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4 years ago

How are Triton and Pluto similar?

Physics

1 answer:

Stells [14]4 years ago

8 0

Answer:

Triton and pluto both have similar surface materials, such as nitrogen, methane and carbon monoxide.

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What is the kinetic energy of a baseball moving at a speed of 40 m/s if the baseball has mass of 0.15 kg?

bonufazy [111]

0.5mv^2 > 0.5*0.15*40*40=120J

4 0

4 years ago

Read 2 more answers

The X-ray source Cygnus X-1 has a mass of at least 11 solar masses and a diameter of only about one-quarter the diameter of the

SOVA2 [1]

Answer:

Answer is It was deduced from the rate at which it glimmers.

Refer below.

Explanation:

The X-ray source Cygnus X-1 has a mass of at least 11 solar masses and a diameter of only about one-quarter the diameter of the Earth. With such a small diameter it must be a compact object, and with such a large mass it can't be a white dwarf or a neutron star, so a black hole is the only possibility remaining. The diameter of Cygnus X-1 found:

It was deduced from the rate at which it glimmers.

7 0

3 years ago

PLEASE HELP FAST

love history [14]

a. Independent variable: <em>TIME </em>

b. Dependent variable: <em>DISTANCE</em>

4 0

4 years ago

Sam is playing football. She kicks the ball with an average force of 75 N.

damaskus [11]

Answer:

22.5J

Explanation:

Here the force is given. Also, the displacement is given as 30cm.

First we should check if all the values are in their standard form.

Here 30cm should be converted to metre by dividing it with 100.

Which would give us 0.3m

Now we use the equation W=force x displacement =75 x 0.3=22.5J

I hope this satisfies you. If u have any further questions please let me know.

I hope u will follow me and make this the brainliest answer.

3 0

3 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago