All categories

3 years ago

If we have several different light bulbs connected to a battery in series, then ___.

Physics

1 answer:

kenny6666 [7]3 years ago

6 0

Answer:

The current will be the same across each bulb.

You might be interested in

An egg is thrown upward with a velocity of 4.5 m/s. How long will it take to reach it's maximum height?

Ray Of Light [21]

Answer:

0.45 seconds

Explanation:

Letting the value of g = 10 m/s/s

final velocity (v) = 0 m/s (since the egg will come to rest at the maximum height)

initial velocity(u) = 4.5 m/s

acceleration = -10 m/s/s (since the gravity is acting against the egg)

time = t seconds

From the first equation of motion:

v = u + at

0 = 4.5 + (-10)t

t = -4.5 / -10

t = 0.45 seconds

3 0

3 years ago

Two resistors, R1 and R2, are

dlinn [17]

The reciprocal of the total resistance is equal to the sum of the reciprocals of the component resistances:

1/(120.7 Ω) = 1/R₁ + 1/(221.0 Ω)

1/R₁ = 1/(120.7 Ω) - 1/(221.0 Ω)

R₁ = 1 / (1/(120.7 Ω) - 1/(221.0 Ω)) ≈ 265.9 Ω

3 0

3 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

4 years ago

( Can someone help? )

Murrr4er [49]

Answer:

Answer would be 0.33

Explanation:

Calculations

8 0

3 years ago

Read 2 more answers

Two charges, each q, are separated by a distance r, and exert mutual attractive forces of F on each other. If both charges becom

jeka57 [31]

Answer:

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

we substitute

F = k 4q² / 9 r²

F = k q² r² 4/9

F = ⅔ F₀

3 0

3 years ago