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polet [3.4K]
3 years ago
12

A 20 kg box is being pulled across a floor by a horizontal rope. The tension in the rope is 99 Newtons. The coefficient of frict

ion is 0.25. What is the force of friction on the box? What is the acceleration of the box?
Physics
2 answers:
fgiga [73]3 years ago
5 0

Answer:

friction = 49 N, a = 2.5 m/s/s

Explanation:

big brain

Burka [1]3 years ago
4 0
Yeah what that guy said lol
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Calculate the speed in m/s at which the moon revolves around the Earth. Note: the orbit is nearly circular.
Vilka [71]
The moon is 384,000 km away from the Earth. It revolves around the earth in 27.3 days. (pi = 3.14)

Speed is rv = r 2pi/T

v = 384,000 2pi/27.3 * 86400

   = 27.9 km/s

Good luck :)
7 0
3 years ago
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Metamorphic rocks are formed at high temperature and medium to low pressure
storchak [24]

Answer: are u kidding

Explanation:

6 0
3 years ago
a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet
ANTONII [103]

Answer:

Explanation:

Given

radius of circular path r=4\ in.

Position is given by

\theta =\cos 2t---1

Differentiate 1  to angular velocity we get

\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2

Differentiate 2 to get angular acceleration

\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3

Net acceleration is the vector summation of tangential and centripetal force

a_t=\alpha \times r

a_t=-4\cos 2t\times 4=-16\cos 2t

a_r=\omega ^2\cdot r

a_r=(-2\sin 2t)^2\cdot 4

a_r=16\sin^2(2t)

a_{net}=\sqrt{a_r^2+a_t^2}

a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}

a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}                                                    

6 0
3 years ago
1)Light of wavelength 588.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 55.5 cm from the slit
Talja [164]

Answer:

These are Diffraction Grating Questions.

Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:  

Given as  

y = nDλ/w                                                       Eqn 1

where  

w = width of slit  

D = distance to screen  

λ = wavelength of light  

n = order number  

Making x the subject of the formula gives,  

w = nDλ/y  

Given  

y = 0.0149 m  

D = 0.555 m  

λ = 588 x 10-9 m  

and n = 3

w = 6.6x10⁻⁵m

Hence, the width of the slit w, in micrometers (μm) = 66μm

Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen

i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx

Recall Eqn 1,     y = nDλ/w  

given, D = 27cm = 0.27m  

λ = 632 x 10-9 m  

w = 0.1mm = 1.0x10⁻⁴m

For the 9th order, n = 9,

y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m

Similarly, for n = 5,

y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m

Recall,  Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m

Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm

8 0
3 years ago
Help please!!! Brainlyist to the person who answers and explains correctly!
postnew [5]

Answer:

answer is box 1

Explanation:

Ok so an equilibrium of a dynamical system is a value of the state variables where the state variables do not change.In other words, an equilibrium is a solution that does not change with time

5 0
3 years ago
Read 2 more answers
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