All categories

3 years ago

True or false Induction works due to the movement of protons

Physics

1 answer:

Anestetic [448]3 years ago

6 0

the answer is false .

induction works due to the mass movement of neutrons actually . pls mark brainliest

You might be interested in

A weightlifter works out at the gym each day. Part of her routine is to lie on her back and lift a 43 kg barbell straight up fro

Tasya [4]

Answer:

A. 231.77 J

B. 5330.71 J

C. 46 donuts

Explanation:

A. To lift the barbell once, she will have to extend it the full length of her arm. The work done will then be:

W = F * d

Where the force is the weight of the barbell.

F = m * g

Hence, the work done in lifting the barbell is:

W = m * g * d

W = 43 * 9.8 * 0.55

W = 231.77 J

B. If she does 23 repetitions, the total energy she expend will be equal to the Potential energy when the barbell is lifted multiplied by 23:

E = 23 * m * g * d

E = 23 * 231.77

E = 5330.71 J

C. 1 Joule = 4.184 calories

5330.71 Joules = 5330.71 * 4.184 = 22303.69

If 1 donut contains 490 calories, the number of donuts she will need will be:

N = 22303.69/490 = 45.5 donuts or 46 donuts

5 0

4 years ago

Kūna veikia jėga kurios momentas 0,4 N × m petys - 5 cm. Koks šios jėgos dydis?

umka21 [38]

Answer:

F = 8 N

Explanation:

The question says "The body is subjected to a force with a moment of 0.4 N × m shoulder - 5 cm. What is the magnitude of this force?

Given that,

Moment/Torque, $\tau=0.4\ N-m$

Distance moved, d = 5 cm = 0.05 m

We need to find the magnitude of this force. We know that, the torque acting on an object is given by :

$\tau=Fd\\\\F=\dfrac{\tau}{d}\\\\F=\dfrac{0.4\ N-m}{0.05\ m}\\\\F=8\ N$

So, the magnitude of force is equal to 8 N.

8 0

3 years ago

What forces act upon an object dropped in a vacuum

Vlada [557]

Gravity is the only one, since there's no air resistance.

8 0

3 years ago

A uniform disk of mass 3.25 kg has a radius of 0.100 m and spins with a frequency of 0.550 rev/s. What is its angular momentum?

shutvik [7]

ANSWER:

0.0562 J

STEP-BY-STEP EXPLANATION:

Angular momentum is expressed in terms of moment of inertia and angular velocity. This is expressed as follows:

$L=I\cdot\omega$

Here, I is the angular momentum and ω is the angular velocity.

Angular momentum is mass time the square of the radius of the object. Moment of inertia for a uniform disk is given as,

$I=\frac{1}{2}m\cdot r^2$

Here, m is the mass of the disk and r is the radius of the disk.

Replacing:

$L=\frac{1}{2}m\cdot r^2\cdot\omega$

Convert the units of angular velocity into rad/s.

$\begin{gathered} \omega=0.55\text{ rev/s}\cdot\frac{2\pi\text{ rad}}{1\text{ rev}} \\ \omega=3.46\text{ rad/s} \end{gathered}$

We replace each data to calculate the angular momentum:

$\begin{gathered} L=\frac{1}{2}\cdot3.25\cdot0.1^2\cdot3.46 \\ L=0.0562 \end{gathered}$

The angular momentum of the uniform disk is 0.0562 J

7 0

1 year ago

Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on

Romashka [77]

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$\frac{1}{4\pi\spsilon_0}=9\times10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35\times10^{-4}$ N

So, from equation (i)

$1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}$

$\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$\frac{q_1+q_2}{2}$

d=20cm=0.2m, and $F_c=1.406\times10^{-4}$ N

So, from equation (i)

$1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}$

$\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}$

$\Rightarrow q_1+q_2=\pm5\times 10^{-8}$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)$