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-BARSIC- [3]
2 years ago
12

A. A cord passing over an easily turned pulley (one that is both massless and frictionless) has 7kg mass hanging from one end an

d a 9kg mass hanging from the other end as shown below. G 7.0 kg 9.0 ks This arrangement is known as Atwood's machine. Calculate the (i). acceleration of the masses (ii). tension in the cord. [2.5 ​
Physics
1 answer:
Zepler [3.9K]2 years ago
6 0

Answer:

M g = M a = a (9 - 7) kg      with the two masses going in different directions

(7 + 9) kg * g = a * 2 kg       2 kg effective in prducing acceleration

a = 16 / 2 * g = 9.8 m/s^2 / 8 = 1.23 m/s^2

M1 a = T - M1 * g        T producing upward acceleration of 1.23 m/s^2

T = 7 (a + g) = 7 * 11 = 77 N

Check:

M2 a =  M2 * g - T          M2 producing downward acceleration

T = M2 (g - a) = 9 * (9.8 - 1.23) = 77 N

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Perform the calculation and report your answer using sig figs. 657.70 - 26.543
Anton [14]

Answer:

The answer is 631.157

Explanation:

The question requested that the answer to the subtraction of 26.543 from 657.70 must be written using significant figures.

Here are a few tips about how to Identify significant figures.

1) It should be noted that <u>the number "0" is what is usually (but not always) affected</u> while trying to identify significant figures. Hence, <u>all other numbers/digits are always significant</u>. For example, 26.543 has five significant figures.

2) The zeros found between these "other numbers/digits" are also significant. For example, 2202 has four significant figures.

3) In the case of a decimal, the tailing zeros or the final zero is also significant. 657.70 and 657.07 have five significant figures.

Now, back to the question

657.70  - 26.543  = 631.157.

Our final answer does not have a zero, hence all the digits (six) are significant.

8 0
3 years ago
Assume you are in the car and the car is moving at a certain speed to
Setler79 [48]
You are at rest with respect to the car.
You are in motion with respect to the School.

6 0
3 years ago
If you live in a very cold area, you may have seen the depth of a bank of snow shrink even though temperatures remain below the
Aleks04 [339]
<span>To answer the question above, if the day sky is clear it collects radiation. If air is dry snow sublimates faster. If both cases overlap it disappears faster where ever coldest. Thank you for posting your question here. I hope my answer helps. </span>
7 0
3 years ago
The sales price of a product is $2 per unit; the variable cost is $1 per unit; and fixed costs total $1,000. how many units must
irakobra [83]

Answer:

<u><em>1000 units for breakeven</em></u>

Explanation:

Let x be the number of units sold at breakeven.

The total sales at the point would be $2x.

Variable costs would be $1x and fixed costs are $1000.

Total costs are = $1x + $1000

At breakeven: Sales = Costs

Sales =m Costs

$2x = $1x + $1000

$1x = $1000

x = 1000 units.

At 1000 units the sales are equal to the costs ("breakeven").

4 0
2 years ago
If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

3 0
3 years ago
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