Justin is riding his bike up Dunmore Hill. The effort force from his feet on his pedals is 1160N.

2. A powerful experimental sewing machine is powered by a mass-spring system. This

Alexus [3.1K]

We have that the Number of stitches per sec and he mass of oscillation motion is mathematically given as

a) Nt=25stitches per sec

b) m=2.033e-5kg

<h3>Number of stitches per sec and he mass of oscillation motion</h3>

Question Parameters:

This sewing machine is capable of stitching 1,500 stiches in one minute.

If the sewing machine has a spring constant of 0.5 N/m,

Generally the equation for the Number of stitches per sec is mathematically given as

Nt=N/t

Therefore

Nt=1500/60

Nt=25stitches per sec

b)

Generally the equation for the Time t is mathematically given as

$T=2\pi\sqrt{\frac{m}{k}}$

Therefore

$0.04=2\pi\sqrt{\frac{m}{0.5}}\\\\m=\frac{0.5*0.04^2}{4\pi^2}$

m=2.033e-5kg

For more information on Mass visit

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7 0

2 years ago

Lol idk what this is

alex41 [277]

Answer:

I know

Explanation:

Physics= hard

3 0

3 years ago

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The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

3 years ago

A cylindrical vessel with water is rotated about its vertical axis with a constant angular velocity co. Find:

Nataly [62]

Answer:

Explanation:

The question is one that examine the physical fundamental of mechanics of a cylindrical vessel .

We would use the Euler' equation and some coriolis and centripetal force formula.

The fig below explains it.

4 0

3 years ago

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To understand the nature of electric fields and how to draw field lines. Electric field lines are a tool used to visualize elect

Brrunno [24]

Explanation:

The electric field is defined as the change in the properties of space caused by the existence of a positively (+) or negatively (-) charged particle. The electric field can be represented by infinitely many lines from a particle, and those lines never intersect each other. Depending on the type of charge we can see different cases:

Let's say we have a positive charge alone (image 1). The field lines are drawn from the centre of the particle outwards to infinity (in other words, they disappear from the edge of the picture). Meaning the direction of the electric field points outwards the particle.

For a negative charge alone (image 2), the lines come from infinity to the centre, and point towards the particle (i.e. lines appear from the edge of the picture).

Let's see what happens if we have two charges together:

Two positive charges (image 3): Since the charges are of the same type (positive), the particles repel each other. Then the field lines will avoid each other so they do not join. The charge is positive, so lines point outwards.
Two negative charges (image 4): Again, the charges are both negative, so they repel. But they are negative, so the field points inwards.
Negative and positive charges (image 5): They are different charges, so the force between them is attractive. This causes the field lines from both to join. They go out of the positive and come into the negative particle.

Image 6:

The lines are passing through infinite points of the space. If we choose a certain point and measure the electric field, we can see to which direction the electric field points. This is the direction of the electric field vector. It does not matter which point we choose; the electric field vector touches the field line only at this point, which means it is tangent to the field line.

7 0

3 years ago