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2 years ago

A clothes dryer in a home has a power of 4,500 watts and runs on a special 220-volt household circuit,

1 answer:

3 0

1). Power = (voltage)² / (Resistance)

 4,500 = (220)² / Resistance

Multiply each side by (resistance) : 4,500 x resistance = (220)²

Divide each side by 4,500 : Resistance = (220)² / 4,500 = 10.76 ohms

2). Power = (voltage) x (Current)

Divide each side by (voltage): Power / voltage = Current

 4,500 / 220 = 20.45 Amperes

3). 4,500 watts = 4.5 kilowatts

 (4.5 kilowatts) x (4 hours) = 18 kilowatt-hours

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A force of 10 newtons toward the right is exerted on a

weeeeeb [17]

Answer:

Explanation:

coefficient of kinetic friction of wooden floor μ = .4

force of friction = μ R , R is reaction force of floor

R = mg = weight of body

R = 25 N

force of friction = .4 x 25 = 10 N

Net force on the crate = 10 - 10 = zero .

Net force on the body will be nil.

6 0

2 years ago

If two waves with identical crests and troughs meet, what is happening?

Nastasia [14]

If two waves with identical crests and troughs meet, what is happening?
C. Constructive interference is occurring.

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2 years ago

Read 2 more answers

Is v2 = v1t+a dimensionally correct? Explain please!

Lady bird [3.3K]

You want v2 = v1 + at
v is measured in m/s, a in m/s2, and t in s.
the dimensions multiply like algebraic quantities.
so because v2 is measured in m/s, then (v1 + at) has to come out in m/s
the units for (v1 + at) are (m/s) + (m/s2)(s)
time "s" cancels out one acceleration "s", so it comes ut to (m/s) + (m/s), which = (m/s).
if you had (v1t + a), then you would have (m/s)(s) + (m/s2) which = (m) + (m/s2), which doesn't work.

4 0

2 years ago

A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl

Firdavs [7]

Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

m (Mass)=3.0 kg

α(incline) =34°

Spring Constant (force constant)=120 N/m

d (distance)=?

Solution

F=mg

F=(3.0)(9.8)

F=29.4 N

As we also know that

Force parallel to the incline=FSinα

F=29.4×Sin(34)

F=16.44 N

d(distance)=F/Spring Constant

d(distance)=16.44/120

d(distance)=0.137 m ⇒13.7 cm

4 0

2 years ago

Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = -2.00 nC and is at x = 5.00 c

tatuchka [14]

Answer:

q₁= +0.5nC

Explanation:

Theory of electrical forces

Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

To solve this problem we apply Coulomb's law:

Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

o solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

F=K*q₁*q₂/d² Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)

d: distance between the charges in meters

Data:

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Data

q₃=+5.00 nC =+5* 10⁻⁹ C

q₂= -2.00 nC =-2* 10⁻⁹ C

d₂= 5.00 cm= 5*10⁻² m

d₁= 2.50 cm= 2.5*10⁻² m

k = 8.99*10⁹ N*m²/C²

Calculation of magnitude and sign of q1

Fn₃=0 : net force on q3 equals zero

F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.

F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.

We propose the algebraic sum of the forces on q₃

F₂₃ - F₁₃=0

$\frac{k*q_{2} *q_{3} }{d_{2}^{2} } -\frac{k*q_{1} *q_{3} }{d_{1}^{2} }=0$

We eliminate k*q₃ of the equation

$\frac{q_{1} }{d_{1}^{2} } = \frac{q_{2} }{d_{2}^{2} }$

$q_{1} =\frac{q_{2} *d_{1} ^{2} }{d_{2}^{2} }$

$q_{1} =\frac{2*10^{-9}*2.5^{2}*10^{-4} }{5^{2}*10^{-4} }$

q₁= +0.5*10⁻⁹ C

q₁= +0.5nC

4 0

2 years ago