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3 years ago

Two appliances are connected in parallel to a 120-v battery and draw currents i1 = 3.0 a and i2 = 3.1

1 answer:

4 0

Initially they are connected in parallel, so they have the same voltage V=120 V at their ends. Therefore we can use Ohm's law to calculate the resistance of each appliance:
$R_1 = \frac{V}{I_1}= \frac{120 V}{3.0 A}=40 \Omega$
$R_2 = \frac{V}{I_2}= \frac{120 V}{3.1 A}=38.7 \Omega$

When they are connected in series, they are crossed by the same current I. The equivalent resistance of the circuit in this case is $R_{eq}=R_1+R_2 = 78.7 \Omega$ , so we can use Ohm's law for the entire circuit to find the current in the circuit:
$I= \frac{V}{R_{eq}}= \frac{120 V}{78.7 \Omega}=1.52 A$

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A machine with a 5-N input force and a 25-N output force has a mechanical advantage of?

jolli1 [7]

So we want to know the mechanical advantage of a machine that has 5 N input force and 25 N out force. Mechanical advantage Ma is the measure of force amplification of some machine. We calculate it by taking the ratio of the output force Fo over the input force Fi. Ma=Fo/Fi=(25 N)/(5 N)=5. So Mechanical advantage for our machine is Ma=5 and the correct answer is the second one.

8 0

3 years ago

A 100 g aluminum calorimeter contains 250 g of water. The two substances are in thermal equilibrium at 10°C. Two metallic blocks

ipn [44]

Answer:

A. 1,950 J/kgºC

Explanation:

Assuming that all materials involved, finally arrive to a final state of thermal equilibrium, and neglecting any heat exchange through the walls of the calorimeter, the heat gained by the system "water+calorimeter" must be equal to the one lost by the copper and the unknown metal.

The equation that states how much heat is needed to change the temperature of a body in contact with another one, is as follows:

Q = c * m* Δt

where m is the mass of the body, Δt is the change in temperature due to the external heat, and c is a proportionality constant, different for each material, called specific heat.

In our case, we can write the following equality:

(cAl * mal * Δtal) + (cH₂₀*mw* Δtw) = (ccu*mcu*Δtcu) + (cₓ*mₓ*Δtₓ)

Replacing by the givens , and taking ccu = 0.385 J/gºC and cAl = 0.9 J/gºC, we have:

Qg= 0.9 J/gºC*100g*10ºC + 4.186 J/gºC*250g*10ºC = 11,365 J(1)

Ql = 0.385 J/gºC*50g*55ºC + cₓ*66g*80ºC = 1,058.75 J + cx*66g*80ºC (2)

Based on all the previous assumptions, we have:

Qg = Ql

So, we can solve for cx, as follows:

cx = (11,365 J - 1,058.75 J) / 66g*80ºC = 1.95 J/gºC (3)

Expressing (3) in J/kgºC:

1.95 J/gºC * (1,000g/1 kg) = 1,950 J/kgºC

3 0

3 years ago

A 750 watt hairdryer is used for 60 seconds. how many joules of energy are used?

daser333 [38]

We know, Power = Energy / time
Here, P = 750 Watt
t = 60 sec

Substitute their values,
750 = E / 60
E = 750 * 60
E = 45,000 J

In short, Your Answer would be 45,000 Joules

Hope this helps!

4 0

4 years ago

Read 2 more answers

An Olympic diver drops from the 10 meter platform with an initial velocity of 0.0m/s. What was the vertical velocity of the dive

sammy [17]

Answer:

v=14.14 m/s

t=1.141 s

Explanation:

Given that

h = 10 m

Initial velocity ,u = 0 m/s

We know that acceleration due to gravity g= 10 m/s²

Lets take final velocity = v

Final velocity v is given as

v² = u²+ 2 g h

v²= 0² + 2 x 10 x 10

v²= 200

v=14.14 m/s

Time taken t is given as

v= u + a t

a=acceleration

t=time

Now by putting the values in the above equation we get

14.14= 0 + 10 x t

14.14 = 10 t

t=1.141 s

4 0

4 years ago

you are riding on a train traveling 80 km/h. if you walk toward the back of the train at a speed of 1.2 km/h relative to the tra

Mazyrski [523]

Answer:

Velocity of the passenger = 78.8[km/h]

Explanation:

In order to solve this problem, we must observe the train from a distant point of the train. In this way we can observe that the train moves at 80[km/h] relative to the ground.

In such a way that the passenger moves in a direction contrary to the movement of the train.

$v_{pass}=80-1.2\\ v_{pass}=78.8[km/h]$

The Observer located outside the train will see how the passenger moves away at 78.8 [km / h] and not at 80[km/h] which is the speed of the train.

6 0

3 years ago