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Alekssandra [29.7K]
3 years ago
7

Help answer all for prize

Physics
1 answer:
marusya05 [52]3 years ago
5 0

Explanation:

Average speed =total distance ÷ total time

speed = d ÷ t

d= 74400

t= 16

then,

speed = 74400÷16

=4650

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The energy associated with the random motion of molecules or atoms within a substance is
Yuliya22 [10]

Answer:

C

technically B too but youre teachers not that smart so there you go

6 0
3 years ago
A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.6 cm before reaching its new equilibrium
Kaylis [27]

Answer:

f = 3.09 Hz

Explanation:

This is a simple harmonic motion exercise where the angular velocity is

         w² = \frac{k}{m}

to find the constant (k) of the spring, we use Hooke's law with the initial data

         F = - kx

where the force is the weight of the body that is hanging

        F = W = m g

we substitute

        m g = - k x

        k = - \frac{m g}{x}

we calculate

        k = - \frac{9.8 m}{- 2.6 \ 10^{-2}}

        k = 3.769 10² m

we substitute in the first equation

       w² = \frac{ 3.769 \ 10^2 \ m }{m}

       w = 19.415 rad / s

angular velocity and frequency are related

       w = 2πf

        f = \frac{w}{2\pi }

        f = 19.415 / 2pi

        f = 3.09 Hz

7 0
3 years ago
Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca
Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*(\sqrt{\frac{r^3}{Gm}})

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

\frac{Tsaturn}{Tearth} = \sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}

Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be \sqrt{10^3}} years, or 31.62 years.

We find the amount of degrees it moves in 1 year:

1year * \frac{360degrees}{31.62years} = 11.38 degrees

or about 12 degrees.

6 0
3 years ago
On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0
3 years ago
PLS HELP ASAP WILL MARK BRAIN and pls explain your answer....
lora16 [44]

Answer:

it is sublimation because There are three ways heat is transferred into and through the atmosphere:

radiation.

conduction.

convection.

Explanation:

please mark me as brainliest as you wrote over there

5 0
3 years ago
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