Which is a sub-atomic particle?

Carla holds a ball 1.5 m above the ground. Daniel, leaning out of a car window, also holds a ball 1.5 m above the ground. Daniel

spayn [35]

Answer:

Carla

Explination: As Daniel's ball is dropped from the car moving at 40 mph in a horizontal direction, at the time the ball is dropped it is also moving at 40 mph in a horizontal direction due to inertia, a property of mass causing resistance to change, Daniel's ball will continue to move in a horizontal direction even after being dropped along with falling due to gravity. Daniel's ball will then fall in a projectile motion curve of sorts which will cause an overall velocity to not be straight down causing it not to fall to the ground as quickly as Carla's ball.

Sorry for the long explanation

8 0

3 years ago

Read 2 more answers

Choose all of the items that are incorrectly matched. A) inactivation gate — closed at restB) activation gate — open at restC) i

Feliz [49]

D) activation gate - opens during depolarization

4 0

4 years ago

A parallel circuit is constructed with three different branches. An ammeter is placed in each of the branches to measure the bra

neonofarm [45]

Answer:

Current in the battery will be 11.7 A

Explanation:

We have given that the there is parallel circuit with three different branches.

Current in each branch is $I_1=2.8A$ , $I_2=4.1A$ , $I_2=4.8A$

We have to find the current I in the battery

It is known that in parallel circuit current is sum of current in each branches

So current in the battery will be equal to

$I=I_1+I_2+I_3$

$I=2.8+4.1+4.8=11.7A$

Therefore current in the battery will be 11.7 A

8 0

3 years ago

A student finds an unlabeled liquid container in his lab. He notices that the container has two liquids. Since the two liquids h

raketka [301]

Answer:

$\rho = 1848.03 kg m^{-3}$

Explanation:

given data:

density of water \rho = 1 gm/cm^3 = 1000 kg/m^3

height of water = 20 cm =0.2 m

Pressure p = 1.01300*10^5 Pa

pressure at bottom

$P = P_{fluid} + P_{h_2 o}$

$P = P_{fluid} + \rho g h$

$P_{fluid} = P - \rho g h$

= 1.01300*10^5 - 1000*0.2*9.8

= 99340 Pa

$p_{fluid} = P_{atm} + \rho g h_{fluid}$ h_[fluid} = 0.307m

$99340 = 104900 + \rho *9.8*0.307$

$\rho = 1848.03 kg m^{-3}$

5 0

4 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago