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2 years ago

What is the difference in KE between a 52.5 kg person running 3.50 m/s and a 0.0200 kg bullet flying 450 m/s?

Physics

1 answer:

rusak2 [61]2 years ago

8 0

Answer:

Ek = 1705.28 [J]

Explanation:

In order to solve this problem, we must remember that kinetic energy can be calculated by means of the following equation.

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

where:

m = mass [kg]

v = velocity [m/s]

Ek = kinetic energy [J] (Units of Joules)

For the person running

$E_{k} =\frac{1}{2}*52.2*(3.5)^{2} \\ E_{k} =319.72[J]$

For the bullet

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

$E_{k} =\frac{1}{2} *0.02*(450)^{2} \\E_{k}=2025 [J]$

The difference in Kinetic energy is equal to:

Ek = 2025 - 319.72

Ek = 1705.28 [J]

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What was the name of the voyages taken by Zheng he during the Ming dynasty on behalf of China

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Answer:

Ming treasure voyages

Explanation:

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3 years ago

Which three characteristics of an object are represented by a motion map

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There are three things that can be represented on a motion map

These three things are:

1)Motion

2)Acceleration

3)Velocity

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3 years ago

Energy of fossil fuels is also derieved from solar energy

Marat540 [252]

Answer:

Explanation:

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2 years ago

Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

$v^2-u^2=2gh$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

$g=-9.8 m/s^2$ is the acceleration of gravity

Solving for u,

$u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s$

The time needed to reach the maximum height can now be found by using the equation

$v=u+gt$

Solving for t,

$t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s$

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

$v'^2-u^2=2gh'$

we find

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s$

So the time to go from h' to h is

$\Delta t = t-t'=0.52-0.32=0.20 s$

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

$\Delta t=2\cdot 0.20 = 0.40 s$

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

$v'^2-u^2=2gh'$

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s$

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

$\Delta t = 2\cdot 0.04 =0.08 s$

8 0

3 years ago

What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?

alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

5 0

3 years ago