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solniwko [45]
3 years ago
10

Use the right-hand rule for magnetic force to determine

Physics
1 answer:
Sphinxa [80]3 years ago
5 0
The charge on the moving particle
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A 50 N object is on Earth. What is its mass?
faust18 [17]

Answer:   5.10 KG

Explanation:  W = 50 N

                       BY FORMULA

                         W = MG

                          M = W/G

                          M = 50 /9.8

                          M=  5.10 KG

7 0
2 years ago
Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin
tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of m_2 would be four times that of m_1.

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is T, then the angular velocity of the SHM would be

\displaystyle \omega = \frac{2\pi}{T}.

Assume that the mass starts with a zero displacement and a positive velocity. If A represent the amplitude of the SHM, then the displacement of the mass at time t would be:

\mathbf{x}(t) = A\sin(\omega\cdot t).

The velocity of the mass at time t would be:

\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t).

The acceleration of the mass at time t would be:

\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t).

Let m represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time t would be:

\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t),

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant k will be equal to:

\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}.

Since \displaystyle \omega = \frac{2\pi}{T}, it can be concluded that:

\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}.

For the first mass m_1, if the time period is T_1, then the spring constant would be:

\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2.

Similarly, for the second mass m_2, if the time period is T_2, then the spring constant would be:

\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Since the two springs are the same, the two spring constants should be equal to each other. That is:

\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Simplify to obtain:

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

6 0
3 years ago
Monochromatic light of 605 nm falls on a single slit, which is located 85 cm from a
Anna [14]

Answer:

mhm

Explanation:

The answer is c 00.095

4 0
3 years ago
Hey guys! Am I right? Thanks!
worty [1.4K]

Answer:

the correct answer is reduce friction

5 0
2 years ago
You are asked to design a spring that will give a 1020 kg satellite a speed of 2.25 m/s relative to an orbiting space shuttle. Y
julsineya [31]

Answer:

(a) 2.45×10⁵ N/m

(b) 0.204 m

Explanation:

Here we have that to have a velocity of 2.25 m/s then the relationship between the elastic potential energy of the spring and the kinetic energy of the rocket must be

Elastic potential energy of the spring =  Kinetic energy of the rocket

\frac{1}{2} kx^2 = \frac{1}{2} mv^2

Where:

k = Force constant of the spring

x = Extension of the spring

m = Mass of the rocket

v =  Velocity of the rocket

Therefore,

\frac{1}{2} kx^2 = \frac{1}{2} \times   1020 \times 2.25^2

or

kx^2 =  1020 \times 2.25^2 = 10,226.25\\So \ that \ the \ force \ on \ the \ satellite\ kx = \frac{10226.25}{x}

(b) Since the maximum acceleration is given as 5.00×g we have

Maximum acceleration = 5.00 × 9.81 = 49.05 m/s²

Hence the force on the rocket is then;

Force = m×a = 1020 × 49.05 = ‭50,031 N

kx = \frac{10226.25}{x} = 50031 \ N

Therefore,

x = \frac{10226.25}{ 50031} = 0.204 \ m

(a) From which

k = \frac{10226.25}{x^2} = \frac{50031}{x} = \frac{50031}{0.204} = 244,772.13 \ N/m or

Force constant of the spring, k = 2.45×10⁵ N/m.

6 0
3 years ago
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