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3 years ago

3. What is the potential difference across a 1 500 12 resistor carrying a

1 answer:

6 0

The potential difference across a 1500 ohm resistor carrying a current of 0.075 A is approx. equal to 113 V.

Explanation:

An ideal electrical circuit obeys Ohm's law. As per this law, the potential difference in any electrical circuit will be directly proportional to the current flowing in that circuit. And the proportionality constant is the resistance offered by the circuit.

$V = I * R$

As here, the current is said to be carried as 0.075 A and the resistance is 1500 ohm, then the potential difference will be

$V = 0.075 * 1500 =112.5 = 113 V$

Thus, the potential difference across a 1500 ohm resistor carrying a current of 0.075 A is approx. equal to 113 V.

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Based on the magnetic field lines shown what is the orientation of the mystery magnet ?

babunello [35]

I think it’s cannot be determined

4 0

3 years ago

The net electric flux through a cubic box with sides that are 22.0 cm long is 4950 N⋅m2/C . What charge is enclosed by the box?

Bond [772]

Answer:

43.83 nC

Explanation:

We know Q = ε₀Ψ where Q = charge enclosed by the cubic box, ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m and Ψ = net electric flux through cubic box = 4950 Nm²/C

So, substituting the values of the variables into the equation, we have

Q = ε₀Ψ

= 8.854 × 10⁻¹² F/m × 4950 Nm²/C

= 43827.3 × 10⁻¹² C = 4.38273 × 10⁻⁸ C

= 43.8273 × 10⁻⁹ C

= 43.8273 nC

≅ 43.83 nC

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3 years ago

A motorcycle accelerates from a stop at a rate of 4m/s 2 for 20 seconds. It then continues at a

8_murik_8 [283]

After 20 s, the motorcycle attains a speed of

$\left(4\dfrac{\rm m}{\mathrm s^2}\right)(20\,\mathrm s)=80\dfrac{\rm m}{\rm s}$

and it continues at this speed for the next 40 s. So at 45 s, its speed is 80 m/s.

4 0

3 years ago

A stone is dropped from the upper observation deck of a tower, 750 m above the ground. (assume g = 9.8 m/s2.) (a) find the dista

Gekata [30.6K]

(a) The stone moves by uniform accelerated motion, with constant acceleration $g=9.81 m/s^2$ directed downwards, and its initial vertical position at time t=0 is 750 m. So, the vertical position (in meters) at any time t can be written as
$y(t)= y_0 - \frac{1}{2}gt^2= 750 - 4.9 t^2$

(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
$750-4.9 t^2 = 0$
from which we find the time t after which the stone reaches the ground:
$t= \sqrt{\frac{750 m}{4.9 m/s^2 }}= 12.37 s$

(c) The velocity of the stone at time t can be written as
$v(t) = -gt$
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
$v(12.37 s)=-(9.81 m/s^2)(12.37 s)=-121.3 m/s$

(d) if the stone has an initial velocity of $v_0 = 6 m/s$ , then its law of motion would be
$y(t)=y_0 - v_0t - \frac{1}{2}gt^2$
and we can find the time it needs to reach the ground by requiring again y(t)=0:
$0=750 - 6t - 4.9 t^2$
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.

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3 years ago

A water hose is used to fill a large cylindrical storage tank of

ludmilkaskok [199]

Answer:

maximum possible speed by solving above equation for 7D is

$v_{max} = \sqrt{\frac{49}{5}gD}$

minimum possible value of speed for solving x = 6D is given as

$v_{min} = \sqrt{9gD}$

Explanation:

Let the nozzle of the hose be at the origin. Then the nearest part of the rim of the tank is at (, ) = (6, 2) and the furthest part of the rim is at (, ) = (7, 2).

The trajectory of the water can be found as follows:

$x = (v_o cos45) t$