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masha68 [24]
2 years ago
14

A pivot at one end supports a beam of uniform density. Archimedes attaches a cable to the beam 1.25 m from its pivoted end. The

cable makes an angle of 25.0° with respect to the beam. The beam is 5.00 m in length and has a mass of 100.0 kg. The pivot exerts significant forces on the end of the beam. Which one of the following choices best describes the magnitude of the tension in the cable?
A. 1855 N.
B. 4638 N.
C. 1960 N.
D. 2319 N.
E. 980 N.
Physics
1 answer:
sergey [27]2 years ago
6 0

Answer:

B.4638 N

Explanation:

We are given that

Distance of cable from its pivot end=1.25 m

\theta=25^{\circ}

Length of beam=5 m

Half length of beam=\frac{5}{2}=2.5 m

Mass =m=100 kg

We have to find the magnitude of tension in the cable .

According to question

Tsin25\times 1.25=100\times 9.8\times 2.5

T=\frac{100\times 9.8\times 2.5}{sin25\times 1.25}

T=4638 N

Hence,option B is true.

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Two gliders collide on a frictionless air track that is aligned along the x axis. Glider A has an initial velocity of +4.0 m/s a
DedPeter [7]

Answer:

As collision is elastic,thus we can use conservation of momentum equation

mA=0.2 kg

(vB)1=0 m/s.......................as it is on rest before collision

(vA)1=4 m/s

(vA)2=-1 m/s

(vB)2=2 m/s

using equation

(mA*vA+mB*vB)1= (mA*vA+mB*vB)2

Where 1 and 2 represents before and after collision

(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)

0.8=-0.2+(2mB)

mass of object B=mB=0.3 Kg

6 0
3 years ago
A spring has a spring constant of 330 n/m. how far is the spring compressed when 150 newtons of force are used?
pishuonlain [190]

Answer:

I think its B

Explanation:

5 0
3 years ago
A 0.50 kg toy is attached to the end of a 1.0 m very light string. The toy is whirled in a horizontal circular path on a frictio
xenn [34]

Answer:

The maximum speed will be 26.475 m/sec

Explanation:

We have given mass of the toy m = 0.50 kg

radius of the light string r = 1 m

Tension on the string T = 350 N

We have to find the maximum speed without breaking the string  

For without breaking the string tension must be equal to the centripetal force

So T=\frac{mv^2}{r}

So 350=\frac{0.5\times v^2}{1}

v^2=700

v = 26.475 m /sec

So the maximum speed will be 26.475 m/sec

6 0
2 years ago
We want to find how much charge is on the electrons in a nickel coin. Follow this method. A nickel coin has a mass of about 4.2
monitta

Answer:

The number of atoms is N = 4.37*10^{22} \ atoms

Explanation:

From the question we are told that

                 The mass of coin  is m_n = 4.2g

                   Number of atom in one mole = n =6.02*10^{23} \ atoms

                   Molar mass of nickel M = 57.8g

Now the relation to obtain the number of atom in  the  nickel coin is

                        N = \frac{Mass \ of Nickel\ coin}{Molar \ mass\ of nickel }  * No\ of\atoms \ in \ \ one\  mole\ of\ nickel

                           = \frac{4.2}{57.8}* 6.02*10^{23}

                           =4.37 *10^{22} atoms

                 

     

8 0
3 years ago
Given:A=6x-2y B:-4x-8y C:-3x+9y. Commute A+B-C
DedPeter [7]
<span>A+B-C
</span><span>A = 6x - 2y
B = -4x - 8y
C = -3x + 9y

(</span>6x - 2y) + (-4x - 8y) - (-3x + 9y)
(6x - 2y) + (-4x - 8y) + (3x - 9y)
2x -10y + (3x - 9y)

5x - 19y
8 0
3 years ago
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