Determine the total momentum before the impact. Picture attached.

How long does it take electrons to get from the car battery to the starting motor? Assume the current is 143 A and the electrons

Ostrovityanka [42]

Answer

the concentration of electron/ volume (n) = 8.5 × 10²⁸ /m³

the drift velocity of electron

$v = \dfrac{I}{neA}$

$v = \dfrac{143}{8.5\times 10^{28}\times 1.6\times 10^{-19}\times 40.2 \times 10^{-6}}$

v = 2.615 × 10⁻⁴ m/s

time require is

$t=\dfrac{L}{v}$

$t=\dfrac{0.845}{2.615\times 10^{-4}}$

t = 3231.35 s

5 0

4 years ago

Archerfish are tropical fish that hunt by shooting drops of water from their mouths at insects above the waterÂs surface to knoc

sergij07 [2.7K]

Answer:

.012

Explanation:

Take the mass of the fish and divide it by the mass of the water:

65/.30=216.667

Divide the given speed by the value we found above:

2.5/216.667=.0115

Answer can be rounded up to .012

3 0

3 years ago

Read 2 more answers

A 128 gram Radium-226 sample with a half-life of 2.25 billion years is found. How many grams will remain after two half-lives?

Bess [88]

Answer:

Option B. 32 g

Explanation:

From the question given above, the following data were obtained:

Original amount (N₀) = 128 g

Half-life (t½) = 2.25 billion years

Number of half-lives (n) = 2

Amount remaining (N) =?

The amount of 128 gram of Radium-226 that will remain after 2 half-lives has elapsed can be obtained as followb

N = 1/2ⁿ × N₀

N = 1/2² × 128

N = 1/4 × 128

N = 0.25 × 128

N = 32 g

Therefore, 32g of the sample will remain.

8 0

3 years ago

Using this illustration, what is the chemical formula

forsale [732]

Answer:la primera o la tercera

Explanation:

Porque son 3 átomos en uno

4 0

3 years ago

Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an

ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

$Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega$

Power factor is given by

$F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802$

The power factor is 0.28802

The average power to the circuit is given by

$P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W$

The average power to the circuit is 2.57162 W

Power to resistor

$P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W$

Power to resistor is 14.28 W

Power to inductor

$P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W$

Power to the inductor is 53.55 W

Power to the capacitor

$P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W$

The power to the capacitor is 6.07142 W

8 0

3 years ago