Ask question

Ask question

All categories

3 years ago

15

3. Ohm’s Law is represented by the equation I=V/R. Explain how the current would change if the amount of resistance decreased an

d the voltage stayed the same. (HINT: try plugging numbers into the equation for R, but make sure V stays the same. Then describe what happens to I(current)

1 answer:

Softa [21]3 years ago

6 0

When the resistance decreases, the current increases.

You might be interested in

An electron is placed at a position on the x-axis where the electric potential is + 10 V. Which idea below best describes the fu

ludmilkaskok [199]

Answer:

option (e)

Explanation:

As the electron is placed at a point where the potential is + 10 V. electrons moves from lower potential to higher potential, so we cannot predict the direction of motion because we don't have any information about the potential at other point.

So, option (e) is correct.

7 0

3 years ago

A metal rod with a length of 22.0cm lies in the xy-plane and makes an angle of 37.9 degrees with the positive x-axis and an angl

Shtirlitz [24]

Answer:

Induced emf in the rod is, $\epsilon=0.08262\ V$

Explanation:

Given that,

Length of the rod, L = 22 cm = 0.22 m

Angle with x axis is 37.9 degrees with the positive x-axis and an angle of 52.1 degrees with the positive y-axis.

$L=0.22(cos37.9\ i)+0.22(sin37.9\ j)$

$L=(0.173\ i+0.135\ j)\ m$

Velocity of the rod, v = 6.8i m/s

Magnetic field, $B=0.170i-0.24j-0.09k$

The formula for the emf induced in the rod is given by :

$\epsilon=(v\times B){\cdot}L$

$\epsilon=(6.8i\times (0.170i-0.24j-0.09k)){\cdot} (0.173\ i+0.135\ j)$

$\epsilon=(0.612j-1.632k){\cdot}(0.173i+0.135j)$

$\epsilon=0.08262\ V$

So, the magnitude of the emf induced in the rod is 0.08262 volts. Hence, this is the required solution.

8 0

4 years ago

A ball is thrown upward with a speed of 40 m/s. Approximately how much time goes it take the ball to travel from the release loc

yulyashka [42]

Answer:

(I). The time at highest point 4.0 sec.

(II). It returns to back to its original height in 8.1 sec

Explanation:

Given that,

Velocity $v=40\ m/s$

(I). We need to calculate the time at highest point

Using equation of motion

$v=u+gt$

Where, v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time

Put the value into the formula

$0=40-9.8t$

$-40=-9.8t$

$t=\dfrac{40}{9.8}$

$t=4.0\ sec$

(II). We know that, when the ball to travel from the initial point and reached at initial point then the displacement is zero.

We need to calculate the total time when it returns to back to its original height

Using equation of motion

$s=ut-\dfrac{1}{2}gt^2$

Where, s = displacement

g = acceleration due to gravity

t = time

u = velocity

Put the value in the equation

$0=40t-\dfrac{1}{2}9.8t^2$

$80t-9.8t^2=0$

$t=0, 8.1 sec$

Hence. (I). The time at highest point 4.0 sec.

(II). It returns to back to its original height in 8.1 sec

5 0

3 years ago

(b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a c

andreyandreev [35.5K]

Answer:

$\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}$

Explanation:

The formula for the surface of the circle is:

$A(r) = \pi\cdot r^{2}$

The rate of change of the spill area is obtained by deriving the previous formula in terms of time:

$\frac{dA}{dt} = 2\pi\cdot r\cdot \frac{dr}{dt}$

Finally, variables are replaced by known data:

$\frac{dA}{dt} = 2\pi\cdot (39\,m)\cdot (1\,\frac{m}{s} )$

$\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}$

4 0

4 years ago

Read 2 more answers

What question did use of a telescope by hubble answer A) Is earth the center of the universe B) Is the sun the center of the uni

cestrela7 [59]

Answer: C

Explanation:

3 0

3 years ago

Read 2 more answers