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PSYCHO15rus [73]

3 years ago

9

Is kirchoff's law applicable for ac circuits?

1 answer:

SIZIF [17.4K]3 years ago

4 0

<span>Kirchhoff's laws apply to AC circuits to either two cases: instantaneousneous values of currents and voltages or to complex values of currents and voltages. However, this never applies to: rms values of currents and voltages. Kirchoff's law relates to the current, voltage and resistance to multiple nodes</span>

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Why is it easier to open a door by holding it from it's edge

Crazy boy [7]

Answer:

leverage

Explanation:

Leverage ...this allows you to use less force through a longer distance

3 0

2 years ago

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

7 0

3 years ago

During the last shot of the game, the basketball goes from rest to 15 m/s and reaches the backboard in 0.41 s. What was the acce

Anna [14]

Answer: a= 37m

Explanation: V= 15 m/s (Velocity) t= 0.41s (time) formula: a= v/t

15 m/s / 0.41 (15 divided by 0.41) = 36.583m

There are 2 significant digits, 36, you look at the third digit, either round up or down in this case up to 36. a= 37m

5 0

3 years ago

If the angular velocity of the pully is -8.4rad/s at a given time, and its anglar acceleration is -2.8rad/s2, what is the angula

snow_lady [41]

Here we know that

$\omega_i = - 8.4 rad/s$

$\alpha = - 2.8 rad/s^2$

$t = 1.5 s$

now from kinematics we have

$\omega_f = \omega_i + \alpha t$

now from above all values we have

$\omega_f = (-8.4 rad/s) + (-2.8 rad/s^2)(1.5)$

$\omega_f = -8.4 + (-4.2)$

$\omega_f = -12.6 rad/s$

so final angular speed is -12.6 rad/s

6 0

3 years ago

If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an

lana [24]

Answer:

Explanation:

Given that,

Mass of star M(star) = 1.99×10^30kg

Gravitational constant G

G = 6.67×10^−11 N⋅m²/kg²

Diameter d = 25km

d = 25,000m

R = d/2 = 25,000/2

R = 12,500m

Weight w = 690N

Then, the person mass which is constant can be determined using

W =mg

m = W/g

m = 690/9.81

m = 70.34kg

The acceleration due to gravity on the surface of the neutron star is can be determined using

g(star) = GM(star)/R²

g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²

g (star) = 8.49 × 10¹¹ m/s²

Then, the person weight on neutron star is

W = mg

Mass is constant, m = 70.34kg

W = 70.34 × 8.49 × 10¹¹

W = 5.98 × 10¹³ N

The weight of the person on neutron star is 5.98 × 10¹³ N

5 0

3 years ago