Types of mechanical waves include

A singly charged ion (q=−1.6×10−19) makes 7.0 rev in a 45 mT magnetic field in 1.29 ms. The mass of the ion in kg is

insens350 [35]

Answer:

$m=1.47\times 10^{-24}\ Kg$

Explanation:

Given that,

Charge, $q=1.6\times 10^{-19}\ C$

Revolution = 7 rev

magnetic field, B = 45 mT

Time, t = 1.29 ms

We need to find the mass of the ion. Let m be the mass. The formula for the mass in terms of time period is given by :

$m=\dfrac{qBT}{2\pi}\\\\m=\dfrac{1.6\times 10^{-19}\times 45\times 10^{-3}\times 1.29\times 10^{-3}}{2\pi}\\\\m=1.47\times 10^{-24}\ Kg$

So, the mass of the ion is equal to $1.47\times 10^{-24}\ Kg$ .

4 0

2 years ago

Two 2.0 kg bodies, A and B, collide. The velocities before the collision are ~vA = (15ˆi + 30ˆj) m/s and ~vB = (−10ˆi + 5.0ˆj) m

AleksandrR [38]

Answer:

Part a)

$10\hat i + 15\hat j = \vec v$

Part b)

$\Delta K = 500 J$

Explanation:

As we know that there is no external force on the system of two masses so here total momentum of the system will remains conserved

so we can say

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$(2kg)(15\hat i + 30 \hat j) + (2 kg)(-10\hat i + 5\hat j) = 2kg(-5\hat i + 20\hat j) + 2\vec v$

$5\hat i + 35\hat j = (-5\hat i + 20\hat j) +\vec v$

$10\hat i + 15\hat j = \vec v$

Part b)

magnitude of the initial speed of A = $\sqrt{15^2 + 30^2} = 33.54 m/s$

magnitude of the initial speed of B = $\sqrt{10^2 + 5^2} = 11.18 m/s$

magnitude of final speed of A = $\sqrt{5^2 + 20^2} = 20.61 m/s$

magnitude of final speed of B = $\sqrt{10^2 + 15^2} = 18.03 m/s$

Now change in total kinetic energy is given as

$\Delta K = (\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2) - (\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2)$

$\Delta K = (\frac{1}{2}2(33.54)^2 + \frac{1}{2}2(11.18)^2) - (\frac{1}{2}2(20.61)^2 + \frac{1}{2}2(18.03)^2)$

$\Delta K = 500 J$

6 0

3 years ago

The change in momentum of an object is equal to the Question 4 options: Force acting on it times its velocity. impulse acting on

lesya [120]

Answer:

impulse acting on it

Explanation:

The impulse is defined as the product between the force applied to an object (F) and the time interval during which the force is applied ( $\Delta t$ ):

$I=F\Delta t$

We can prove that this is equal to the change in momentum of the object. In fact, change in momentum is given by:

$\Delta p = m \Delta v$

where m is the mass and $\Delta v$ is the change in velocity. Multiplying and dividing by $\Delta t$ , we get

$\Delta p = m \frac{\Delta v}{\Delta t} \Delta t$

and since $\frac{\Delta v}{\Delta t}$ is equal to the acceleration, a, we have

$\Delta p = ma \Delta t$

And since the product (ma) is equal to the force, we have

$\Delta p = F \Delta t$

which corresponds to the impulse.

5 0

3 years ago

What would we need to do to make an electromagnet strong enough to move cars and trains

SVEN [57.7K]

Answer:

The combined magnetic force of the magnetized wire coil and iron bar makes an electromagnet very strong. In fact, electromagnets are the strongest magnets made. An electromagnet is stronger if there are more turns in the coil of wire or there is more current flowing through it.

5 0

2 years ago

Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec

Mnenie [13.5K]

Answer:

ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

5 0

2 years ago