Making the material you are trying to memorize personally meaningful to you is called the ___________

A wire 3.22 m long and 7.32 mm in diameter has a resistance of 11.9 mΩ. A potential difference of 33.7 V is applied between the

Scorpion4ik [409]

Answer:

(a) Current is 2831.93 A

(b) $8.40A/m^2$

(c) $\rho =15.52\times 10^{-9}ohm-m$

Explanation:

Length of wire l = 3.22 m

Diameter of wire d = 7.32 mm = 0.00732 m

Cross sectional area of wire

$A=\pi r^2=3.14\times 0.00366^2=4.20\times 10^{-5}m^2$

Resistance $R=11.9mohm=11.9\times 10^{-3}ohm$

Potential difference V = 33.7 volt

(A) current is equal to

$i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A$

(B) Current density is equal to

$J=\frac{i}{A}$

$J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2$

(c) Resistance is equal to

$R=\frac{\rho l}{A}$

$11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}$

$\rho =15.52\times 10^{-9}ohm-m$

4 0

2 years ago

A bumblebee darts past at 3 m/s. The frequency of the hum made by its wings is 152 Hz. Assume the speed of sound to be 342 m/s.

Veronika [31]

It'll be 152 Hz at the exact instant the bumblebee
is right at the tip of your nose, on his way past you.

Before he gets there, while he's coming at you,
he sounds like a frequency higher than 152 Hz.

After he passes by, and is going away from you,
he sounds like a frequency lower than 152 Hz.

8 0

3 years ago

Read 2 more answers

A car is moving in the positive direction along a straight highway and accelerates at a constant rate while going from point A t

rusak2 [61]

Answer:

The time where the avergae speed equals the instaneous speed is T/2

Explanation:

The velocity of the car is:

v(t) = v0 + at

Where v0 is the initial speed and a is the constant acceleration.

Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

$v_{ave} = \frac{1}{T}\int\limits^T_0 {v(t)} \, dt$

v_ave = v0+a(T/2)

We can esaily note that when t=T/2

v(T/2)=v_ave

Now we want to know where the car should be, the osition of the car is:

$x(t) = x_A + v_0 t + \frac{1}{2}at^2$

Where x_A is the position of point A. Therefore, the car will be at:

x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2

8 0

2 years ago

Two children, Ferdinand and Isabella, are playing with a water hose on a sunny summer day. Isabella is holding the hose in her h

mojhsa [17]

Answer:

Isabella will not be able to spray Ferdinand.

Explanation:

We'll begin by calculating the time taken for the water to get to the ground from the hose held at 1 m above the ground. This can be obtained as follow:

Height (h) = 1 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =.?

h = ½gt²

1 = ½ × 9.8 × t²

1 = 4.9 × t²

Divide both side by 4.9

t² = 1/4.9

Take the square root of both side

t = √(1/4.9)

t = 0.45 s

Next, we shall determine the horizontal distance travelled by the water. This can be obtained as follow:

Horizontal velocity (u) = 3.5 m/s

Time (t) = 0.45 s

Horizontal distance (s) =?

s = ut

s = 3.5 × 0.45

s = 1.58 m

Finally, we shall compare the distance travelled by the water and the position to which Ferdinand is located to see if they are the same or not. This is illustrated below:

Ferdinand's position = 10 m

Distance travelled by the water = 1.58 m

From the above, we can see that the position of the water (i.e 1.58 m) and that of Ferdinand (i.e 10 m) are not the same. Thus, Isabella will not be able to spray Ferdinand.

8 0

2 years ago

about how much more energy is released in a 6.5 richter magnitude earthquake than in one with magnitude 5.5?

OverLord2011 [107]

Answer:

For example, an earthquake of magnitude 5.5 releases about 32 times as much energy as an earthquake measuring 4.5. Another way to look at this is that it takes about 900 magnitude 4.5 earthquakes to equal the energy released in a single 6.5 earthquake.

Explanation:

8 0

2 years ago

Making the material you are trying to memorize personally meaningful to you is called the ____________.