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2 years ago

5

A person travels by car from Tucson to Phoenix at a constant speed of 75 km/hr. They then return from Phoenix to Tucson at a con

stant speed of 65 km/hr. What was their average velocity?

1 answer:

kompoz [17]2 years ago

7 0

Answer:

$v=0$

Explanation:

Knowing that the formula for average velocity is:

$v=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}$

Being said that, we know that the person's displacement is zero because it returns to its starting point

$x_{2}=x_{1}$

That means $x_{2}-x_{1}=0$

$v=\frac{0}{t_{2}-t_{1}}=0$

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1 year ago

Read 2 more answers

(6) A 75 kg human total footprint area is 0.05 m2 when wearing winter boots. Suppose that you want to walk on snow that can at m

fredd [130]

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0.25m²

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We know that the summation of forces in the vertical direction is zero

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PA-mg=0

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2 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$

Generally from the law energy conservation we have that

$T__{E}} = KE_p$

So

$2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

4 0

2 years ago