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3 years ago

What is the law of conservation of energy.

1 answer:

3 0

The law of conservation of energy implies that energy can neither be created nor destroyed, but can be changed from one form to another.

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A current of 1.4 A flows in a conductor for 7.0 s. How much charge passes a given point in the conductor during this time?

bija089 [108]

Answer:

The charge passes a given point in the conductor during this time is 9.8 C.

Explanation:

Given that,

Current = 1.4 A

Time = 7.0 sec

We need to calculate the charge during this time

Charge :

Charge is the product of current and time.

In mathematically form,

$Q = i\times t$

Where, i = cirrent

t = time

Put the value into the formula

$Q =1.4\times7.0$

$Q=9.8\ C$

Hence, The charge passes a given point in the conductor during this time is 9.8 C.

5 0

3 years ago

Organ pipe A, with both ends open, has a fundamental frequency of 360 Hz. The third harmonic of organ pipe B, with one end open,

iragen [17]

Answer:

The length of the pipe A is $L_A$ = 0.4763 m & the length of the pipe B is $L_B =$ 0.357 m

Explanation:

Fundamental frequency = 360 Hz

Velocity = 343 $\frac{m}{s}$

(a). Length of the pipe is given by

$L = \frac{V}{2 f}$

Put all the values in above equation we get

$L = \frac{343}{2 (360)}$

$L_A$ = 0.4763 m

(b). Given that

The third harmonic of organ pipe B = the second harmonic of pipe A

$\frac{n_B}{4L_B} = \frac{n_A}{2L_A}$

Thus

$L_B = \frac{2 n_{B} L_A }{4n_A}$

Put all the values in above formula we get

$L_B = \frac{2 (3)(0.4763) }{4(2)}$

$L_B =$ 0.357 m

Therefore the length of the pipe A is $L_A$ = 0.4763 m & the length of the pipe B is $L_B =$ 0.357 m

3 0

3 years ago

After 24.0 days 2.00 milligrams of an original 128.0. Milligram sample remain what is the half life of the sample

alina1380 [7]

Answer:

4 days

either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2

n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2

Explanation:

5 0

3 years ago

A 1.4-kg block slides freely across a rough surface such that the block slows down with an acceleration of â1.25 m/s2. what is t

Marianna [84]

Mass of the block = 1.4 kg

Weight of the block = mg = 1.4 × 9.8 = 13.72 N

Normal force from the surface (N) = 13.72 N

Acceleration = 1.25 m/s^2

Let the coefficient of kinetic friction be μ

Friction force = μN

F(net) = ma

μmg = ma

μg = a

μ = $\frac{a}{g}$

μ = $\frac{1.25}{9.8}$

μ = 0.1275

Hence, the coefficient of kinetic friction is: μ = 0.1275

6 0

4 years ago

A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 41.5-gram mass is attached at the 28.5-cm

zzz [600]

Answer:

The value is $M = 42.3 \ kg$

Explanation:

From the question we are told that

The first position of the fulcrum is x = 49.7 cm

The mass attached is $m = 41.5 \ g$

The position of the attachment is $x_1 = 28.5 \ cm$

The second position of the fulcrum is $x_2 = 39.2 \ cm$

Generally the sum of clockwise torque = sum of anti - clockwise torque

$CWT = m (x_2 - x_1)$

Here CWT stands for clockwise torque

$ACWT = M ( x - x_2)$

$m (x_2 - x_1) = M ( x - x_2)$

=> $41.5 (39.2 - 28.5 ) = M ( 49.7 -39.2 )$

=> $M = 42.3 \ kg$

3 0

3 years ago