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2 years ago

This graph indicates that the acceleration between 2.00 and 4.00 seconds is the same as that between 12.00 and 14.00 seconds.

Physics

1 answer:

adoni [48]2 years ago

5 0

I believe it’s false I’m probably wrong I ju at need points for my test

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From the mass spectrum of the compound, the molecule was determined to have a molar mass of 907 g/mol. what is its molecular for

Molodets [167]

Please elaborate more on your question so I can help you

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3 years ago

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole huma

gayaneshka [121]

Explanation:

Below is an attachment containing the solution.

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3 years ago

A proton moving at 8.9 × 106 m/s through a magnetic field of 0.96 T experiences a magnetic force of magnitude 3.8 × 10−13 N. Wha

goblinko [34]

Answer: 15.66 °

Explanation: In order to solve this proble we have to consirer the Loretz force for charge partcles moving inside a magnetic field. Thsi force is given by:

F=q v×B = qvB sin α where α is teh angle between the velocity and magnetic field vectors.

From this expression and using the given values we obtain the following:

F/(q*v*B) = sin α

3.8 * 10^-13/(1.6*10^-19*8.9*10^6* 0.96)= 0.27

then α =15.66°

8 0

3 years ago

1. Charges acquired by rubbing is called_____

Leokris [45]

Answer:

static electricity and then lightning rod

4 0

2 years ago

Read 2 more answers

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

3 years ago