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Marina CMI [18]
2 years ago
7

a 5.00kg crate is on a 21.0° hill. Using X-Y axes tilted down the plane, what is the x-component of the weight?

Physics
2 answers:
Elodia [21]2 years ago
8 0

M g = 5.00 (9.81) = 49.05

sin 21 =  x149.05

x = 17.6N

monitta2 years ago
7 0

Answer:

The x- component of the weight is 17.56 N.

Explanation:

It is given that,

Mass of the crate, m = 5 kg

X-Y axes tilted down a plane. We have to find the x- component of the weight.  The weight of the crate is acting vertically downward.

From the attached figure, the x- component of the crate is given by

W_x=mg\ sin\ \theta

So, W_x=5\ kg\times 9.8\ m/s^2\times sin(21)

W_x=17.56\ N

So, the x- component of the weight is 17.56 N. Hence, this is the required solution.                                             

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An organ pipe open at both ends has a radius of 4.0 cm and a length of 6.0 m. what is the frequency (in hz) of the third harmoni
Marysya12 [62]

When air is blown into the open pipe,

L = \frac{nλ}{2}

where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation

⇒λ=\frac{2L} {n}

Note here that n=1 is for fundamental, n=2 is first harmonic and so on..

⇒ third harmonic will be n=4

Given L=6m, n=4, solving for λ we get:

λ=\frac{(2)*(6)}{4} =3m

Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:

c=f.λ Or f= \frac{c}{λ}

⇒f=\frac{344}{3}

≈115 Hz

8 0
2 years ago
Every winter I fly home to Chicago. It takes 3 hours. What is my average speed?
Tanya [424]

It depends on where you live when you're not visiting Chicago. We need to know the distance of the trip.

Your average speed on the trip is . . .

(total distance in miles) / (3 hours)

miles per hour

5 0
2 years ago
A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit
makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

            2.87 = -1.442 + 1.8 u

               u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0
2 years ago
A stunt car driver testing the use of air bags drives a car at a constant speed of25 m/s for a total of 100m. He applies his bra
PIT_PIT [208]

Answer:

The graphs are attached

Explanation:

We are told that he starts with a constant speed of 25 m/s for a distance of 100 m.

At constant velocity, v = distance/time

time(t) = distance(d)/velocity(v)

t1 = 100/25

t1 = 4 s

Now, we are told that he applies his brakes and accelerates uniformly to a stop just as he reaches a wall 50m away.

It means, he decelerate and final velocity is zero.

Thus;

v² = u² + 2as

0² = 25² + 2a(50)

25² = - 100a

625 = - 100a

a = - 625/100

a = - 6.25 m/s²

v = u + at

0 = 25 + (-6.25t)

25 = 6.25t

t = 25/6.25

t = 4 s

With the values gotten, kindly find attached the distance-time and velocity-time graphs.

4 0
2 years ago
Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the
Stells [14]

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0
2 years ago
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