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3 years ago

a 5.5kg box us pushed across the lunch table. the net force applied to the box is 9.7n. what is the acceleration of the box?

Physics

1 answer:

lys-0071 [83]3 years ago

4 0

Answer:

1.76 m/s^2

Explanation:

The formula to find the force is F = ma (reference from 2nd law of Newton's Laws of Motion). Since you need to find the acceleration of the box, you need to use the formula a = f/m

a = 9.7 n divided by 5.5 kg = 1.76 m/s^2

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The human eye can readily detect wavelengths from about 400 nm to 700 nm. If white light illuminates a diffraction grating havin

nadya68 [22]

Answer:

The range of angles is from 17.50° to 31.76°

Explanation:

The diffraction grid equation is as follows:

$dsen\theta=m\lambda$

Clearing for $\theta$

$sen\theta=\frac{m\lambda}{d}$

$\theta=sen^{-1}(\frac{m\lambda}{d})$

where $\theta$ is the angle, $m$ is the order, in this case $m=1$ , $\lambda$ is the wavelength, and $d$ is defined as follows:

$d=\frac{1}{resolution}$

and since the resolution is 750 lines/mm wich is the same as $750lines/1x10^{-3}m$

$d$ will be:

$d=\frac{1}{750lines/1x10^{-3}m}=\frac{1x10^{-3}m}{750lines}=1.33x10^{-6}m$

wich is the distance between each line of the diffraction grating.

substituting the values for $m$ and $d$ :

$\theta=sen^{-1}(\frac{(1)\lambda}{(1.33x10^{-6}m)})$

And we need to find two angle values: one for when the wavelength is 400nm and one for when it is 700 nm. So we will get the angle range

$\theta=sen^{-1}(\frac{(400x10^{-9})}{(1.33x10^{-6}m)})=17.50$

and

$\theta=sen^{-1}(\frac{(700x10^{-9})}{(1.33x10^{-6}m)})=31.76$

The range of angles is from 17.50° to 31.76°

3 0

3 years ago

A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume th

Katyanochek1 [597]

Answer:

9.88 milivolt

Explanation:

Given: diameter d = 5.2 cm

magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T

t = 0.29 sec.

we know emf = - dΦ/dt

and flux Φ = BA

A= area

therefore emf ε = -A(B_2-B_1)/Δt

$=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV$

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3 years ago

What do you mean by velocity ratio of a wheel and axle

IgorC [24]

Answer:

Explanation:

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3 years ago

The electric current in a copper wire is composed of what

Tanzania [10]

Answer:

A copper wire current consists of electrons appropriately called conduction electrons.

Explanation:

This answer came from quizlet.com. I hope that this helps you and good luck!

8 0

3 years ago

To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel

alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0

3 years ago

a 5.5kg box us pushed across the lunch table. the net force applied to the box is 9.7n. what is the acceleration of the box? ​

a 5.5kg box us pushed across the lunch table. the net force applied to the box is 9.7n. what is the acceleration of the box?