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3 years ago

If you double the frequency of a vibrating object, its period:

Physics

2 answers:

Furkat [3]3 years ago

8 0

Answer:

becomes halved.

Explanation:

trust

lilavasa [31]3 years ago

6 0

Answer:

becomes halved. or is reduced to half

Explanation:

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Three capacitors with capacitances:

mr_godi [17]

Answer:

A. 80.0microColoumbs

B.120.0 microcoloumbs

C.37.3v

Explanation:

See attached file

6 0

4 years ago

An electric car has a battery that can hold 16 kwh of energy. if the battery is designed to operate at 340 v, how many coulombs

Zanzabum

charge needed= 169412 C

Explanation:

energy stored in the battery= 16 KWh= 16 (1000 W)(3600 s)= 5.76 x 10⁷ J

Energy stored is given by U= q V

q= charge

V= voltage= 340 V

so 5.76 x 10⁷ = q (340)

q= 169412 C

4 0

4 years ago

What feature on the periodic table separates the metals and non metals?

sertanlavr [38]

Answer: the amphoteric line, mental- nonmetal line, metalloid line, staircase, semimetal line. it can be called any of these names

7 0

3 years ago

What will be the final velocity of a 5.0 g bullet starting from rest, if a net force of 45 N is applied over a time of 0.02s?

Wewaii [24]

We know there’s a change in momentum due to a force applied over a time interval. Ft= m[v(final)-v(initial)]. Now simply plug in know values: (45)(0.02)=.005[v(final)-0]. Remember converting grams to kilograms. Solve for v final

3 0

4 years ago

How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter to p

katrin [286]

Answer:

Required charge $q=2.6\times 10^{9}C$ .

$n=1.622\times 10^{10}\ electrons$

Explanation:

Given:

Diameter of the isolated plastic sphere = 25.0 cm

Magnitude of the Electric field = 1500 N/C

now

Electric field (E) is given as:

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

where,

k = coulomb's constant = 9 × 10⁹ N

q = required charge

r = distance of the point from the charge where electric field is being measured

The value of r at the just outside of the sphere = $\frac{25.0}{2}=12.5cm=0.125m$

thus, according to the given data

$1500N/C=\frac{9\times 10^{9}N\times q}{(0.125m)^2}$

$q=\frac{0.125^2\times 1500}{9\times 10^{9}}$

Required charge $q=2.6\times 10^{9}C$ .

Now,

the number of electrons (n) required will be

$n=\frac{required\ charge}{charge\ of\ electron}$

$n=\frac{2.6\times 10^{-9}}{1.602\times 10^{-19}}$

$n=1.622\times 10^{10}\ electrons$

6 0

4 years ago