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3 years ago

The current in a hair dryer measures 11 amps. The resistance of the hair dryer is 12 ohms. What is the voltage?

Physics

1 answer:

Valentin [98]3 years ago

8 0

Voltage is given by the formula

V = IR (Ohms law)

where V is the Voltage

I is the current

and R is the Resistance

Here it is given that the current is I=11

Resistance is R =12

so plugging this in the formula

V = IR

V= 11 * 12

V= 132 Volts

So the Voltage for the given dryer is 132 Volts

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Review the Four Social Errors and Biases in the Highlights area on page 127 in Ch. 4 of THiNK: Critical Thinking and Logic Skill

AysviL [449]

Answer: Provided in the explanation

Explanation:

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4 0

3 years ago

Betty (mass 40 kg), standing on slippery ice, catches her leaping dog (mass 15 kg) moving horizontally at 3.0 m/s. Show that the

MA_775_DIABLO [31]

Answer:

v = 2.18m/s

Explanation:

In order to calculate the speed of Betty and her dog you take into account the law of momentum conservation. The total momentum before Betty catches her dog must be equal to the total momentum after.

Then you have:

$Mv_{1o}+mv_{2o}=(M+m)v$ (1)

M: mass Betty = 40kg

m: mass of the dog = 15kg

v1o: initial speed of Betty = 3.0m/s

v2o: initial speed of the dog = 0 m/s

v: speed of both Betty and her dog = ?

You solve the equation (1) for v:

$v=\frac{Mv_{1o}+mv_{2o}}{M+m}=\frac{(40kg)(3.0m/s)+(15kg)(0m/s)}{40kg+15kg}\\\\v=2.18m/s$

The speed fo both Betty and her dog is 2.18m/s

7 0

3 years ago

A 2 kg box of taffy candy has 40j of potential energy relative to the ground. Its height above the ground is

lord [1]

Answer:

2.04m

Explanation:

PE=mgh

h= PE/mg

h= 40/2(9.82)

h=40/19.64

h=2.04

8 0

3 years ago

Read 2 more answers

A golf ball is struck with a five iron on level ground. it lands 100.0 m away 4.60 s later. what was the magnitude and direction

finlep [7]

consider the motion in x-direction

$v_{ox}$ = initial velocity in x-direction = ?

X = horizontal distance traveled = 100 m

$a_{x}$ = acceleration along x-direction = 0 m/s²

t = time of travel = 4.60 sec

Using the equation

X = $v_{ox}$ t + (0.5) $a_{x}$ t²

100 = $v_{ox}$ (4.60)

$v_{ox}$ = 21.7 m/s

consider the motion along y-direction

$v_{oy}$ = initial velocity in y-direction = ?

Y = vertical displacement = 0 m

$a_{y}$ = acceleration along x-direction = - 9.8 m/s²

t = time of travel = 4.60 sec

Using the equation

Y = $v_{oy}$ t + (0.5) $a_{y}$ t²

0 = $v_{oy}$ (4.60) + (0.5) (- 9.8) (4.60)²

$v_{oy}$ = 22.54 m/s

initial velocity is given as

$v_{o}$ = sqrt(( $v_{ox}$ )² + ( $v_{oy}$ )²)

$v_{o}$ = sqrt((21.7)² + (22.54)²) = 31.3 m/s

direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg

6 0

3 years ago

At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?

sergey [27]

1) Initial upward acceleration: $6.0 m/s^2$

2) Mass of burned fuel: $0.10\cdot 10^4 kg$

Explanation:

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude $F_g = mg$ , in the downward direction, where

$m=1.9\cdot 10^4 kg$ is the rocket's mass

$g=9.8 m/s^2$ is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

$T=3.0\cdot 10^5 N$

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

$T-mg=ma$ (1)