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Lisa [10]
3 years ago
12

1) The current in a light bulb is 0.335 Amps. How long does it take for a total charge of 2.76 C to

Physics
1 answer:
uysha [10]3 years ago
7 0

Answer:

Time=8.23880597 seconds

Explanation:

Quantity of charge(q)=2.76c

Current(I)=0.335A

Time(t)=?

t=q/I

t=2.76/0.335

t=8.23880597seconds

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6 0
3 years ago
The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,
kondaur [170]

Answer:

a= -2\ m/s^2

a=-12.5\ m/s^2

x=2.17 m

x=8.4 m

Explanation:

Given that

v=A+Bt^{-1}

v=2+0.25t^{-1}

To find acceleration :

we know that

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=0-0.5t^{-2}

a=-0.5t^{-2}

Acceleration at t= 2 s

a=-0.5\times 2^{-2}

a= -2\ m/s^2

Acceleration at t= 5 s

a=-0.5\times 5^{-2}

a=-12.5\ m/s^2

We know that

v=\dfrac{dx}{dt}

dx=\left(2+\dfrac{1}{4t}\right)dt

Position at t= 2 s:

\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{2}

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{5}

x=8+0.25 ln5

x=8.4 m

4 0
2 years ago
A test charge is placed at a distance of 2.5 × 10-2 meters from a charge of 6.4 × 10-5 coulombs. What is the electric field at t
Novosadov [1.4K]
Using the formula: E = kQ / d² where E is the electric field, Q is the test charge in coulomb, and d is the distance. 

E = kQ / d²

k = 9 x 10^9 N-m²/C²
Q = 6.4 x 10^-5 C
d = 2.5 x 10^-2 m

Substituting the given values to the equation, we have:
E = (9 x 10^9)(6.4 x 10^-5) / (2.5 x 10^-2) ²

Electric field at the test charge is 921600000 N/C
8 0
3 years ago
Read 2 more answers
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