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3 years ago

Charges may be transferred by:

2 answers:

5 0

There are three methods by which charges can be transferred to build up static electricity: charging by friction, by conduction, and by induction. Charging by Friction When two uncharged objects rub together, some electrons from one object can move onto the other object

Ket [755]3 years ago

3 0

Friction, conduction and induction. Those are the answers.

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Which compound is most likely a covalent compound? Melting Boiling

mote1985 [20]

Answer:

compound A

Explanation:

because covalent compounds can't conduct electricity and because they also have low boiling points

7 0

3 years ago

Question 1

frez [133]

B 20 m/s
It should go to 100 that fast nor 40

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3 years ago

What are the ways of heat transfer?

AnnZ [28]

Answer:

Conduction, Convection and Radiation

Explanation:

Hope this helps!

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3 years ago

Read 2 more answers

Peter designed a road with a curve of radius 30 m that is banked so that a 950 kg car traveling at 40.0 km/h can round it even i

spayn [35]

Answer:

$v = 15.56 m/s$

$v = 56 km/h$

Explanation:

When coefficient of friction is approximately zero then we have

$F_ncos\theta = mg$

$F_n sin\theta = \frac{mv^2}{R}$

$tan\theta = \frac{v^2}{Rg}$

here we know that

$v = 40 km/h = 11.11 m/s$

R = 30 m

$tan\theta = \frac{11.11^2}{30\times 9.81}$

$\theta = 22.75 degree$

now when friction coefficient is 0.30 then we have

$F_n cos\theta = mg + F_f sin\theta$

$F_f cos\theta + F_n sin\theta = \frac{mv^2}{R}$

now we have

$v = \sqrt{Rg(\frac{\mu + tan\theta}{1 - \mu tan\theta})}$

$v = \sqrt{30(9.81)(\frac{0.30 + tan22.75}{1 - (0.30) tan22.75})}$

$v = 15.56 m/s$

$v = 56 km/h$

3 0

3 years ago

A printer is connected to a 1.0 m cable. if the magnetic force is 9.1 × x10-5 n, and the magnetic field is 1.3 × 10-4 t, what is

Valentin [98]

The magnetic force on a current-carrying wire due to a magnetic field is given by
$F=ILB$
where
I is the current
L the wire length
B the magnetic field strength

In our problem, L=1.0 m, $F=9.1 \cdot 10^{-5} N$ and $B=1.3 \cdot 10^{-4}T$ , so we can re-arrange the formula to find the current in the wire:
$I= \frac{F}{LB}= \frac{9.1 \cdot 10^{-5} N}{(1.0 m)(1.3 \cdot 10^{-4} T)} =0.7 A$

5 0

3 years ago