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3 years ago

A physics student has a battery and three equal resistors. If she uses all of the

Physics

2 answers:

Soloha48 [4]3 years ago

4 0

Answer:

she can arrange the resistors and battery like a butterfly so that you know.... she can do that

Korvikt [17]3 years ago

3 0

Question: A physics student has a battery and an assortment of resistors. If she uses all of the resistors, how should she arrange them in a circuit to obtain the maximum current flow through the battery and the circuit?

Ans: she should arrange them in parallel

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Which action involves the most efficient conversion of electrical energy into thermal energy?

azamat

Answer:

the answer is option d

because in oven which works through electricity ,it will bake the cake (heat energy is produced)

7 0

3 years ago

Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a

raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

3 0

2 years ago

A car travels up a hill at a constant speed of 38 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the a

mojhsa [17]

Answer:

Average speed will be 48.23 km/h

Explanation:

Let the distance up to hill is = d km

Speed when car goes to hill = 38 km/h

So time required $t=\frac{distance}{speed}=\frac{d}{38}hour$

Speed when car return from hill = 66 km/h

So time required to return fro hill $t=\frac{d}{66}h$

Total time $t_{total}=\frac{t}{38}+\frac{t}{66}$

Total distance = d+d =2d

So average speed $=\frac{total\ distance}{total\ time}=\frac{2d}{\frac{d}{38}+\frac{d}{66}}=48.23km/h$

8 0

3 years ago

M/s

SashulF [63]

Answer:

a. Final velocity, V = 2.179 m/s.

b. Final velocity, V = 7.071 m/s.

Explanation:

Given the following data;

Acceleration = 0.500m/s²

a. To find the velocity of the boat after it has traveled 4.75 m

Since it started from rest, initial velocity is equal to 0m/s.

Now, we would use the third equation of motion to find the final velocity.

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.

Substituting into the equation, we have;

$V^{2} = 0^{2} + 2*0.500*4.75$

$V^{2} = 4.75$

Taking the square root, we have;

$V^{2} = \sqrt {4.75}$

Final velocity, V = 2.179 m/s.

b. To find the velocity if the boat has traveled 50 m.

$V^{2} = 0^{2} + 2*0.500*50$

$V^{2} = 50$

Taking the square root, we have;

$V^{2} = \sqrt {50}$

Final velocity, V = 7.071 m/s.

8 0

3 years ago

Why is there a wet climate on the windward side of a tall mountain and a dry climate on the leeward side?

Eduardwww [97]

One side of the mountain that has constant wind and rain blowing onto it, is more likely to catch what is falling than the other side leaving it dryer.

3 0

3 years ago

Read 2 more answers