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3 years ago

Energy of motion is known as what energy

Physics

2 answers:

GaryK [48]3 years ago

3 0

Kinetic Energy = Energy of Motion

asambeis [7]3 years ago

3 0

Answer:

kinetic energy

Explanation:

USA testprep

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What is the current in mA through a 500 Ω resistor that is connected to a 1.5 V battery? Show all calculations using Ohm’s law (

Ludmilka [50]

Answer:

3 mA.

Explanation:

The following data were obtained from the question:

Resistor (R) = 500 Ω

Potential difference (V) = 1.5 V

Current (I) =.?

Using the ohm's law equation, we can obtain the current as follow:

V = IR

1.5 = I x 500

Divide both side by 500

I = 1.5 / 500

I = 3×10¯³ A.

Therefore, the current in the circuit is 3×10¯³ A.

Finally, we shall convert 3×10¯³ A to milliampere (mA).

This can be obtained as follow:

Recall:

1 A = 1000 mA

Therefore,

3×10¯³ A = 3×10¯³ × 1000 = 3 mA

Therefore, 3×10¯³ A is equivalent to 3 mA.

Thus, the current in mA flowing through the circuit is 3 mA.

3 0

3 years ago

The diffusion constant for oxygen diffusing through tissue is 1.0 × 10-11 m2/s. In a certain sample oxygen flows through the tis

Eduardwww [97]

Answer:

m' = 1 x 10⁻⁶ kg/s

Explanation:

Given that

Diffussion constant = 1 x 10⁻¹¹

Mass flow rate ,m = 2 x 10⁻⁶ kg/s

The diffusion is inversely proportional to the thickness of the membrane and therefore when the thickness is doubled, the mass flow rate would become half.

So new flow rate m'

$m'=\dfrac{m}{2}$

$m'=\dfrac{2\times 10^{-6}}{2}\ kg/s$

m' = 1 x 10⁻⁶ kg/s

6 0

3 years ago

A motorcycle is travelling at a constant velocity of 30ms. The motor is in high gear and emits a hum of 700Hz. The speed of soun

timurjin [86]

Answer:

a) $T=1.43\times 10^{-3}\ s$

b) $d=0.0429\ m$

c) $\lambda=0.4857\ m$

d) $f_o=767.7\ Hz$

Explanation:

Given:

velocity of the sound from the source, $S=340\ m.s^{-1}$

original frequency of sound from the source, $f_s=700\ Hz$

speed of the source, $v_s=30\ m.s^{-1}$

(a)

We know time period is inverse of frequency:

Mathematically:

$T=\frac{1}{f}$

$T=\frac{1}{700}$

$T=1.43\times 10^{-3}\ s$

(b)

Distance travelled by the motorcycle during one period of sound oscillation:

$d=v_s.T$

$d=30\times 1.43\times 10^{-3}$

$d=0.0429\ m$

(c)

The distance travelled by the sound during the period of one oscillation is its wavelength.

$\lambda=\frac{S}{f}$

$\lambda=\frac{340}{700}$

$\lambda=0.4857\ m$

(d)

observer frequency with respect to a stationary observer:

<u>According to the Doppler's effect:</u>

$\frac{f_o}{f_s}= \frac{S+v_o}{S-v_s}$ ...........................(1)

where:

$f_o\ \&\ v_o$ are the observed frequency and the velocity of observer respectively.

Here, observer is stationary.

$\therefore v_o=0\ m.s^{-1}$

Now, putting values in eq. (1)

$\frac{f_o}{700}= \frac{340+0}{340-30}$

$f_o=767.7\ Hz$

5 0

3 years ago

Suppose that the voltage is reduced by 10 percent (to 90 V). By what percentage is the power reduced

skad [1K]

Answer:

percent decrease in power = 20 % .

Explanation:

Power P = v² / R where v is voltage and R is resistance which is constant .

taking log on both sides

log P = 2 log v - log R

differentiating on both sides

dP/P = 2 dv/v

-dP/P x 100 = 2 x -dv/v x 100

percent decrease in power = 2 x percent decrease in v ( voltage )

percent decrease in power = 2 x 10 = 20 %

3 0

2 years ago

Distance travelled by a free falling object in the first second is: a) 4.9m b) 9.8m c) 19.6m d) 10m

zvonat [6]

Time=1s=t
Acceleration due to gravity=g=9.8m/s^2
Distance=s

In free fall

$\boxed{\sf s=-\dfrac{1}{2}gt^2}$

$\\ \sf\longmapsto s=-\dfrac{1}{2}\times 9.8(1)^2$

$\\ \sf\longmapsto s=-4.9(1)$

$\\ \sf\longmapsto s=-4.9m$

Take it positive

$\\ \sf\longmapsto s=4.9m$

Option a is correct

6 0

2 years ago