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grin007 [14]
3 years ago
14

What is the current through a 11v bulb with a power of 99w

Physics
1 answer:
Murrr4er [49]3 years ago
8 0

Answer:

9.01amp

Explanation:

Power = V^2/R

Given that v = 11volts, P = 99watts

99 = 11^2/R

11×11 = 99R

121= 99R

R = 121/99

R= 1.22ohms

From ohms Law; V = IR

11volts = I × 1.22ohms

I = 11/1.23

I = 9.01 amp

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Answer:

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At high pressure- \mu_{k}=0.00445

Explanation:

Initial speed, V_{i}=3.3 m/s

Final speed, V_{f}=3.3/2= 1.65 m/s

Net horizontal force due to rolling friction F_{net}=\mu_{k} mg where m is mass, g is acceleration due to gravity, \mu_{k} is coefficient of rolling friction

From kinematic relation, V_{f}^{2}- V_{i}^{2}=2ad

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V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd

Making \mu_{k} the subject

\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}

Under low pressure of 40 Psi, d=18 m

\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315

Therefore, \mu_{k}=0.02315

At a pressure of 105 Psi, d=93.7

\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445

Therefore, \mu_{k}=0.00445

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