1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Alecsey [184]
3 years ago
9

Which is the correct unit for electrical power? a) amp b) volt c) watt d) coulomb

Physics
2 answers:
musickatia [10]3 years ago
8 0
T<span>he correct unit for electrical power is "watt".
</span>

That<span>’s actually the unit that measures the rate per time that electric energy is transferred.</span>

<span>
</span>

<span>Have a nice day! :)</span>

Alex Ar [27]3 years ago
6 0
C. watt hope this helps
You might be interested in
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
All organisms have adaptations. The list below describes how different adaptations might affect different species.
VMariaS [17]
I think A, because rodents already live in winter months when little food is available, but I'm not sure.
3 0
3 years ago
Read 2 more answers
A box is hanging from two strings. String #1 pulls up and left, making an angle of 50° with the horizontal on the left, and stri
creativ13 [48]

Answer:

mb = 3.75 kg

Explanation:

System of forces in balance

ΣFx =0  

ΣFy = 0

Forces acting on the box

T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left

T₂  = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.

Wb :Weightt of the box (vertical downward)

x-y T₁ and T₂ components

T₁x= T₁cos50°

T₁y= T₁sin50°

T₂x= 30*cos75° = 7.76 N

T₂y= 30*sin75° = 28.98 N

Calculation of the Wb

ΣFx = 0  

T₂x-T₁x = 0

T₂x=T₁x

7.76 = T₁cos50°

T₁ = 7.76 /cos50° = 12.07 N

ΣFy = 0  

T₂y+T₁y-Wb = 0

28.98 + 12.07(cos50°) = Wb

Wb = 36.74 N

Calculation of the mb ( mass of the box)

Wb = mb* g

g: acceleration due to gravity = 9.8 m/s²

mb = Wb/g

mb = 36.74 /9.8

mb = 3.75 kg

8 0
3 years ago
What tension would you need to make a middle c (261.6 hz) fundamental mode on a 1 m string (for example, on a harp)? the linear
Allisa [31]
The frequency of middle C on a string is
f = 261.6 Hz.

The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
   = 0.002 kg/m

The length of the string is L = 1 m.

Let T =  the tension in the string (N).
The velocity of the standing wave is
v= \sqrt{ \frac{T}{\rho} }

In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
\sqrt{ \frac{T}{\rho} } =f \lambda = fL \\\\ \frac{T}{\rho} = (fL)^{2} \\\\ T = \rho (fL)^{2}

From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
   = 136.87 N

Answer: 136.9 N (nearest tenth)

4 0
3 years ago
How can x-rays make pictures of the inside of solid objects
Mariana [72]
X-rays have high energy and can penetrate matter that light cannot.
6 0
3 years ago
Other questions:
  • Atoms with a low ionization energy hold tight to their outer valence electrons.
    11·2 answers
  • Consider a long cylindrical charge distribution of radius R with a uniform charge density rho. Find the electric field at distan
    9·1 answer
  • What is your understanding of repetition and replication in science?
    11·1 answer
  • What are the large, dark basaltic plains on the moon called?
    5·2 answers
  • How are mass and inertia related
    8·1 answer
  • You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of 0.200 rev/s^2. Th
    11·1 answer
  • A hammer has a mass of 1 kg. What is its weight (i) on Earth (ii) on the
    15·1 answer
  • A laser emits a beam of light whose photons all have the same frequency. When the beam strikes the surface of a metal, photoelec
    13·1 answer
  • In a model generator, a 310-turn rectangular coil 0.075 m by 0.18 m rotates with an angular frequency of 12.2 rad/s in a uniform
    7·1 answer
  • 4
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!