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3 years ago

11

A planet with a mass one-half that of Earth has a radius that is 3 times that of Earth's radius. What is the gravitational field

strength of the planet?

1 answer:

stiv31 [10]3 years ago

6 0

The gravitational acceleration of a planet is proportional to the planet's mass, and inversely proportional to square of the planet's radius.

So when you stand on the surface of this particular planet, you feel a force of gravity that is

(1/2) / (3²)

of the force that you feel on the surface of the Earth.

That's <em>(1/18)</em> as much as on Earth.

The acceleration of gravity there would be about <em>0.545 m/s²</em>.

This is about 12% less than the gravity on Pluto.

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What is the wavelength of a wave that has a frequency of 841 Hz if the speed of the wave is 457 m/s?

Dvinal [7]

Answer:

<h3>The answer is 0.54 m</h3>

Explanation:

The wavelength of a wave can be found by using the formula

$\lambda = \frac{c}{f} \\$

where

c is the velocity

f is the frequency

So we have

$\lambda = \frac{457}{841} \\ = 0.543400713...$

We have the final answer as

<h3>0.54 m</h3>

Hope this helps you

6 0

3 years ago

If a semi-truck decreases its speed from 10 m/s to 5 m/s, its new kinetic energy is:

Otrada [13]

Answer is: <span>1/4 its old kinetic energy .
</span>V₁ = 10 m/s.
V₂ = 5 m/s.
m₁ = m₂ = m.
E₁ = 1/2 · m₁ · V₁², E₁ = 1/2 · m · (10 m/s)² = 50 · m.
E₂ = 1/2 · m₂ · V₂², E₂ = 1/2 · m · (5 m/s)² = 12,5 · m.
E₂/E₁ = 12,5m / 50m = 0,25.
V - speed of semi-truck.
m - mass of semi-truck.
E - kinetic energy of semi-truck.

5 0

4 years ago

Read 2 more answers

How many moons does Venus have?<br> a. 3<br> b. 6<br> c. 0<br> d. 12

baherus [9]

<span>The question is asking how many moons Venus, the planet, has and the answer is 0 -Venus doesn't have any moons. Another planet that doesn't have any moons is Mercury. Earth has one moon and Mars has two moons. Jupiter has 4 moons and the most impressive is Saturn, with its 53 confirmed moons (as claimed by Nasa)!</span>

3 0

3 years ago

Read 2 more answers

You are standing on top of the Empire State Building (height 330 [m]) when Superman flies past. He is headed straight down with

tensa zangetsu [6.8K]

Answer:

The ball will be overtake superman after travelled distance 248.5 m.

Explanation:

Given that,

Speed of superman = 35 m/s

Height = 330 m

Mass of ball = 10 kg

Let the height at which they are at same position be S.

For superman,

We need to calculate the time

Using equation of motion

$S=ut$ ...(I)

For ball,

$S=u't+\dfrac{1}{2}gt^2$

From equation (I) and (II)

$ut=u't+\dfrac{1}{2}gt^2$

$ut=0+\dfrac{1}{2}\times g\times t^2$

$u=\dfrac{1}{2}\times g\times t$

$t=\dfrac{2u}{g}$

Put the value into the formula

$t=\dfrac{2\times35}{9.8}$

$t=7.1\ sec$

We need to calculate the distance travelled

Using formula of distance

$d=v\times t$

Put the value into the formula

$d=35\times 7.1$

$d=248.5\ m$

Hence, The ball will be overtake superman after travelled distance 248.5 m.

6 0

3 years ago

Two identical air-filled parallel-plate capacitors C1 and C2, each with capacitance C, are connected in series to a battery that

solmaris [256]

Answer:

a) The ratio is $\frac{U}{U_{0}} =\frac{2}{(1+\frac{1}{\kappa})}$ .

b) The total stored energy increases after the dielectric is inserted.

Explanation:

<em><u>Before the dielectric is inserted:</u></em>

We have two parallel plate capacitors ( $C_{1}$ and $C_{2}$ ), each with a capacitance C, connected in series to a battery that has voltage V.

The two capacitors can be replaced by an equivalent one. The equivalent capacitance for two capacitors in series is

$\frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{2}{C}$

The potential differences across the two capacitors are:

$\Delta V_{1}=\frac{Q}{C_{1}}=\frac{Q}{C}$ and $\Delta V_{2}=\frac{Q}{C_{2}}=\frac{Q}{C}$

The total potential difference $\Delta V$ is the sum of the potential differences across each capacitor:

$\Delta V=\Delta V_{1}+ \Delta V_{2}=2\frac{Q}{C}$

The initial stored energy $U_{0}$ is:

$U_{0}=\frac{1}{2}C_{eq} \Delta V^{2}=\frac{1}{2}\frac{C}{2}(2\frac{Q}{C})^{2} \ \Longrightarrow\ U_{0}=\frac{Q^{2}}{C}$

Using $\Delta V=V$ instead we have that $U_{0}=\frac{1}{4}CV^{2}$ .

<u><em>After the dielectric is inserted:</em></u>

A dielectric with $\kappa > 1$ is inserted between the plates of one of the capacitors while the two capacitors are connected to the battery. The total potential difference reamins the same. If Q is the charge in the plates without the dielectric, in the capacitor with the dielectric the charge increases to $\kappa Q$ and the capacitance to $\kappa C$ .

Now we have that the equivalent capacitance is:

$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{\kappa C}=\frac{1}{C} (1+\frac{1}{\kappa})$

So the stored energy after the dielectric is inserted $U$ becomes:

$U=\frac{1}{2}C_{eq}\Delta V^{2}=\frac{1}{2} \frac{C}{(1+\frac{1}{\kappa})}V^{2} \Longrightarrow\ U=\frac{1}{2(1+\frac{1}{\kappa})}CV^{2}$

So, the ratio $\frac{U}{U_{0}}$ is:

$\frac{U}{U_{0}}=\frac{\frac{1}{2(1+\frac{1}{\kappa})}CV^{2}}{\frac{1}{4}CV^{2}} \Longrightarrow\ \frac{U}{U_{0}}=\frac{2}{(1+\frac{1}{\kappa})}$

Because $\frac{U}{U_{0}} >1$ if $\kappa >1$ this means that the total stored energy increases after the dielectric is inserted.

8 0

3 years ago