Why do you think nutrition experts recommend that young people<br> eat foods high in calcium?

In which of the following situations is the Doppler effect absent?

Bumek [7]

The source and the observer are moving towards each other. The observer is moving toward the source. The source is moving away from the observer

4 0

3 years ago

For this situation to demonstrate a balanced force, how much force must Omar apply? (In the picture, Sam is pushing a box to the

11111nata11111 [884]

The answers 10N as it equals it out

3 0

3 years ago

Read 2 more answers

Calculate the potential energy of a 6kg bowling ball suspended 3m above the surface of the earth

Valentin [98]

Answer: 176.4 Joules

Explanation: Potential energy of a body is defined as the energy possessed by virtue of position.

Potential energy $=m\times g\times h$

m = mass of the body = 6 kg

g = acceleration due to gravity = $9.8 ms^{-2}$

h = height of body = 3 m

Potential energy $=6 kg\times 9.8ms^{-2}\times 3m=176.4kgm^2s^{-2}$

= 176.4Joules { $1kgm^2s^{-2}=1Joule$ }

8 0

3 years ago

Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =

Troyanec [42]

Answer:

The magnitude of the electrostatic force on $\mathrm{+32\,\mu C}$ is: $\mathbf{12\;N}$

Explanation:

Consider

$q_1=+32\;\mu\text{C}$ is at position $x_1=0\;\text{m}$
$q_2=+20\;\mu\text{C}$ is at position $x_2=40\;\text{cm}$
$q_3=-60\;\mu\text{C}$ is at position $x_3=60\;\text{cm}$

The sum of all horizontal forces on $q_1$ is given as

$\sum F_x=F_{12}+F_{13}$

where

$F_{12}$ is the force exerted by $q_2$
$F_{13}$ is the force exerted by $q_3$

The force on $q_1$ exerted by $q_2$ is repulsive, so the direction of the force $F_{12}$ is to the left (negative direction). Thus

$F_{12}=-\frac{K\,|q_1|\,|q_2|}{r_{12}^2}\\F_{12}=-\frac{K\,|q_1|\,|q_2|}{(x_2\;-\;x_1)^2}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,|+32\;\mu C|\,|+20\;\mu C|}{(40\;cm\;-\;0\;cm)^2}}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,(32\;\mu C)\,(20\;\mu C)}{(40\;cm)^2}}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,(32\times10^{-6}\;C)\,(20\times10^{-6}\;C)}{(0.40\;m)^2}}\\F_{12}=\mathrm{-36\;N}$

The force on $q_1$ exerted by $q_3$ is attractive, so the direction of the force $F_{13}$ is to the right (positive direction). Thus

$F_{13}=+\frac{K\,|q_1|\,|q_3|}{r_{13}^2}\\F_{13}=+\frac{K\,|q_1|\,|q_3|}{(x_3\;-\;x_1)^2}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,|+32\;\mu C|\,|-60\;\mu C|}{(60\;cm\;-\;0\;cm)^2}}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,(32\;\mu C)\,(60\;\mu C)}{(60\;cm)^2}}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,(32\times10^{-6}\;C)\,(60\times10^{-6}\;C)}{(0.60\;m)^2}}\\F_{13}=\mathrm{+48\;N}$

Therefore

$\sum F_x=\mathrm{-36\;N+48\;N} \;\;\;\;\;\Rightarrow\;\;\;\;\; \mathbf{\sum F_x=+12\;N}$

the electrostatic force on $q_1$ is to the right and has a magnitude of $\mathbf{12\;N}$

8 0

3 years ago

How many octants are in the three-dimensional coordinate system?

Akimi4 [234]

There are eight octants in the three-dimensional coordinate system. From the word octants you can easily say or associate it to the number 8. Hope this answers the question. Have a nice day. Feel free to ask more questions.

8 0

4 years ago

Why do you think nutrition experts recommend that young people eat foods high in calcium?