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3 years ago

You connected the 5 Ω, 10 Ω, 15 Ω resistors in series with a 90 V battery. What is the current?

Physics

1 answer:

kykrilka [37]3 years ago

3 0

Answer:

Explanation:

Rtoal=R1+R2+R3=5+10+15=30

I=V/R 90/30

I=3

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A machinist turns the power on to a grinding wheel, at rest at tome t=0 s. the wheel accelerates uniformly for 10 s and reaches

Assoli18 [71]

To convert 2030 rad into rev, divide 2030 by 2pie. So final answer will be 2030/2 pie =323.08 revolutions.

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3 years ago

Jenny was applying her makeup when she drove into the student parking lot last Friday morning . Unaware that Cheryl was stopped

Akimi4 [234]

Answer: F = 102141N

Explanation: Newton's 2nd Law states that a force can change the motion of a body. The relation is given by

F = m.a

whose units are:

[F] = N

[m] = kg

[a] = m/s²

Jenny's car, at the moment of the break, had acceleration:

$a=\frac{\Delta v}{\Delta t}$

$a=\frac{11}{0.14}$

a = 78.57 m/s²

Then, Force is

F = 1300*78.57

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3 years ago

A mass m = 14 kg is pulled along a horizontal floor with NO friction for a distance d =5.7 m. Then the mass is pulled up an incl

frosja888 [35]

Answer:

W ≅ 292.97 J

Explanation:

1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)

Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;

W = (Fcosθ)d

Given that:

Tension of the force = 62 N

angle of incline θ = 34°

distance d =5.7 m.

Then;

W = 62 × cos(34) × 5.7

W = 353.4 cos(34)

W = 353.4 × 0.8290

W = 292.9686 J

W ≅ 292.97 J

Hence, the work done by tension before the block goes up the incline = 292.97 J

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3 years ago

Halley's comet orbits the sun roughly once every 76 years. It comes very close to the surface of the Sun on its closest approach

Licemer1 [7]

Answer:

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

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3 years ago

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the

aleksandrvk [35]

Answer:

θ= 5 radian

Explanation:

Given data:

Radius r = 0.70 m

Initial angular speed ω_i = 2rev/s

Time t = 5 s

Final angular speed ω_f =0

so we have angular displacement

$\theta= \frac{\omega_f-\omega_i}{2}\times t$

putting values

$\theta= \frac{0-2}{2}\times5$ = 5 rad

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3 years ago

You connected the 5 Ω, 10 Ω, 15 Ω resistors in series with a 90 V battery. What is the current?​

You connected the 5 Ω, 10 Ω, 15 Ω resistors in series with a 90 V battery. What is the current?