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KonstantinChe [14]
3 years ago
14

You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c

onstant speed of 1.6 m/s. The rope makes an angle of 35 degrees with the horizontal. What is the net force on the block
Physics
1 answer:
Sergeeva-Olga [200]3 years ago
6 0

Answer:

Fnet = 0

Explanation:

  • Since the block slides across the floor at constant speed, this means that it's not accelerated.
  • According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

       F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N  (1)

  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

⇒    169 N + Fn = Fg = 216 N  (3)

  • This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
  • Fn = 216 N - 169 N = 47  N (4)

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6. DRAW 3 3/8° LINE<br> 7. DRAW A 5* LINE
Fiesta28 [93]

Answer:

Each part so obtained will represent the fraction 1/8 and the number line obtained will be of the form: To mark 3/8; move three parts on the right-side of zero. To mark 5/8; move five parts on the right-side of zero. To mark -1 3/8 i.e. -11/8; move eleven parts on the left-side of zero.

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4 years ago
A 4 kg box is on a frictionless 35° slope and is connected via a massless string over a massless, frictionless pulley to a hangi
Anarel [89]

Answer:

(a) 19.62 N

(b) Box moves down the slope

(c) 24.43 N

Explanation:

(a)  

2 Kg box  causes tension

T=mgwhere m is mass, g is gravitational force taken as 9.81T=2*9.81 =19.62 N  (b)  Block mass of 4 Kg  [tex]T'-mg sin \theta=0 hence T'=mg sin \theta where m is mass and g is gravitational force  

T'=4*9.81 sin 35= 22.5071 N  

Since T' is greater than mg sin\theta , then the box moves down the slope  

(c)  

Acceleration a= \frac {forward   force-backward   force}{Total mass}= \frac {mg sin \theta -mg}{m1 + m2}  

a= \frac {22.51-19.62}{2+4}=0.48

When moving, the box will exert force T"= mgsin \theta + ma  

T"= 4*9.81 sin 35 +(4*0.48)= 24.43 N

7 0
3 years ago
Read 2 more answers
7. You start to walk toward the east towards home at a constant speed of 4 km/hr. At the same time, Someone else leaves your hom
amm1812

A) position time graph for both is shown

here one of the graph is of lesser slope which means it is moving with less speed while other have larger slope which shows larger speed

At one point they intersects which is the point where they both will meet

B) Let the two will meet after time "t"

now we can say that

if they both will meet after time "t"

then the total distance moved by you and other person will be same as the distance between you and home

so it is given as

v_1t + v_2t = d

4*t + 28*t = 3.2 km

t = \frac{3.2}{32} = 0.1 hr

so they will meet after t = 6 min

so from position time graph we can see that two will meet after t = 6 min where at this position two graphs will intersect


4 0
4 years ago
A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. Even without knowing how bi
k0ka [10]

Answer:

a) F₁₂₀ = 1.34 pa A  , b)  F₂₀ = 0.746 pa A

Explanation:

Part. A .    The definition of pressure is

         P = F / A

As the air can approach an ideal gas we can use the ideal gas equation

        P V = n R T

Let's write this equation for two temperatures

       P₁ V = n R T₁

       P₂2 V = n R T₂

       P₁ / P₂ = T₁ / T₂

point 1 has a pressure of P₁ = pa and a temperature of (20 + 273) K, point 2 is at (120 + 273) K, we calculate the pressure P₂

       P₂ = P₁ T₂ / T₁

       P₂ = pa 393/293

       P₂ = 1.34 pa

We calculate the strength

       P₂ = F₁₂₀ / A

       F₁₂₀ = 1.34 pa A

Part B

In this case the data is

Point 1 has a temperature of 393K and an atmospheric pressure (P₁ = pa), point 2 has a temperature of 293K, let's calculate its pressure

        P₁ / P₂ = T₁ / T₂

       P₂ = P₁ T₂ / T₁

       P₂ = pa 293/393

       P₂ = 0.746 pa

Let's calculate the force (F20), from this point

      F₂₀ / A = 0.746 pa

     F₂₀ = 0.746 pa A

6 0
3 years ago
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