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Daniel [21]
3 years ago
5

Three resistors of resistances, R1=10Ω, R2=5Ω, R3=20Ω are connected in a circuit in such way that same amount of current flows t

hrough them. Draw a circuit diagram showing the correct connections of the three resistors, a cell of 2.5V, an ammeter and a voltmeter across R3. Calculate the potential difference across R3.
Physics
1 answer:
Nezavi [6.7K]3 years ago
5 0

Answer

Explanation:

As the three resistors are connected in series, the expression to be used for the  

calculation of RT equivalent resistance

is:  

RT = R1 + R2 + R3

We replace the data of the statement in the previous expression and it remains:  

5 10 15 RT + R1 + R2 + R3 + +

We perform the mathematical operations that lead us to the result we are looking for:  

RT - 30Ω

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A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0
3 years ago
A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along
babymother [125]

Answer:

Explanation:

height of pole = 15 ft

height of man = 6 ft

Let the length of shadow is y .

According to the diagram

Let at any time the distance of man is x.

The two triangles are similar

\frac{y-x}{y}=\frac{6}{15}

15 y - 15 x = 6 y

9 y = 15 x

y=\frac{5}{3}x

Differentiate with respect to time.

\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}

As given, dx/dt = 4 ft/s

\frac{dy}{dt}=\frac{5}{3}\times 4

\frac{dy}{dt}=\frac{20}{3} ft/s

6 0
4 years ago
What group does paracetamol come under
agasfer [191]
Paracetamol is grouped within the 'painkillers' category. 
6 0
3 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
the solubility product of lead fluoride is 3.6 x 10–8. what is its solubility in 0.10M NaF solution, in grams per liter
BartSMP [9]

Answer:

8.8 × 10⁻³ g/L

Explanation:

NaF is a strong electrolyte that ionizes according to the following reaction.

NaF(aq) → Na⁺(aq) + F⁻(aq)

Then, the concentration of F⁻ will also be 0.10 M.

In order to find the solubility of PbF₂ (S), we will use an ICE Chart.

        PbF₂(s) ⇄ Pb²⁺(aq) + 2 F⁻(aq)

I                           0                0.10

C                         +S               +2S

E                          S              0.10 + 2S

The solubility product (Kps) is:

Kps = 3.6 × 10⁻⁸ = [Pb²⁺].[F⁻]² = S . (0.10 + 2S)²

In the term 0.10 + 2S, 2S is negligible in comparison with 0.10 and we can omit it to simplify calculations.

Kps = 3.6 × 10⁻⁸ = S . (0.10)²

S = 3.6 × 10⁻⁵ M

The molar mass of PbF₂ is 245.20 g/mol. The solubility of PbF₂ in g/L is:

3.6 × 10⁻⁵ mol/L × 245.20 g/mol = 8.8 × 10⁻³ g/L

4 0
3 years ago
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