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3 years ago

13

a spring scale is constructed so that each 0.10-n increment is separated by 0.50 cm. what spring constant is necessary for the s

cale to be accurate?

1 answer:

Goryan [66]3 years ago

7 0

Answer:

5.00

Explanation:

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Please help on this one somebody?

coldgirl [10]

7.5 x 10⁻¹¹m. An electromagnetic wave of frecuency 4.0 x 10¹⁸Hz has a wavelength of 7.5 x 10⁻¹¹m.

Wavelength is the distance traveled by a periodic disturbance that propagates through a medium in a certain time interval. The wavelength, also known as the space period, is the inverse of the frequency. The wavelength is usually represented by the Greek letter λ.

λ = v/f. Where v is the speed of propagation of the wave, and "f" is the frequency.

An electromagnetic wave has a frecuency of 4.0 x 10 ¹⁸Hz and the speed of light is 3.0 x 10⁸ m/s. So:

λ = (3.0 x 10⁸ m/s)/(4.0 x 10¹⁸ Hz)

λ = 7.5 x 10⁻¹¹m

8 0

3 years ago

1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant

Vladimir79 [104]

Answer:

The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.

Explanation:

Given:

Speed of jumbo jet in southwesterly direction $(v_j)$ = 550 mph

Velocity of jet stream from west to east direction $(v_s)=80\ mph$

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

$\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph$

Similarly, the velocity of the stream is, $\vec{v_s}=80\vec{i}$

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

$\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}$

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

$|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph$

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle $\theta$ with the x axis in the third quadrant.

The direction is given as:

$\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)$

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

5 0

3 years ago

A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find

Ludmilka [50]

Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

m = mass of the ladder, 7.98 kg

L = length of the ladder, 4.15 m

On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

I = 137.44 / 12

I = 11.45 kg·m²

That is the moment of inertia about the center.

Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that

τ = 11.453 kg·m² * 0.395 rad/s²

τ = 4.535 N·m

8 0

3 years ago

Bullets spin when shot from a rifle or handgun. What causes this spinning?

tamaranim1 [39]

Answer:

The spark from the primer ignites the gunpowder. Gas converted from the burning powder rapidly expands in the cartridge. The expanding gas forces the bullet out of the cartridge and down the barrel with great speed. The rifling in the barrel causes the bullet to spin as it travels out of the barrel.

Explanation:

7 0

2 years ago

The magnetic flux through a loop:

dexar [7]

Answer:

The magnetic flux through a loop is zero when the B field is perpendicular to the plane of the loop.

Explanation:

Magnetic flux are also known as the magnetic line of force surrounding a bar magnetic in a magnetic field.

It is expressed mathematically as

Φ = B A cos(θ) where

Φ is the magnetic flux

B is the magnetic field strength

A is the area

θ is the angle that the magnetic field make with the plane of the loop

If B is acting perpendicular to the plane of the loop, this means that θ = 90°

Magnetic flux Φ = BA cos90°

Since cos90° = 0

Φ = BA ×0

Φ = 0

This shows that the magnetic flux is zero when the magnetic field strength B is perpendicular to the plane of the loop.

4 0

3 years ago

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