All categories

3 years ago

What is the equivalent resistance of the circuit?

Physics

2 answers:

Sonbull [250]3 years ago

8 0

Answer:

80Ω

Explanation:

Resistance in a series circuit (when one resistor is after another in the same line) is calculated with the following equation

Total resistance = R1 + R2 + R3...Rn

Substitute in your values and solve for total resistance

Total resistance = 60 + 20 = 80Ω

Nikolay [14]3 years ago

3 0

Answer:

Explanation:

You might be interested in

An electromagnet is created using a battery, an insulated copper wire and an iron nail. The wire is wrapped around the nail 30 t

Veseljchak [2.6K]

Answer:

Explanation:

If one assume that each turn is like a strand of electromagnet, which can then be added up. Therefore, increase in the number of turns will yield to increase in the magnetic strength. Also if the current increases, then there will be increase in the magnetic field strength.

From Ohm's law

V = IR

I = V/R

That is a direct increase in voltage will lead to increase in current.

Increase the voltage of the battery and increases the number of turns of the coil. Will suit the situation

4 0

3 years ago

1. What is the wave speed of a wave that has a frequency of 100 Hz and a wavelength of 0.30 m?

aivan3 [116]

Answer:

1. v = 30 m/s

2. v = 5 m/s

3. f = 40 Hz

4. f = 400 Hz

5. f = 300 Hz

6. λ = 0.772 m

7. λ = 0.386 m

8. λ = 0.625 m

9. v = 100 m/s

10. v = 50 m/s

Explanation:

The relationship between frequency, wavelength, and speed of a wave is given by the following formula:

$v = f\lambda$

where,

v = speed of wave

f = frequency of wave

λ = wavelength

f = 100 Hz

λ = 0.3 m

Therefore,

v = (100 Hz)(0.3 m)

v = 30 m/s

f = 50 Hz

λ = 0.1 m

v = (50 Hz)(0.1 m)

v = 5 m/s

v = 20 m/s

λ = 0.5 m

$f = \frac{v}{\lambda} = \frac{20\ m/s}{0.5\ m}$

f = 40 Hz

v = 80 m/s

λ = 0.2 m

$f = \frac{v}{\lambda}=\frac{80\ m/s}{0.2\ m}$

f = 400 Hz

v = 120 m/s

λ = 0.4 m

$f = \frac{v}{\lambda}=\frac{120\ m/s}{0.4\ m}$

f = 300 Hz

v = 340 m/s

f = 440 Hz

$\lambda = \frac{v}{f}=\frac{340\ m/s}{440\ Hz}\\$

λ = 0.772 m

v = 340 m/s

f = 880 Hz

$\lambda = \frac{v}{f}=\frac{340\ m/s}{880\ Hz}\\$

λ = 0.386 m

v = 250 m/s

f = 400 Hz

$\lambda = \frac{v}{f}=\frac{250\ m/s}{400\ Hz}\\$

λ = 0.625 m

f = 50 Hz

λ = 2 m

v = (50 Hz)(2 m)

v = 100 m/s

10.

f = 100 Hz

λ = 0.5 m

v = (100 Hz)(0.5 m)

v = 50 m/s

6 0

3 years ago

A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m.

strojnjashka [21]

Answer:

$D=99.4665307m \approx 99.5m$

Explanation:

From the question we are told that

Mass $m=6kg$

Velocity of mass $V_m=16$

Force of Tunnel $F_t=8N$

Length of Tunnel $L_t=1.6$

Height of frictional incline $H_i=2.9$

Angle of inclination $\angle =16 \textdegree$

Acceleration due to gravity $g=9.8m/s^2$

First Frictional surface has a coeﬃcient $\alpha_1 =0.21\ for\ d_c=1$

Second Frictional surface has a coeﬃcient $\alpha _2=0.1$

Generally the initial Kinetic energy is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}(6)(16)^2$

$K.E=768$

Generally the work done by the Tunnel is mathematically given as

$w_t=F_t*d_t$

$w_t=8*1.6$

$w_t=12.8J$

Therefore

$Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t$

$E_t=K.E+E_t\\E_t=768J+12.8J$

$E_t=780.8J$

Generally the energy lost while climbing is mathematically given as

$E_c=mgh$

$E_c=(6)(9.8)(2.9)$

$E_c=170.52J$

Generally the energy lost to friction is mathematically given as

$E_f=\alpha *m*g*cos\textdegree*d_c$

$E_f=0.21*6*9.8*cos16*1$

$E_f=11.86965942 \approx 12J$

Generally the energy left in the form of mass $Em$ is mathematically given as

$E_m=E_t+E_c+E_f$

$E_m=(768J)-(170.52)-(12)$

$E_m=585.48J$

Since

$E_m=\alpha_2*g*m*d$

Therefore

It slide along the second frictional region

$D=\frac{585.46}{0.1*9.81*6}$

$D=99.4665307m \approx 99.5m$

6 0

3 years ago

During what stage of mitosis are the chromatid pairs now visible and the spindle fibers are beginning to form

Mekhanik [1.2K]

The stage of mitosis is prophase

5 0

3 years ago

2 76 Consider a household that uses 14,000 kWh of elec- (tricity per year and 3400 L of fuel oil during a heating sea- son. The

NNADVOKAT [17]

Answer:

The total amount of CO₂ produced will be = 20680 kg/year

The reduction in the amount of CO₂ emissions by that household per year = 3102 kg/year

Explanation:

Given:

Power used by household = 14000 kWh

Fuel oil used = 3400 L

CO₂ produced of fuel oil = 3.2 kg/L

CO₂ produced of electricity = 0.70 kg/kWh

Now, the total amount of CO₂ produced will be = (14000 kWh × 0.70 kg/kWh) + (3400 L × 3.2 kg/L)

⇒ The total amount of CO₂ produced will be = 9800 + 10880 = 20680 kg/year

Now,

if the usage of electricity and fuel oil is reduced by 15%, the reduction in the amount of the CO₂ emission will be = 0.15 × 20680 kg/year = 3102 kg/year

6 0

4 years ago