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3 years ago

7

As a general rule, how does mass affect the force of friction?

1 answer:

Setler79 [48]3 years ago

5 0

Mass affects it usually due to weight. The more mass something has, the heavier it will be. This will cause friction with gravity and the surface it is on :)

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Select all that apply

borishaifa [10]

a,b,c is your answer light and sound are not considered matter and heat is energy created from matter and electricity is particles moving basically therefore electricity is matter hope this helps

5 0

3 years ago

I wonder if people that say that there ugly is to get attention (not to be mean...)

alexandr1967 [171]

They do. mostly because they want validation from others

5 0

2 years ago

Read 2 more answers

Experiments and investigations must be ____. a. approved b. repeatable c. not reproducible d. accepted

deff fn [24]

Experiments and investigations must be B. Repeatable.

7 0

3 years ago

Read 2 more answers

A coil consists of 180 turns of wire. Each turn is a square of side d=30 cm, and a uniform magnetic field directed perpendicular

alexira [117]

Answer:

Emf induced i equal to 329.4 volt

Explanation:

Note : Here i think we have to find emf induced in the coil

Number of turns in the coil N= 180

Sides of square d = 30 cm = 0.3 m

So area of the square $A=0.3\times 0.3=0.09m^2$

Magnetic field is changes from 0 to 1.22 T

Therefore $dB=1.22-0=1.22T$

Time interval in changing the magnetic field dt = 0.06 sec

Induced emf is given by

$e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}$

$e=-180\times 0.09\times \frac{1.22}{0.06}=329.4volt$

8 0

3 years ago

Capacitor C1 is initially charged to V1 and capacitor C2 is initially charged to V2. The capacitors are then connected to each o

o-na [289]

Answer:

<em>20.08 Volts</em>

Explanation:

<u>Parallel Connection of Capacitors</u>

The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is

$\displaystyle V_1=\frac{Q_1}{C_1}$

$\displaystyle V_2=\frac{Q_2}{C_2}$

They are both the same after connecting them, thus

$\displaystyle \frac{Q_2}{C_2}=\frac{Q_1}{C_1}$

Or, equivalently

$\displaystyle Q_2=\frac{C_2Q_1}{C_1}$

The total charge of both capacitors is

$\displaystyle Q_t=Q_1\left(1+\frac{C_2}{C_1}\right)$

We can compute the total charge by using the initial conditions where both capacitors were disconnected:

$Q_t=V_{10}C_1+V_{20}C_2=25\cdot 24+13\cdot 11=743\ \mu C$

Now we compute Q1 from the equation above

$\displaystyle Q_1=\frac{Q_t}{\left(1+\frac{C_2}{C_1}\right)}=\frac{743}{\left(1+\frac{13}{24}\right)}=481.95\ \mu C$

The final voltage of any of the capacitors is

$\displaystyle V_1=V_2=\frac{481.95}{24}=20.08\ V$

7 0

3 years ago