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3 years ago

14

a child is riding a bike at a speed of 0.6m/s with a total kinetic energy of 12.4J.If the mass of the child is 30kg, what is the

mass of the bike?

2 answers:

Nesterboy [21]3 years ago

8 0

Ek=kinetic energy (J)
m=mass of object (Kg)
v=speed of object (m/s)

We have the following formula:

Ek=(1/2)mv² ⇒ m=2(Ek)/v²

In this case:
m= mass of the child + mass of the bike
m=30 kg + m₁ (m₁=mass of the bike).

We would have to modify our formula like this:

m=2(Ek)/v²
m=30 kg + m₁

30 kg +m₁=2(Ek)/v²
m₁=2(Ek)/v²-30 kg

We have the following data:

v=0.6 m/s
Ek=12.4J
m=30 Kg + m₁

Therefore:

m₁=2(Ek)/v² - 30 kg
m₁=2(12.4 J)/(0.6 m/s)² - 30 kg
m₁=24.8 J/(0.36 m²/s²) -30 kg
m₁=68.89 Kg - 30 kg
m₁=38.89 Kg

Answer: the mass of the bike is 38.89 Kg.

vova2212 [387]3 years ago

8 0

The mass of the bike is 38.89 Kg.

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Answer:

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Thus, the magnitude of the angular momenta of both solar systems are given by:

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By taking into account that T1=3T2, we have

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HOPE THIS HELPS!!

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2 years ago

If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be

il63 [147K]

Answer:

Explanation:

We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .

∫ E ds = q / ε

Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so

∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .

This also represents total flux coming out of the charge q on all sides .

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3 years ago

A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera

yanalaym [24]

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

$\Delta V_{max}$ = 240 V

frequency, f = 50Hz

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To calculate the parameters of the given circuit series RLC circuit:

angular frequency, $\omega$ = $2\pi f = 2\pi \times50 = 100\pi$

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$Z = 2400\Omega$

d). Resistance, R is given by:

$Z = \sqrt {R^{2} + (X_{L} - X_{C})}$

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$R = \sqrt {5757839.238}$

$R = 2399.5\Omega$

e). Phase angle between current and the generator voltage is given by:

$tan\phi = \frac{X_{L} - X_{C}}{R}$

$\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})$

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$\phi = 1.11^{\circ}$

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ololo11 [35]

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