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The following three appliances are connected to a 120 volt house circuit: (i) a toaster, 1200 Watts, (ii) a coffee pot, 750 Watts, and (iii) a microwave, 600 Watt. If all operate at the same time, what total current would they draw

Answers

Answer 1

Answer:

Answer:

They would draw a total of 21.25 amperes

Explanation:

The total power consumed is

1200 W+ 750 W + 600 W= 2550 Watts

The formula relating the power consumed, the voltage and the current is given as

$P=IV$ ---------------1

given that the voltage supply is 120V

$2550=I*120\n\I=(2550)/(120) \n\nI= 21.25amps$

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To determine the tension in the string that connects M2 and M3, we can follow these steps:

Step 1: Identify the necessary variables. Given data (for example) could be:
- Mass of M2, which is 5 kg
- Mass of M3, which is 10 kg
- The acceleration due to gravity, which is approximately 9.8 m/s²
- The angle at which the string pulls on M2, which is 30 degrees
- Assume the system is in equilibrium, meaning there is no net acceleration, so the acceleration is 0 m/s²

Step 2: Calculate the weight of M3, which is its mass times the acceleration due to gravity. This is because weight is the force exerted by gravity on an object, which equals the object's mass times the acceleration due to gravity.

For M3, this calculation would be M3 * g = 10 kg * 9.8 m/s² = 98 N (Newtons).

Step 3: Determine the force exerted by M2 that acts along the line of the string. This won't be the full weight of M2, because the string pulls at an angle. This component of the force can be calculated using the sine of the angle, because sine gives us the ratio of the side opposite the angle (here, the force along the string) to the hypotenuse (here, the full weight of M2) in a right triangle.

The horizontal component of the force of M2 is then M2 * g * sin(30deg) = 5 kg * 9.8 m/s² * sin(30deg) = 24.5 N.

Step 4: The tension in the string is the force M3 exerts on it, which is its weight, minus the component of M2's weight that acts along the string. This is because M2 and M3 are pulling in opposite directions, so they subtract from each other.

The tension in the string is then the weight of M3, 98 N, minus the horizontal (along the string) component of M2's weight, 24.5 N.

So, the tension in the string is 98 N - 24.5 N = 73.5 N.

This is the force that the string needs to exert in order to keep M2 and M3 connected and in equilibrium.

Learn more about Tension in a string here:

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You are exiting a highway and need to slow down on the off-ramp in order to make the curve. It is rainy and the coefficient of static friction between your tires and the road is only 0.4. If the radius of the off-ramp curve is 36 m, then to what speed do you need to slow down the car in order to make the curve without sliding?

Answers

Answer:

11.87m/s

Explanation:

To solve this problem it is necessary to apply the concepts related to frictional force and centripetal force.

The frictional force of an object is given by the equation

$F_r = \mu N$

Where,

$\mu =$ Friction Coefficient

N = Normal Force, given also as mass for acceleration gravity

In the other hand we have that centripetal force is given by,

$F_c= (mv^2)/(R)$

The force experienced to stay on the road through friction is equal to that of the centripetal force, therefore

$F_r = F_c$

$\mu mg = (mv^2)/(R)$

Re-arrange to find the velocity,

$V = √(R\mu g)$

$V = √(36*0.4*9.8)$

$V = 11.87m/s$

Therefore the speed that it is necessaty to slow down the car in order to make the curve without sliding is 11.87m/s

a trcuk weighs four times as much as a stationary car. if teh truck coasts into the car at 12 km/s and they stick toegther, what is their final velocity

Answers

Answer:

v=9.6 km/s

Explanation:

Given that

The mass of the car = m

The mass of the truck = 4 m

The velocity of the truck ,u= 12 km/s

The final velocity when they stick = v

If there is no any external force on the system then the total linear momentum of the system will be conserve.

Pi = Pf

m x 0 + 4 m x 12 = (m + 4 m) x v

0 + 48 m = 5 m v

5 v = 48

$v=(48)/(5)\ km/s$

v=9.6 km/s

Therefore the final velocity will be 9.6 km/s.

You spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.111 liters per second and the diameter of the nozzle you hold is 5.79 mm. At what speed does the water exit the nozzle

Answers

Answer:

29.5 m/s

Explanation:

Volumetric flowrate = (average velocity of flow) × (cross sectional area)

Volumetric flowrate = 0.111 liters/s = 0.000111 m³/s

Cross sectional Area of flow = πr²

Diameter = 0.00579 m,

Radius, r = d/2 = 0.002895 m

A = π(0.002895)² = 0.0000037629 m²

Velocity of flow = (volumetric flow rate)/(cross sectional Area of flow)

v = 0.000111/0.0000037629

v = 29.5 m/s

Given Information:

diameter of the nozzle = d = 5.79 mm = 0.00579 m

flow rate = 0.111 liters/sec

Required Information:

Velocity = v = ?

Answer:

Velocity = 4.21 m/s

Explanation:

As we know flow rate is given by

Flow rate = Velocity*Area of nozzle

Where

Area of nozzle = πr²

where

r = d/2

r = 0.00579/2

r = 0.002895 m

Area of nozzle = πr²

Area of nozzle = π(0.002895)²

Area of nozzle = 2.6329x10⁻⁵ m²

Velocity = Flow rate/area of nozzle

Divide the litters/s by 1000 to convert into m³/s

0.111/1000 = 1.11x10⁻⁴ m³/s

Velocity = 1.11x10⁻⁴/2.6329x10⁻⁵

Velocity = 4.21 m/s

Therefore, the water exit the nozzle at a speed of 4.21 m/s

Two charged particles are located on the x axis. The first is a charge +Q at x = −a. The second is an unknown charge located at x = +3a. The net electric field these charges produce at the origin has a magnitude of 2keQ/a2 . Explain how many values are possible for the unknown charge and find the possible values.

Answers

Answer:

-9Q

Explanation:

Electric field at origin is:

$E=(2keQ)/(a^2)$

Electric field due to first charge at origin would be:

$E_1=(keQ)/(a^2)$

Electric field due to second charge would be:

$E_2=E-E_1\nE_2=(2keQ)/(a^2)-(keQ)/(a^2) = (keQ)/(a^2)$

If the second charge is Q', then $E_2$ should be:

$E_2=(keQ')/((3a)^2)=(keQ')/(9a^2)$

compare the above two values to find the possible values of Q':

$(|Q'|)/(9)=Q\n |Q'|=9Q$

The net electric field at origin is greater than the one due to first charge. It means the second charge adds on to the electric field at the origin. Thus, it should be a negative charge.

Thus, Q' = -9Q

One value is possible as the location of the second charge is given to be on the positive x-axis.

Final answer:

The possible values for the unknown charge are 1/9 of the magnitude of the known charge.

Explanation:

To find the possible values for the unknown charge, we need to use the principle of superposition. The net electric field at the origin is given by the sum of the electric fields due to each charge. We know that the magnitude of the net electric field is 2keQ/a^2, so we can set up the equation:

2keQ/a^2 = keQ/(-a)^2 - keq/(3a)^2

By solving this equation, we can find the possible values for the unknown charge. Simplifying the equation, we get:

2 = 1 - 1/9

1/9 = 1

After solving the equation, we find that the possible value for the unknown charge is 1/9 of the magnitude of the known charge.