You find a measurement or 34 kg written in a laboratory notebook. What physical characteristic is this a measurement of?

A father racing his son has 1/4 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.2 m

sergejj [24]

Answer:

Explanation:

KE_s: Kinetic Energy Son

KE_f: Kinetic Energy Father.

Relationship

KE_f: = (1/4) KE_s

m_s: = (1/3) m_f

v_f: = velocity of father

v_s: = velocity of the son

Relationship

1/2 mf (v_f + 1.2)^2 = 1/2 m_s (v_s)^2 Multiply both sides by 2.

mf (v_f + 1.2)^2 = m_s * (v_s)^2 Substitute for the mass of the m_s

mf (v_f + 1.2)^2 = (m_f/3) * (v_s)^2 Divide both sides by father's mass

(v_f + 1.2)^2 = 1/3 * (v_s)^2 multiply both sides by 3

3*(v_f + 1.2)^2 = (v_s) ^2 Take the square root both sides

√3 * (v_f + 1.2) = v_s

Note

You should work your way through all the cancellations to find the last equation shown about
We have another step to go. We have to use the first relationship to get the final answer.

KE_f = (1/4) KE_s Multiply by 4

4* KE_f = KE_s Substitute (again)

4*(1/2) m_f (v_f + 1.2)^2 = 1/2* (1/3)m_f *v_s^2 Divide by m_f

2* (v_f + 1.2)^2 = 1/6 * (v_s)^2 multiply by 6

12*(vf + 1.2)^2 = (v_s)^2 Take the square root

2*√(3* (v_f + 1.2)^2) = √(v_s^2)

2*√3 * (vf + 1.2) = v_s

Use the second relationship to substitute for v_s so you can solve for v_f

2*√3 * ( v_f + 1.2) = √3 * (v_f + 1.2) Divide by sqrt(3)

2(v_f + 1.2) = vf + 1.2

Edit

2vf + 2.4 = vf + 1.2

2vf - vf + 2.4 = 1.2

vf = 1.2 - 2.4

vf = - 1.2

This answer is not possible, but 2 of us are getting the same answer. The other person is someone whose math I would never question. She rarely makes an error. And I do mean rarely. Could you check to see that you have copied this correctly?

5 0

4 years ago

The velocity of an object includes its speed and what

SCORPION-xisa [38]

I believe that the correct answer you are looking for is the distance traveled

5 0

3 years ago

Read 2 more answers

Each gas sample has the same temperature and pressure. which sample occupies the greatest volume? each gas sample has the same t

Anit [1.1K]

Answer:

20 gram of H as the greastest volume

Explanation:

Each gas sample has the same temperature and pressure. which sample occupies the greatest volume? each gas sample has the same temperature and pressure. which sample occupies the greatest volume? 40.0 gar 20.0 gne 20.0 gh2 4.0 ghe

Given that,

40.0 g of Ar

20.0 g of Ne

20.0 g of h

24.0 g of he

The ideal gas equation ; Pv = nRT

where P is the pressure, V is the volume, R is the gas constant, T is the temperature, n is the number of mole

which is equal to

n = m / M

where m is the given mass

M is the molar mass

The given gas sample are at the same temperature and pressure,

So, the temperature, pressure and gas constant are fixed.

Therefore the volume is proportional to the given mass and molar mass

V ∝ m/M

Substitute the given mass and molar mass of given masses for comparison

No of mole of Ar = mass / molar mass

= 40 / 40 = 1 mole

No of mole of Ne = mass / molar mass

= 20 / 20 = 1 mole

No of mole of h = mass / molar mass

= 20 / 1 = 20 mole

No of mole of he = mass / molar mass

= 24 / 4 = 6 mole

Hence, 20 gram of H as the greastest volume

3 0

3 years ago

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

atroni [7]

(a) $2.79 rev/s^2$

The angular acceleration can be calculated by using the following equation:

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha$ is the angular acceleration

$\theta=50.0 rev$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\alpha$ , we find

$\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2$

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s$

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s$

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

$\theta=?$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\theta$ , we find

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev$

4 0

3 years ago

A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take t

marin [14]

Answer:

$T=66262.4s$

Explanation:

From the question we are told that:

Altitude $A=2.90 *10^7$

Mass $m=5.97 * 10^{24} kg$

Radius $r=6.38 *10^6 m.$

Generally the equation for Satellite Speed is mathematically given by

$V=(\frac{GM}{d} )^{0.5}$

$V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}$

$V=3354.83m/s$

Therefore

Period T is Given as

$T=\frac{2 \pi *a}{V}$

$T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}$

$T=66262.4s$

4 0

3 years ago