Can an object have positive acceleration and decreasing speed?.

Hydrometer is a devise used to measure the volume of a liquid. *<br> True<br> False

RSB [31]

Answer:

false

Explanation:

3 0

2 years ago

A proton is located at <3 x 10^-10> m. What is r, the vector from the origin to the location of the proton

Eduardwww [97]

Complete Question

A proton is located at <3 x 10^{-10}, -5*10^{-10} , -5*10^{-10}> m. What is r, the vector from the origin to the location of the proton

Answer:

The vector position is $\=r=$

Explanation:

From the question we are told that

The position of the proton is $m$

Generally the vector location of the proton is mathematically represented as

$\= r =$

So substituting values

$\=r=$

4 0

3 years ago

Help me please please?

Rainbow [258]

Answer: I looked it up and it says something about the waves traveling in a solid but I don’t know if that’s correct.

4 0

3 years ago

Read 2 more answers

A 30 ohm resistor and a 20 ohm resistor are

Reptile [31]

The current that would pass through the 30 ohms resistor is 2 A.

<h3>What is electric current?</h3>

Electric current is the rate of flow of electric charge round a conductor.

To calculate the electric current that would pass through the 30 ohms resistor, we use the formula below

Formula:

I = V/Rt........... Equation 1

Where:

I = Electric current passing through the 30 ohms resistor
V = Voltage
Rt = Total or effective resistance of the resistors.

From the question,

Given:

V = 100 volts
Rt = (30+20) ohms (since both resistors are connected in series)

Rt = 50 ohms

Substitute these values into equation 1

I = 100/50
I = 2 A

Hence, The current that would pass through the 30 ohms resistor is 2 A.

Learn more about electric current here: brainly.com/question/1100341

8 0

2 years ago

A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

3 0

3 years ago