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Ad libitum [116K]
2 years ago
6

A.) Review the graphical data above and create a data table that reflects the graph (10 pts.)

Physics
1 answer:
Artist 52 [7]2 years ago
7 0
No shushhhhhhhhhhhhhhhhhhh
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Kahalagahan ng diagram​
timama [110]

  • <em>m</em><em>ahalaga ito upang mas mapaliwanag mo ng maayos at klaro ang isang bagay o pangyayari.</em>

<em>sana </em><em>makatulong</em><em> </em><em>hehe</em>

5 0
3 years ago
Read 2 more answers
What do comets and asteroids have in common? Check all that apply.
Sidana [21]

Answer:

- They are orbiting the Sun

- They are small compared to the planets

Explanation:

- Comets are small bodies made of rock and ice that have a highly elliptical orbit, and that warm when they pass close to the Sun, releasing gas due to the sublimation of the ice, producing a characteristic "tail" that can extend even for hundreds of kilometers. They are generally located outside the Solar System, in the Oort cloud.

- Asteroids are irregular bodies whose size range from a few km to even a few hundreds of km, that orbit the Sun in an area of the Solar System located between Mars and Jupiter, called "Asteroid belt". They are made mainly of rock and metal debris

So now we  can analyze the statements given:

- They are made of ice: FALSE, because this is only valid for comets

- They are orbiting the Sun: TRUE for both  comets and asteroids

- They are made of metal debris: FALSE, this is only valid for asteroids

- They are small compared to the planets: TRUE for both (the biggest asteroid, Ceres, has a radius of approx 500 km, much smaller than the smallest planet, Mercury)

- They are at the center of the solar system: FALSE for both

7 0
3 years ago
Read 2 more answers
Which of the following is not a part of a wave?
Alisiya [41]
B is the correct answer
y=Asin(wt-kx)
A=amplitude
f=frequency
x=wavelength
since refraction is not on the wave formula,then option B is the correct answer
5 0
3 years ago
Read 2 more answers
A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w
cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

\displaystyle P=\frac{W}{t}

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

W=F.x

The second Newton's law gives us the net force as

F=m.a

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

v_f=v_o+at

v_f=at

Solving for a

\displaystyle a=\frac{v_f}{t}={5.4}{1}

a=5.4\ m/s^2

The distance traveled in the interval is given by

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Since vo=0

\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}

x=2.7\ m

The force is given by

F=m.a

We don't know the value of m, so the force is

F=2.7m

Computing the work done by the sprinter

W=F.x=2.7m(5.4)

W=14.58m

The power is finally computed

\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}

P=14.58m

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}

a=8.8\ m/s^2

The distance is

\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}

x=3.8\ m

The net force is

F=m(8.8)=8.8m

The work done by the sprinter is now computed as

W=8.8m(3.8)=33.44m

At last, the output power is

\displaystyle P=\frac{33.44m}{0.5}=66.88m

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0
3 years ago
A box of mass m = 1.80 kg is dropped from rest onto a massless, vertical spring with spring constant k = 2.00 ✕ 102 N/m that is
Rudik [331]

Answer:

spring compressed is 0.724 m

Explanation:

given data

mass = 1.80 kg

spring constant k = 2 × 10²  N/m

initial height = 2.25 m

solution

we know from conservation of energy is

mg(h+x)  = 0.5 × k × x²       ...................1

here x is compression in spring

so put here value in equation 1 we get

1.8 × 9.8 × (2.25+x)  = 0.5 × 2× 10² × x²

solve it we get

x = 0.724344

so spring compressed is 0.724 m

3 0
3 years ago
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