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3 years ago

A stationary block of mass 35 Kg is suspended on a string.What is the tension in the string ?( Neglect the mass of the string)

Physics

1 answer:

SVEN [57.7K]3 years ago

8 0

Answer:

Tstring = 343.35 [N].

Explanation:

Since the block is in equilibrium, we can say that the sum of the forces on the Y-axis or vertical is equal to zero.

∑Fy = 0

$T_{string}-W=0$

where:

Tstring = tension of the string [N]

W = weight of the block [N]

And the weight can be calculated multiplying the mass of the block by the gravity acceleration.

Now replacing:

$T_{string}=m*g\\T_{string}=35*9.81\\T_{string}=343.35[N]$

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Your diagram is a muscle that helps you breathe. do you think this muscle is voluntary or involuntary? Explain.

PIT_PIT [208]

The diaphragm is primarily an involuntary muscle. Although voluntary means can be achieved, such as when you hold your breath while swimming, this is temporary as the diaphragm will eventually work on its own to supply your body with oxygen.

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3 years ago

A lunar module weighs 12 metric tons on the surface of the Earth. How much work is done in propelling the module from the surfac

zhuklara [117]

Answer:

W=76.55 miles.metric tons

Explanation:

Given that

Weight on the earth = 12 tons

So weight on the moon =12/6 = 2 tons

( because at moon g will become g/6)

As we know that

$F=\dfrac{K}{x^2}$

Here x= 1100 miles

F 2 tons

$2=\dfrac{K}{1100^2}$

$K=2.4\times 10^6$

We know that

Work = F. dx

$W=\int_{x_1}^{x_2}F.dx$

$W=\int_{1100}^{1140}\dfrac{2.4\times 10^6}{x^2}.dx$

$W=-2.4\times 10^6\left[\dfrac{1}{x}\right]_{1100}^{1140}$

$W=-2.4\times 10^6\left[\dfrac{1}{1140}-\dfrac{1}{1100}\right]$

W=76.55 miles.metric tons

6 0

3 years ago

If A > B, under what condition is |A-BI=|A|- IB|? a. Vectors A and B are in opposite directions b. Vectors A and B are in the

kupik [55]

Answer:

b) Vectors A and B are in the same direction.

Explanation:

To understand this problem we will say that vector A has a magnitude of 5 units and vector B a magnitude of 3 units. In the subtraction of vectors the initial parts of vectors always bind together. And the vector resulting from the subtraction is traced from the end of the second vector (B) to the end of the first vector (A).

The length of the resultant vector will be 5 - 3 = 2

In the attached image, we analyze case a), b), and d)

For a)

As we can see in the attached image the resultant vector has a length of 8 units.

For d)

As we can see in the attached image the resultant vector has a length of 5.83 units.

For b)

The resultant vector has a length of 2 units.

Therefore the case given in b) is true