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madam [21]
3 years ago
14

The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on

the planet’s surface? Round your answer to the nearest whole number.
Physics
2 answers:
choli [55]3 years ago
7 1

Answer:

25 times the average speed

Allisa [31]3 years ago
6 0

Answer:

Tangential speed of objects on the planet’s surface = 241 m/s

Explanation:

Speed is given by the ratio of distance traveled to the time.

Radius of mars, r = 3397 km = 3.397 x 10⁶ m

Distance traveled = 2πr = 2 x π x 3.397 x 10⁶ = 2.13 x 10⁷ m

Time taken = 24.6 hours = 24.6 x 60 x 60 = 88560 s

\texttt{Tangential speed =}\frac{2.13\times 10^7}{88560}=240.51m/s

Tangential speed of objects on the planet’s surface = 240.51 m/s = 241 m/s

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Answer:

Left

Explanation:

The force is applied opposite of the acceleration.

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The complete combustion of a small wooden match produces approximately 512 cal of heat. How many kilojoules are produced? Expres
k0ka [10]

We have to covert 512 cal of heat in kilo joules.

As, 1 cal = 0.004184 kJ = 4.184 joules.

Therefore,

512 \ cal = 512 \times 4.184 \ J = 2142.208 \ J \\\\\ =  2.142 \times 10 ^3 J = 2.142 \ kJ

Thus, combustion of a small wooden match produces approximately ( in kilo joules ) is 2.142 kJ .

5 0
3 years ago
What will a spring scale read for the weight of a 75.0-kg woman in an elevator that moves upward with constant speed of 5.8 m/s
Sholpan [36]
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7 0
3 years ago
When the frequency of an ac circuit is decreased, the current in the circuit increases. which combination of elements is most li
zaharov [31]

Answer:

The combination of elements most likely to comprise the circuit are resistor, inductor and capacitor

Explanation:

The impedance of an LCR circuit shown as

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     Z​ = √R² + (2π∨L - 1/2π∨c)²

Variation of Z with respect to υ is shown in the figure.

As υ increases, Z decreases and so the current increases.

At υ = υ↓r

Z is minimum, current is maximum. Beyond

    υ = υ↓r

​Z increases and so current decreases.

so the combination of circuit elements that is most suitable to comprise

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brainly.com/question/13140756

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5 0
1 year ago
A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?a.What is the speed of the mountain lio
Arisa [49]

Answer:

What is the speed of the mountain lion as it leaves the ground?

9.98m/s

At what angle does it leave the ground?

50.16°

Explanation:

This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.

To solve this, you will need use two kinematic equations and SOHCAHTOA:

d = v_it + \dfrac{1}{2}at^{2}\\\\vf = vi + at

With these formulas, we can derive formulas for everything you need:

Things you need to remember:

  • A projectile at an angle has a x-component (horizontal movement) and y-component (vertical movement), which is the reason why it creates an angle.
  • Treat them separately.
  • At maximum height, the vertical final velocity is always 0 m/s going up. And initial vertical velocity is 0 m/s going down.
  • Horizontal movement is not influenced by gravity.
  • acceleration due to gravity (a) on Earth is constant at 9.8m/s

First we need to take your given:

10.0 m long (horizontal) and maximum height of 3.0m (vertical).

d_x=10.0m\\d_y=3.0m

What your problem is looking for is the initial velocity and the angle it left the ground.

Vi = ?     Θ =?

Vi here is the diagonal movement and do solve this, we need both the horizontal velocity and the vertical velocity.

Let's deal with the vertical components first:

We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.

d_y=V_i_yt+\dfrac{1}{2}at^{2}

Since it is at maximum height at this point, we can assume that the lion is already making its way down so the initial vertical velocity would be 0 m/s. So we can reduce the formula:

d_y=0+\dfrac{1}{2}at^{2}

d_y=\dfrac{1}{2}at^{2}

From here we can derive the formula of time:

t=\sqrt{\dfrac{2d_y}{a}}

Now we just plug in what we know:

t=\sqrt{\dfrac{(2)(3.0m}{9.8m/s^2}}\\t=0.782s

Now that we know the time it takes to get from the highest point to the ground. The time going up is equal to the time going down, so we can use this time to solve for the intial scenario of going up.

vf_y=vi_y+at

Remember that going up the vertical final velocity is 0m/s, and remember that gravity is always moving downwards so it is negative.

0m/s=vi_y+-9.8m/s^{2}(0.782s)\\-vi_y=-9.8m/s^{2}(0.782s)\\-vi_y=-7.66m/s\\vi_y=7.66m/s

So we have our first initial vertical velocity:

Viy = 7.66m/s

Next we solve for the horizontal velocity. We use the same kinematic formula but replace it with x components. Remember that gravity has no influence horizontally so a = 0:

d_x=V_i_xt+\dfrac{1}{2}0m/s^{2}(t^{2})\\d_x=V_i_xt

But horizontally, it considers the time of flight, from the time it was released and the time it hits the ground. Also, like mentioned earlier the time going up is the same as going down, so if we combine them the total time in flight will be twice the time.

T= 2t

T = 2 (0.782s)

<em>T = 1.564s</em>

<em>So we use this in our formula:</em>

<em>d_x=V_i_xT\\\\10.0m=Vi_x(1.564s)\\\\\dfrac{10.0m}{1.564s}=V_i_x\\\\6.39m/s=V_i_x</em>

Vix=6.39m/s

Now we have the horizontal and the vertical component, we can solve for the diagonal initial velocity, or the velocity the mountain lion leapt and the angle, by creating a right triangles, using vectors (see attached)

To get the diagonal, you just use the Pythagorean theorem:

c²=a²+b²

Using it in the context of our problem:

Vi^{2}=Viy^2+Vix^2\\Vi^2=(7.66m/s)^2+(6.39m/s)^2\\\sqrt{Vi}=\sqrt{(7.66m/s)^2+(6.39m/s)^2}\\\\Vi=9.98m/s

The lion leapt at 9.98m/s

Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:

Tan\theta=\dfrac{O}{A}\\\\Tan\theta=\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{7.66m/s}{6.39m/s}\\\\\theta=50.17

The lion leapt at an angle of 50.16°.

6 0
2 years ago
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