving by circular motion (uniform circular motion if its tangential speed is constant).

In order to keep a circular motion, an object must have a force that pushes it towards the centre of the circular trajectory: this force is called centripetal force, and its magnitude is given by

$F=m\frac{v^2}{r}$

where m is the mass of the object, v its tangential speed, r the radius of the trajectory. This force always points towards the centre of the circular path.

3 0

3 years ago

A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T

Serga [27]

Answer:

a) $k_{e}$ = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m, e) v = 15.23 m / s

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

$k_{e}$ = ½ k x²

$k_{e}$ = ½ 2900 0.80²

$k_{e}$ = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

U = m h and

U = 8 9.8 (-0.80)

U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

K = ½ m v²

K = 0

d) write the energy at two points, maximum compression and maximum height

Em₀ = ke = ½ m x²

$E_{mf}$ = mg y

Emo = $E_{mf}$

½ k x² = m g y

y = ½ k x² / m g

y = ½ 2900 0.8² / (8 9.8)

y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

Y = y- 0.80

Y = 11.8367 -0.80

Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

Emo = ½ k x² = 928 J

$E_{mf}$ = ½ m v²

Emo = $E_{mf}$

Emo = ½ m v²

v =√ 2Emo / m

v = √ (2 928/8)

v = 15.23 m / s

8 0

3 years ago