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Diano4ka-milaya [45]
3 years ago
7

Physics!!! Please help!!!!!!!

Physics
1 answer:
Marianna [84]3 years ago
8 0

Answer:

it is a very good morning amor de g the first paragraph of the first paragraph of the first paragraph of the first paragraph of the first paragraph of the first paragraph of 88th paragraph of the first paragraph

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The figure shows two 1.0 kg blocks connected by a rope. a second rope hangs beneath the lower block. both ropes have a mass of 2
lutik1710 [3]
We need first to use the formula  F=m(a+g), m iis the total mass, a is the acceleration, g is gravity pulling the blocks. So the procedure will be 
<span>m=2kg(both blocks)+500g(both ropes) → m=2.5kg </span>
<span>a=3.00m/s^2 </span>
<span>g=9.8m/s^2 </span>
<span>F=m(a+g) → F=2.5kg (3.00m/s^2 + 9.8m/s^2) → F=2.5kg (12.8m/s^2) → F=32 N 
To calculate the tension at the top of rope 1 you need to use the formula </span>T=m(a+g) so it will be <span>T=m(a+g) → T=1.5kg(12.8m/s^2) → T=19.2N 
</span>We can now calculate the tension at the bottom of rope 1 using the formula: <span>T=m(a+g) → T=1.25kg(12.8m/s^2) → T=16N 
</span>Now to find the tension at the top of rope 2 we do it like this: 
<span>T=m(a+g) → T=.25kg(12.8m/s^2) → T=3.2</span>
7 0
3 years ago
Read 2 more answers
Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the
Gnesinka [82]

Answer:

F = 233.52 N,  θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

        = 80.5 N

Jill

       cos 45 = F_{2x} / F₂2

       sin 45 = F_{2y} / F₂2

       F_{2x} = F₂ cos 45

       F_{2y} = F₂ sin 45

       F_{2x} = 81.7 cos 45 = 57.77 N

       F_{2y} = 81.7 sin 45 = 57.77 N

Jane

      cos (270 + 45) = F_{3x} / F₃3

      sin 315 = F_{3y} / F₃

      F_{3x} = 131 cos 315 = 92.63 N

      F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

         F_{x} = F_{1x} + F_{2x} + F_{3x}

         F_{x} = 80.5 +57.77 + 92.63

         F_{x} = 230.9 N

Axis y

         F_{y} = F_{1y} + F_{2y} + F_{3y}

         F_{y} = 0 + 57.77 -92.63

         F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

         F = √(Fₓ² + F_{y}²

        F = √(230.9² + 34.86²)

        F = 233.52 N

let's use trigonometry for the angle

        tan θ = \frac{F_y}{F_x} }

        θ = tan⁻¹ (\frac{F_y}{F_x} })

        θ = tan⁻¹ (-34.86 / 230.9)

        θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

          θ' = 360 -θ

          θ‘= 360- 8.59

          θ' = 351.41º

5 0
3 years ago
Jonas and his family are moving to another part of the city. As Jonas, his brother, and his Dad were driving one of the trucks f
tiny-mole [99]

Answer:newtons second law

Explanation:

To get something to accelerate you have to apply a pull force . if the mass increase a grater pull is required

6 0
3 years ago
You are outdoors when you hear the constant chirp of a still cricket. You start walking toward the cricket and at some point you
givi [52]

Answer:

4) True, the distance was reduced by a de factor 2

Explanation:

To analyze the statements, let's first see how the sound is produced

The cricket makes the sound rubbed its legs, the energy needed to produce it is constant, after the sound is produced it expands in all directions and this wave is subject to the law of energy conservation.

By expanding in all directions we can assume that the energy for the time that is the power of the sound is distributed over a sphere with different radii each time. The perceived intensity is defined as the power per unit area

       I = P / A

Where P is the power, A are the areas and I are the intensities heard by the person

Let's clear the power and match, for two different radii (distances)    

      P = I₁ A₁ = I₂ A₂

The area of ​​a sphere is

     A = 4π R₂

     I₂ = I₁ A₁ / A₂

     I₂ = I₁ 4π R₁² / 4π R₂²

     I₂ = I₁ R₁² / R₂²

Let's analyze this last equation for the situation presented if the intensity increases 4 times, substitute

    I₂ = 4 I₁

    4 I₁ = I₁ R₁² / R₂²

   R₂ = R₁ / 2

We see that to increase the intensity 4 times the distance was reduced twice

Now we can analyze the expressions given

2) False, the power is constant, the cricket always does the same job

3) False, the speed of the wave is constant in the air

4) True, the distance was reduced by a de factor 2

3 0
3 years ago
If Hubble's constant is 73 km/s/Mpc then the age of the universe is 13.4 billion years, assuming a uniform expansion. Suppose it
ki77a [65]

Answer:

It would decrease the calculated age of the universe

Explanation:

Since the age of the universe is the reciprocal of Hubble's constant, it's therefore if Hubble's constant is increased the age decreases but if the Hubble's constant is decreased, the age of universe increases. Therefore, the age of universe and Hubble's constant are inversely proportional. Conclusively, any attempt to increase Hubble's constant would imply the calculated age of the universe decreases.

3 0
3 years ago
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