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Nat2105 [25]
3 years ago
8

If an antelope can run 14.1 km in 39 min, how fast is it km/h(for reference , 110 km/h is 65 mph )

Physics
2 answers:
shtirl [24]3 years ago
8 0
It might make more sense putting it another way but this is basically it. you just take the minutes and divide them by 60 to convert them to hours. then simplify the ratio

adelina 88 [10]3 years ago
3 0
Should be 21.7 km/hr. Convert minutes to hours by dividing 39/60. then divide 14.1 by .65 which is 39 minutes converted to hours.
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One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet
Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

               \frac{G M m}{r^{2}}=m \omega^{2} r

The orbit’s period is given by,

               T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}

Where,

T_{e} = Earth’s period

T_{p} = planet’s period

M_{s} = sun’s mass

r_{e} = earth’s radius

Now,

             T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}

As, planet mass is equal to 0.7 times the sun mass, so

            T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}

Taking the ratios of both equation, we get,

             \frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}

            \frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}

           \frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}

Given T_{p}=9.5 \text { days } and T_{e}=365 \text { days }

          \frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}

         r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}

7 0
3 years ago
A radio receiver has an effective resistance of 300 Ω to the input signal on the antenna downlead. The signal voltage is 700 µV.
White raven [17]

Answer:

I = 2.33 µA

Explanation:

given,

Effective resistance (R)= 300 Ω

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current flow in the antenna = ?

Using ohm's law

V = I R

where I is the current flow

I = \dfrac{V}{R}

I = \dfrac{700\times 10^{-6}}{300}

      I = 2.33 x 10⁻⁶ A

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Hence, the current flow in the antenna downloaded is equal to I = 2.33 µA

6 0
3 years ago
Read 2 more answers
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Harman [31]
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4 0
3 years ago
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zhuklara [117]

Explanation:

Density = mass / volume

ρ = 5200 g / 1400 mL

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6 0
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den301095 [7]

Answer and Explanation:

I can't write the essay for you, but I can give you useful tips on what information should be in your essay.

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- Making sure that you have some sort of protection, whether that would be with a weapon, a tool, or something else.

- Making sure that you have safe water to drink that you brought with you.

- A back-up plan in case there is a reason that you need to leave.

- A first aid kit in case of injury.

- How to create a fire and use it safely.

- A way to get back home, with a plan and possibly even a compass.

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I hope that this helps.

7 0
2 years ago
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