All categories

2 years ago

The magnitude of the electric current is inversely proportional to the _____ of the circuits in R2D2.

1 answer:

4 0

All you need to know for this question is Ohm’s Law:

V=IR
I=V/R
R=V/I

So, the answer is (3) Resistance, since it is inversely proportional to Current (I=V/R)

You might be interested in

A 54 kg man holding a 0.65 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 12.1 m

Inessa [10]

Answer:

The velocity of the man is 0.144 m/s

Explanation:

This is a case of conservation of momentum.

The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.

Mass of ball = 0.65 kg

Mass of the man = 54 kg

Velocity of the ball = 12.1 m/s

Before collision, momentum of the ball = mass x velocity

= 0.65 x 12.1 = 7.865 kg-m/s

After collision the momentum of the man and ball system is

(0.65 + 54)Vf = 54.65Vf

Where Vf is their final common velocity.

Equating the initial and final momentum,

7.865 = 54.65Vf

Vf = 7.865/54.65 = 0.144 m/s

4 0

2 years ago

A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3

4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

$B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT$

8 0

2 years ago

The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the ob

givi [52]

Answer:

Charge, $Q=1.56\times 10^{-8}\ C$

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

$E=\dfrac{kQ}{r^2}$

Q is the charge on an object

$Q=\dfrac{Er^2}{k}$

$Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}$

$Q=1.56\times 10^{-8}\ C$

So, the charge on the object is $1.56\times 10^{-8}\ C$ . Hence, this is the required solution.

7 0

2 years ago

The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the

xz_007 [3.2K]

Answer:

a) -1.14 rev/min²

b) 9900 rev

c) -9.92×10⁻⁴ m/s²

d) 30.8 m/s²

Explanation:

First, convert hours to minutes:

2.2 h × 60 min/h = 132 min

a) Angular acceleration is change in angular velocity over change in time.

α = (ω − ω₀) / t

α = (0 rev/min − 150 rev/min) / 132 min

α = -1.14 rev/min²

b) θ = θ₀ + ω₀ t + ½ αt²

θ = 0 rev + (150 rev/min) (132 min) + ½ (-1.14 rev/min²) (132 min)²

θ = 9900 rev

c) The tangential component of linear acceleration is:

a_t = αr

First, convert α from rev/min² to rad/s²:

-1.14 rev/min² × (2π rad/rev) × (1 min / 60 s)² = -1.98×10⁻³ rad/s²

Therefore:

a_t = (-1.98×10⁻³ rad/s²) (0.50 m)

a_t = -9.92×10⁻⁴ m/s²

d) The magnitude of the net linear acceleration can be found from the tangential component and the radial component:

a² = (a_t)² + (a_r)²

The radial component is the centripetal acceleration:

a_r = v² / r

a_r = ω² r

First, convert 75 rev/min to rad/s:

75 rev/min × (2π rad/rev) × (1 min / 60 s) = 7.85 rad/s

Find the radial component:

a_r = (7.85 rad/s)² (0.50 m)

a_r = 30.8 m/s²

Now find the net linear acceleration:

a² = (-9.92×10⁻⁴ m/s²² + (30.8 m/s²)²

a = 30.8 m/s²

5 0

2 years ago

A pump operating at steady state receives liquid water at 508C, 1.5 MPa. The pressure of the water at the pump exit is 15 MPa. T

Irina-Kira [14]

Answer:

$n=76\%$

Explanation:

From the question we are told that:

Temperature $T=50 \textdegree C$

Intake Pressure $P=1.5MPa$

Exit Pressure $P=15MPa$

Work $W=18kJ$

Therefore

From Table

V_f=1.01*10^{-4}

Generally the equation for isentropic Work is mathematically given by

$W_{iso}=V(P_2-P_1)$

$W_{iso}=1.01*10^{-4}(15-1.5)$

$W_{iso}=13.622$

Since

Efficiency n

$n=\frac{W_{iso}}{W}$

$n=\frac{W_{iso}}{18}$

$n=0.76$

Therefore

$n=76\%$

3 0

2 years ago