All categories

2 years ago

2 Ten identical lengths of wire are laid closely side-by-

Physics

1 answer:

Mice21 [21]2 years ago

7 0

Answer:

(a) 0.71 mm

(b) 0.158 cubic cm

Explanation:

The width of one wire is the diameter of the wire.

(a) Let the diameter of each wire is d.

So, 10 d = 14.2 mm

d = 1.42 mm

radius of each wire, r = d/2 = 1.42/2 = 0.71 mm

(b) Length, L = 10 cm

The volume of the single wire is given by

$V =\pi\times r^2\times h \\\\V =3.14\times 0.071^2\times 10\\\\V =0.158 cm^3$

You might be interested in

A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg.

rodikova [14]

Answer:

the moment of inertia of the merry go round is 38.04 kg.m²

Explanation:

We are given;

Initial angular velocity; ω_1 = 37 rpm

Final angular velocity; ω_2 = 19 rpm

mass of child; m = 15.5 kg

distance from the centre; r = 1.55 m

Now, let the moment of inertia of the merry go round be I.

Using the principle of conservation of angular momentum, we have;

I_1 = I_2

Thus,

Iω_1 = I'ω_2

where I' is the moment of inertia of the merry go round and child which is given as I' = mr²

Thus,

I x 37 = ( I + mr²)19

37I = ( I + (15.5 x 1.55²))19

37I = 19I + 684.7125

37I - 19 I = 684.7125

18I = 684.7125

I = 684.7125/18

I = 38.04 kg.m²

Thus, the moment of inertia of the merry go round is 38.04 kg.m²

7 0

3 years ago

Read 2 more answers

A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t

Morgarella [4.7K]

Answer:

(A). The work done is $1.50\times10^{-6}\ J$ .

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy $K.E=1.50\times10^{-6}\ J$

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

$W = \Delta K.E$

$W=K.E_{f}-K.E_{i}$

Put the value into the formula

$W= 1.50\times10^{-6}-0$

$W=1.50\times10^{-6}\ J$

The work done is $1.50\times10^{-6}\ J$ .

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

$\Delta P.E=\Delta K.E$

So, $U = 1.50\times10^{-6}$

Using formula of potential

$V=\dfrac{U}{q}$

Put the value into the formula

$V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}$

$V=357.14\ V$

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

$W=F\times r$ ....(I)

Using formula of force

$F=qE$

Put the value in the equation (I)

$W=qE\times r$

$E=\dfrac{W}{q\times r}$

Put the value into the formula

$E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}$

$E=5952.38\ N/C$

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

7 0

3 years ago

Which statement best describes balancing equations and the law of conservation of mass?

Dmitrij [34]

Answer:

The number of atoms is the same in the reactants and in the products, and the total mass is the same in the reactants and in the products.

5 0

1 year ago

A battery of emf 5V and internal resistance 2ohm is joined to a resistor of 8ohm.Calculate the terminal potential difference.

nata0808 [166]

Answer:

Explanation:

First, we calculate the total resistance to the given battery cell of emf 5V. The total resistance is the sum of all the resistance in the cell i.e.

Total resistance = 2Ω + 8Ω = 10Ω

Using ohms law equation to calculate the current passing through the battery cell:

V = IR

Where; V = voltage, I = current, R = resistance

5V = I × 10Ω

I = 5/10

I = 0.5A

Terminal voltage is calculated by the us of the following equation:

V=emf−IR

Where; R is internal resistance

V = 5 - (0.5 × 2)

V = 5 - 1

V = 4V

Therefore, the potential difference across the terminals of the battery cell is 4V

3 0

2 years ago

Calculate the first and second velocities of the car with four washers attached to the pulley, using the formulas v1 = 0.25 m /

Blizzard [7]

Answer:

0.28 that is for the 25 and for 50 is 0.56

Explanation:

7 0

3 years ago

Read 2 more answers