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Harlamova29_29 [7]
3 years ago
6

Can someone help me please

Physics
1 answer:
stealth61 [152]3 years ago
4 0
B hope this helps you!!!
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A block with a mass of 1kg moving at a velocity of 3m/s collides and sticks to a block of mass of 4kg initially at rest. What is
Vsevolod [243]

Answer:using Newton third law

Let initial velocity of block be u1=3m/s

Mass of moving block m1 =1kg

Final velocity of block =V

Mass of stationary block m2= 4kg

Since they stick together, their final velocity will be the same.

m1u1 + m2u2=(m1+m2)v

(1*3)+(0*4)=(1+4)v

3=5v

Divide both sides by 5

V=0.6

Final velocity is 0.6m/s

Explanation:

8 0
3 years ago
A force of 7.50 N is applied to a spring whose spring constant is .298 N/cm. Find it’s length
gizmo_the_mogwai [7]

Answer:

to find the length just divide 7.50 and . 298

5 0
3 years ago
A steel plate weighing 180 lb with center of gravity at point G is supported by a roller at point A, a bar DE, and a horizontal
melamori03 [73]

Answer:

Therefore, the force supported  by the hydraulic cylinder is = -70.01 lb

Thus, the force supported by the bar = -233.84 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Explanation:

The attached diagrams is shown in the file below:

From there; taking moments about point A

\sum M__A }=0

- 40 (180) - (80) F_E Cos 37° + 55  

-7200 + F_E (-80 Cos 37° + 55 sin 37° ) = 0

= - 7200 - 30.79  F_E

- 30.79  F_E = 7200

F_E = -\frac{7200}{30.79}

F_E  = -233.84 lb

Thus, the force supported by the bar = -233.84 lb

Taking the equilibrium of forces in the vertical direction

\sum f_y = 0

F_E sin 37 + F_A cos 20 - F_G = 0

- 233.84 sin 37 + F_A cos 20 - 180 = 0

F_A cos 20  = 320.73

F_A = \frac{320.73}{cos 20}

F_A = 341.31 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Taking the equilibrium of forces on the horizontal direction.

\sum f_y = 0

F_E cos 37 - F_{CB} + F_A sin 20° = 0

-233.84 cos 37 - F_{CB} + 341.31 sin 20° = 0  

-186.75 -  F_{CB} + 116.74 = 0

-  F_{CB} -70.01 = 0

-  F_{CB} = 70.01

F_{CB} = - 70.01 lb

Therefore, the force supported  by the hydraulic cylinder is = -70.01 lb

7 0
4 years ago
Solve using correct significant figures and indicating maximum absolute uncertainty.
Vera_Pavlovna [14]

We use the criterion of significant figures to find the result with reliable figures

          X = 9.2 10-5

now with the propagation of errors we obtain the result with its uncertainty

         X ± ΔX = (9.2 ± 0.5) 10⁻⁵

given Parameter

     * expression values ​​with their absolute errors

to find

     * the result with the correct significant figures

     * the absolute error of the expression

Significant figures are defined with the number of decimals that give information, the number of figures in a quantity gives information about the uncertainty of this quantity.

There are two criteria for applying significant figures:

     * Add and subtract the result of going with the number of decimal places of the figure that has the least

    * Product and division as a result of going with the least number of significant figures than the value that has the least.

Remember that the zero to the left do not form a pair of the significant figures

Let's apply this belief to the case presented, let's write the precaution

 

              x = \frac{a-b}{c}

where in this case they are worth

         a = 0.0336 ± 0.0002

         b = 0.010 ± 0.001

         c = 255.4 ± 0.4

We see that the significant figures of each parameterize (a, b, c) and their absolute errors are correct.

Let's apply the criteria to the operation

          a-b = 0.0336 - 0.010

          a- b = 0.0236

we apply the criterion of significant figures for the subtraction, the result must be with 3 decimal places

        a - b = 0.024

let's do the other operation

         X = \frac{a-b}{c}

         X = 0.024 / 255.4

         X = 9.24 10⁻⁵

We apply the criterion of significant figures for the division, in this case the result is left with two significant figures

         X = 9.2 10⁻⁵

The uncertainty or error of the measurements is of most importance as it determines how many significative figures are reliable at a given magnitude.

If the magnitudes are measured with some type of instrument, the absolute error is given by the appreciation of the instrument, if the magnitude is calculated using some equation, the errors must be propagated using the variations of each parameter in the worst case.

           

the uncertainty of the calculated quantity (X) is

        \Delta X = | \frac{dX}{da}| \Delta a + | \frac{dX}{db} | \Delta b + | \frac{dX}{dc}| \Delta c

let's perform the derivatives

        \frac{dX}{da} = \frac{1}{c}

        \frac{dX}{db} = - \frac{1}{c}

        \frac{dX}{dc} = - \frac{a-b}{c^2}

we substitute

remember that the bulk value guarantees that we tune the worst case. So all the mistakes add up

          ΔX = \frac{1}{c}  Δa + \frac{1}{c} Δb + \frac{a-b}{c^2}  Δc

          ΔX = \frac{1}{c} (Δa + Δb) + \frac{a-b}{c^2} Δc

we substitute

         ΔX = \frac{1}{255.4}  (0.0002 + 0.001) + \frac{0.0336-0.010}{255.4^2}  0.4

         ΔX = 4.698 10⁻⁶ + 1.45 10⁻⁷

         DX = 4.8 10-6

Absolute errors must be given with a single significant figure

         ΔX = 5 10⁻⁶

The result of the requested quantity using the criterion of significant figures and propagation of errors is

          X ± ΔX = (9.2 ± 0.5) 10⁻⁵

learn more about   significative figure here:

brainly.com/question/18955573

8 0
3 years ago
What is the relevant similarity in the following argument?
Tasya [4]
Your answer would be C
7 0
3 years ago
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