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Bumek [7]
3 years ago
6

Please answer this question I have uploaded the picture of question please answer​

Physics
1 answer:
8090 [49]3 years ago
5 0

Answer:

20 mangintiude beacuse

Explanation:

mt everst is in mounatin region we eat rice we eat pizza burger sandwich go to thamel for a good reason doing lamo lamo hw so be obtidnet you 4 kaccha fail boy

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How can a mass be hot enough to flow but still act like a solid
Lunna [17]
The inner core is solid because it is made of very dense, or heavy, materials like iron and nickel. Even though it is very hot, these materials don't
6 0
2 years ago
g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction betwe
olga_2 [115]

Answer:

The minimum compression is  x= 0.046m

Explanation:

From the question we are told that

              The mass of the block is m_b = 1.45 kg

               The spring constant is  k = 860 N/m

               The coefficient of static friction is  \mu = 0.36

For the the block not slip it mean the sum of forces acting on the  horizontal axis is equal to the forces acting on the vertical axis

     Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

                   F_y = m_b*g

And the force acting on the horizontal axis is  force due to the spring which is mathematically represented as

                   F_x = k *x * \mu

where x is the minimum compression to keep the block from slipping

        Now equating this two formulas and making x the subject

                      x = \frac{m_b * g}{k * \mu}

substituting values we have

                     x = \frac{1.45 * 9.8}{860 *0.36}

                        x= 0.046m

 

3 0
3 years ago
The vector position of an object is given by r what is the torque acting on the object about the origin when a force f = (−12.5i
attashe74 [19]
Let the vector position of the object in the (x-y) plane be 
\vec{r} = x \hat{i} + y \hat{j}

The applied force is
\vec{f} = -12.5 \hat{i}


By definition, the applied torque is
\vec{T} = \vec{r} \times \vec{f} = (x\hat{i} + y\hat{j}) \times (-12.5y \hat{i}) = 12.5\hat{k}

Answer: 12.5y \, \hat{k}

7 0
3 years ago
A cord of mass 0.65 kg is stretched between two supports 8.0 m apart. If the tension in the cord is 120 N, how long will it take
DedPeter [7]

The time taken by the pulse to travel from one support to the other is 0.208 s.

<h3>Given:</h3>

The mass of the cord is m = 0.65 kg.

The distance between the supports is, d = 8.0 m.

The tension in the cord is T = 120 N.

The time taken by the pulse to travel from one support to the other is given as,

v=\frac{d}{t}

t=\frac{d}{v}

Here, v is the linear velocity of a pulse. Its value is,

v=\sqrt{\frac{T d }{m} }

v=\sqrt{\frac{120 * 8}{0.65} }

v= 38.43 m/s

Then,

t=\frac{8}{38.43}

t=0.208 s

Thus, the time taken by the pulse to travel from one support to the other is 0.208 s.

Learn more about tension here:

brainly.com/question/24994188

#SPJ4

7 0
1 year ago
A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
3 years ago
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