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AlladinOne [14]

3 years ago

10

A duck is flying around and its velocity v as a function of time t is given in the graph below where rightwards is the positive

velocity direction. What time t does the duck have the same position as time t = 0 secs?

1 answer:

Gwar [14]3 years ago

6 0

Answer:

t= 2.0s

Explanation:

Khan Academy

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A ball is dropped from rest at height of 20m. If it loses 25% of its kinetic energy when it strikes the ground. What is the heig

victus00 [196]

Answer:

15 meters

Explanation:

The inicial energy of the ball is just potencial energy, and its value is:

E = m * g * h = m * g * 20,

where m is the ball mass, and g is the value of gravity.

In the moment that the ball strickes the ground, all potencial energy transformed into kinetic energy, and 25% of this energy is lost, so the total energy at this moment will be:

E' = 0.75 * E = 0.75 * m * g * 20 = 15*m*g

This kinetic energy will make the ball goes up again, and at the maximum height, all kinetic energy is transformed back into potencial energy.

So, as the mass and the gravity are constants, we can calculate the height the ball will reach:

E' = m*g*h = 15*m*g -> h = 15 meters

6 0

3 years ago

Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.

Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

a)

We can solve this first part of the problem by applying the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

$f=\frac{R}{2}$

For this mirror, R = 20 cm, so its focal length is

$f=\frac{20}{2}=+10 cm$ (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm$

b)

We can solve this part by using the magnification equation:

$M=-\frac{y'}{y}=\frac{q}{p}$

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

$y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm$

and the negative sign means that the image is inverted.

#LearnwithBrainly

6 0

3 years ago

What hot springs in the ocean floor

LuckyWell [14K]

<span>Its when lava shoots right out of the ocean floors. Known as Hydrothermal vents </span>

6 0

3 years ago

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A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.400 rev

saw5 [17]

Answer:

The final angular velocity is rev/s is 0.293 rev/s.

Explanation:

Given;

mass of the merry-go-round, m₁ = 120 kg

radius of the merry-go-round, r = 1.8 m

initial angular velocity, ω = 0.4 rev/s

mass of the child, m₂ = 22 kg

Apply the principle of conservation angular momentum to determine the final angular velocity;

$I_i= I_f\\\\\frac{1}{2} m_1r^2 \omega _i = \frac{1}{2} m_1r^2 \omega _f + m_2r^2 \omega _f\\\\ \frac{1}{2} m_1r^2 \omega _i =( \frac{1}{2} m_1r^2 + m_2r^2 )\omega _f\\\\\omega _f = \frac{ \frac{1}{2} m_1r^2 \omega _i}{\frac{1}{2} m_1r^2 + m_2r^2} \\\\\omega _f = \frac{ \frac{1}{2} m_1 \omega _i}{\frac{1}{2} m_1 + m_2}\\\\\omega _f = \frac{0.5 \ \times \ 120\ kg \ \times \ 0.4\ rev/s}{0.5 \ \times 120\ kg \ \ + \ \ 22 \ kg} \\\\\omega _f = 0.293 \ rev/s\\$

Therefore, the final angular velocity is rev/s is 0.293 rev/s.

3 0

3 years ago

The process by which high energy radiation can generate free radicals is called

Anettt [7]

I believe the answer is ionizing radiation

4 0

3 years ago

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