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olya-2409 [2.1K]
4 years ago
10

The power dissipated in each of two resistors is the same. The current across resistor A is triple that across resistor B. If th

e resistance of resistor B is R, what is the resistance of A?

Physics
1 answer:
Svetllana [295]4 years ago
5 0

Answer:

Explanation:

this is the answer to your question

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If the length of the ramp is 2m and the height of the ramp is 1m , what is the mechanical advantage?
slamgirl [31]

Answer:

Mechanical Advantage is 2

Explanation:

M.A = length of ramp / height of ramp

= 2/1

=2

5 0
4 years ago
Determine the frequency of a sound wave if it has a speed of 350 m/s and a wavelength of 3.80 m.
Eva8 [605]
Since we have , v=f×lambda (wavelength). Where v equals 350m/s and wavelength equals 3.80. so it will become f = v/lambda=350/3.80=92.1052Hz
7 0
3 years ago
A viscous liquid is sheared between two parallel disks of radius �, one of which rotates with angular speed Ω, while the other i
Alexus [3.1K]

Answer:

Upper disk rotates at a constant angular velocity. The velocity at any height from stationery disk, say at x metres

U_o=v(\frac {x}{h}) where v is tangential velocity at radius r from the centre of disk

U_o=r\omega (\frac {x}{h})

The radial component of velocity is given as

U_r=0

The z component of velocity is also given as  

W=0

Total velocity, v= r\omega (\frac {x}{h})\hat e_{o}

5 0
3 years ago
A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse
Elanso [62]

Answer:

v_{f} = 0.51 \frac{m}{s}

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg :  crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight  (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)   

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W = 0

N = W

N = 882 N

We replace the  data in the formula (2)

Ff= μk*N  = 0.351* 882 N

Ff=  309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax    , ax=0

282 N - 309.58 N = 90*a  

a=  (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the  data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}

v_{f} = 0.51 \frac{m}{s}

8 0
3 years ago
The diagram shows how air circulation forms a sea breeze on a warm day. A beach meets the sea in the middle of the image with a
nadezda [96]
Yes it was then a red left label
7 0
3 years ago
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