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2 years ago

10

You plan to take your hair blower to europe, where the electrical outlets put out 240 v instead of the 120 v seen in the united

states. the blower puts out 1600 w at 120 v .

1 answer:

Eddi Din [679]2 years ago

3 0

P = U × I
I = P / U = 1600W / 120V = 13.4A
P = 240V × 13.4A = 3216W

If your hair blower isnt rated for 220- 230V(this is the voltage in EU) you are most likely going to burn it.

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Two children, A and B, fire identical 10.0g all bearings from a catapult. the elastic band of each catapult extended by 0.10m an

sergiy2304 [10]

Answer:

The speed decreases and the frequency remains the same.

Explanation:

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6 0

2 years ago

Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.

kobusy [5.1K]

With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees

ELECTRIC FORCE (F)

F = $\frac{KQq}{d^{2} }$

Where K = 9 x $10^{9}$ N $m^{2}$ / $C^{2}$

The distance between $q_{1}$ and $q_{3}$ can be calculated by using Pythagoras theorem.

d = $\sqrt{33^{2} + 33^{2} }$

d = 46.7 cm = 0.467 m

For force $F_{1}$ , substitute all the parameters into the formula above

$F_{1}$ = (9 x $10^{9}$ x 3 x 1)/ $0.467^{2}$

$F_{1}$ = 2.7 x $10^{10}$ /0.218

$F_{1}$ = 1.24 x $10^{11}$ N

For force $F_{4}$ , substitute all the parameters into the formula above

$F_{4}$ = (9 x $10^{9}$ x 3 x 4)/ $0.33^{2}$

$F_{4}$ = 1.08 x $10^{11}$ /0.1089

$F_{4}$ = 9.92 x $10^{11}$ N

For force $F_{2}$ , substitute all the parameters into the formula above

$F_{2}$ = (9 x $10^{9}$ x 3 x 2)/ $0.33^{2}$

$F_{2}$ = 5.4 x $10^{10}$ /0.1089

$F_{2}$ = 4.96 x $10^{11}$ N

Summation of forces on Y component will be

$F_{y}$ = $F_{4}$ - $F_{1}$ Sin 45

$F_{y}$ = 9.92 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{y}$ = 9.04 x $10^{11}$ N

Summation of forces on X component will be

$F_{x}$ = $F_{2}$ - $F_{1}$ Cos 45

$F_{x}$ = 4.96 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{x}$ = 4.08 x $10^{11}$ N

Net Force = $\sqrt{F_{x} ^{2} + F_{y} ^{2} } }$

Net force = $\sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2} }$

Net force = 9.9 x $10^{11}$ N

The direction will be

Tan ∅ = $F_{y}$ / $F_{x}$

Tan ∅ = 9.04 x $10^{11}$ / 4.08 x $10^{11}$

Tan ∅ = 2.216

∅ = $Tan^{-1}$ (2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

8 0

1 year ago

Read 2 more answers

If an object is thrown upward at 128 feet per second from a height of 76 feet, its height S after t seconds is given by the foll

faltersainse [42]

<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>

Explanation:

a) Given S(t) = 76 + 128t − 16t²

s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft

s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

Displacement in 4 seconds = 332 - 76 = 256 ft

Time = 4 - 0 = 4 s

$\texttt{Velocity = }\frac{256}{4}=64ft/s$

Average velocity in first 4 seconds is 64 ft/s upward

a) Given S(t) = 76 + 128t − 16t²

s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft

Displacement in 4 seconds = 78 - 332 = -254 ft

Time = 4 - 0 = 4 s

$\texttt{Velocity = }\frac{-254}{4}=-63.5ft/s$

Average velocity in second 4 seconds is 63.5 ft/s downward

3 0

2 years ago

A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of

OLEGan [10]

Answer:

c) $F_{net} = 0.640 kN$

Explanation:

As we know that resultant force is the net force that is acting on the system

As per Newton's II law we know that net force is product of mass and acceleration

so we will have

$F_{net} = ma$

here we know

m = 80 kg

for circular motion acceleration is given as

$a_c = \frac{v^2}{R}$

$a_c = \frac{40^2}{200} = 8 m/s^2$

now we have

$F_{net} = 80 \times 8$

$F_{net} = 640 N$

$F_{net} = 0.640 kN$

7 0

2 years ago

Compare and contrast the energy transfer of a roller coaster to that of a pendulum

Softa [21]

When the pendulum and roller coaster move to the top, its has more potential energy whereas when comes to the bottom has more kinetic energy.

<h3>Compare and contrast the energy transfer of a roller coaster to that of a pendulum:</h3><h3>What is the transfer of energy in a roller coaster?</h3>

The transfer of potential energy to kinetic energy occur when the roller coaster move along the track. As the motor pulls the cars to the top, the body has more potential energy whereas when the body comes to the bottom , it has kinetic energy in the object.

<h3>What is the energy transfer in a pendulum?</h3>

As a pendulum swings, its potential energy changes to kinetic energy and kinetic energy changes into potential energy. At the top more potential energy is present.

So we can conclude that When the pendulum and roller coaster move to the top, its has more potential energy whereas when comes to the bottom has more kinetic energy.

Learn more about energy here: brainly.com/question/13881533

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8 0

1 year ago