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2 years ago

4

Blender does 5000 J of work on the food in its bowl. During the time the blender runs, 2000 J of heat is transferred from the wa

rm food to the cooler environment. What is the change in the thermal energy of the food?

1 answer:

adelina 88 [10]2 years ago

4 0

Answer:

The total change is 3,000 J.

Explanation:

Hope this helps. (:

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A person drops a stone down a well and hears the echo 8.9 s later. if it takes 0.9 s for the echo to travel up the well, approxi

Temka [501]

Total time in between the dropping of the stone and hearing of the echo = 8.9 s

Time taken by the sound to reach the person = 0.9 s

Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds

Initial speed (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s^2

Time taken (t) = 8 seconds

Let the depth of the well be h.

Using the second equation of motion:

$h = ut + \frac{1}{2}\times a \times t^2$

$h = 0 \times 8 + \frac{1}{2} \times 9.8 \times 8^2$

h = 313.6 m

Hence, the depth of the well is 313.6 m

4 0

2 years ago

What is the direction of the magnetic field b⃗ net at point a? Recall that the currents in the two wires have equal magnitudes.

andrew11 [14]

Answer:

Explanation:

The direction of a magnetic field indicates where the magnetic inluence on the electric charges are directed to.

From the given question, we are to determine the direction of the magnetic field bnet at a point A.

Also, having the notion that the currents in the two wires have equal magnitudes, Then:

$\bar{B_{net}} = \bar{B_1} + \bar{B_2}$

$\bar{B_{net}} = \frac{\mu_oI}{2 \pi r } \bar {k}+ \frac{\mu_oI}{2 \pi r } \bar {k}$

$\bar{B_{net}} = \frac{2 \mu_oI}{2 \pi r } \bar {k} \ out$

Thus; $\bar{B_{net}}$ points out of the screen at A.

6 0

2 years ago

A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a sma

konstantin123 [22]

Answer:

$a=2.5\ m/s^2$

Explanation:

Given that,

Initial speed, u = 5 m/s

Final speed, v = 10 m/s

Time, t = 2 s

The radius of the tire of the bike, r = 35 cm

We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.

$a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2$

So, the required angular acceleration of the pebble is equal to $2.5\ m/s^2$ .

8 0

1 year ago

A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov

vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

$Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}$

$Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V$

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

$R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200$

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.

if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

$Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V$

$Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V$

$R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140$

$R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260$

so.

$V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}$

4 0

2 years ago

An airplane is delivering food to a small island. It flies 100 m above the ground at a speed of 150 m/s .

miss Akunina [59]

Answer:

The airplane should release the parcel $6.7*10^2$ m before reaching the island

Explanation:

The height of the plane is $y_0=100m$ , and its speed is v=150 m/s

When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is

$y=y_0 - \frac{gt^2}{2}$ [1]

And the distance X is

x = V.t [2]

Being t the time elapsed since the release of the parcel

If we isolate t from the equation [1] and replace it in equation [2] we get

$X = V . \sqrt{\frac{2y_0}{g}}$

Using the given values:

$x = 150 m/s \sqrt{\frac{2\times 100m}{9.8 m/sec^2}}$

x = $6.7*10^2$ m

4 0

2 years ago