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Levart [38]
3 years ago
13

A girl on a bike is moving at a speed of 1.40 m/s at the start of a 2.15 m high and 12.4 m long incline. The total mass is 57.0

kg, air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, and the speed at the lower end of the incline is 6.80 m/s. Determine the work done (in J) by the girl as the bike travels down the incline.
Physics
1 answer:
musickatia [10]3 years ago
5 0

Answer:

568.16J

Explanation:

V = 1.4 m/s

h = 2.15m

H = 12.4m

M = 57.0kg

Fr = 41.0N

Vb = 6.80m/s

g = 9.8m/s²

K.E at the top + P.E at the top + Work done = K.E at the bottom + frictional force.

½mv² + mgh + W = ½mVb + Fr.H

(½* 57 * 1.4²) + (57 * 9.8 * 2.15) + W = (½ * 57 * 6.8²) + (41 * 12.4)

55.86 + 1202.22 + w = 1317.84 + 508.4

1258.08 + w = 1828.24

W = 1826.24 - 1258.08

W = 568.16J

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The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.

They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.

The given data in the problem is;

q₁  is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C

F is the repulsive force = 36 mN =36 ×10⁶ N

d is the distance = 0.70 m

The Coulomb force is found as;

\rm F = \frac{Kq_1q_2}{r^2}\\\\\ \rm 36\times 10^6 = \frac{9 \times 10^9 }{(0.7)^2} \times 4.4 \times 10^{-9} \times q_2\\\\\ q_2 = 8.6241  \times 10^{-19 } \ C

Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.

To learn more about Coulomb's law, refer to the link;

brainly.com/question/1616890

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2 years ago
In what part of the earth does convection occur?
n200080 [17]
The Asthenosphere is where the convection currents in the Earth occor
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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
Most people can detect frequencies as high as 20 000 Hz. Assuming the speed of sound in air is 345 m/s, determine the wavelength
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Wavelength = speed / frequency

(345 m/s) / (20,000 Hz) = 0.017 m

6 0
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Afina-wow [57]

Answer:

5760 J

Explanation:

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

Mass of block = 48 kg

Height (h) = 12 m

Gravitational field strength (g) = 10 N/Kg

Gravitational potential energy (PE) =?

The gravitational potential energy stored by the block can simply be obtained as follow:

PE = mgh

PE = 48 × 10 × 12

PE = 5760 J

Therefore, the gravitational potential energy stored by the block is 5760 J

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