What formula could be used to find distance if you know the speed an the time

Physics

1 answer:

gogolik [260]3 years ago

8 0

Answer: d = st

Explanation:

We know that the distance is equal to the rate (speed) times the time

d = st

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A coffee filter of mass 1.5 grams dropped from a height of 3 m reaches the ground with a speed of 0.7 m/s. How much kinetic ener

Mademuasel [1]

The kinetic energy gained by the air molecules is 0.0437 J

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Given:

Mass of a coffee filter, m = 1.5 g

Height from which it is dropped, h = 3 m

Speed at ground, v = 0.7 m/s

Initially, the coffee filter has potential energy. It is given by :

$P =mgh$

P = 1.5 × 10⁻³ kg × 9.8 m/s² × 3m

P = 0.0441 J

Finally, it will have kinetic energy. It is given by :

$E= \frac{1}{2} mv^{2}$

$E= \frac{1}{2}$ × $1.4$ × 10⁻³ × (0.7)²

E = 0.000343 J

The kinetic energy Kair did the air molecules gain from the falling coffee filter is :

E = 0.000343 - 0.0441

= 0.0437 J

So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.0437 J

Learn more about kinetic energy here:

brainly.com/question/8101588

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8 0

2 years ago

Satellite A is orbiting Earth at an altitude of 500 km and Satellite B is orbiting 800 km above the surface . How does the veloc

sp2606 [1]

The velocity of Satellite A is 2% greater than velocity of satellite B.

The given parameters;

<em>Altitude of Satellite A = 500 km</em>
<em>Altitude of Satellite B = 800 km</em>

The forces acting on the Satellites are given as follows;

$F_c = \frac{mv^2}{r} \\\\F_g = \frac{GMm}{r^2} \\\\\frac{v^2}{r} = \frac{GM}{r^2}\\\\v^2 = \frac{GM}{r} \\\\v^2 r = GM\\\\v_1^2 r _1 = v_2^2 r_2\\\\v_A^2r_A = v_B^2 r_B\\\\(\frac{v_A}{v_B} )^2 = \frac{r_B}{r_A} \\\\\frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A} } \\\\\frac{v_A}{v_B} = \sqrt{\frac{(800,000\ + \ 6.4 \times 10^6}{(500,000\ + \ 6.4 \times 10^6} )} \\\\\frac{v_A}{v_B} = 1.02 \\\\v_A = 1.02 \ v_B$