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zepelin [54]
3 years ago
10

Consider the two-body system at the right. A 22.7-N block is placed upon an inclined plane which is inclined at a 17.2 degree an

gle. The block is attached by a string to a 34.5-N block which is suspended over a pulley. The coefficient of friction is 0.219. Determine the acceleration of the block and the tension in the stringAnswers:

Physics
1 answer:
artcher [175]3 years ago
3 0

Answer:

Given that

For A weight Wt= 22.7 N

m₁ = 22.7/10 = 2.27 kg

Force alone inclined plane

Wt₁ = m₁ g sin θ

Wt₁  =22.7 sin 17.2°

Wt₁ = 6.7 N

For B weight Wt₂= 34.5-N

m₂ = 3.45 kg

coefficient of friction ,μ= 0.219

θ = 17.2 degree

The friction force on the block A

Fr= μ m₁ g cos θ

Fr= 0.216 x 22.7 x cos 17.2°

Fr= 4.68 N

Lets take acceleration of system is a m/s²

Tension = T

From Newtons law

Wt₂ - Wt₁ - Fr = (m₁ +m₂) a

34.5 -  6.7 - 4.68 = (2.27 + 3.45 ) a

a= 4.05 m/s²

Block B

Wt₂ - T = m₂ a

T = 34.5 - 4.05 x 3.45

T= 20.52 N

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Let's examine over the next few questions an NFL kick as described in the 3rd Law video. The announcer claimed that the ball is
hichkok12 [17]

Answer:

8451.62109367671 Newtons

Explanation:

1\ kg=1\times 9.8066500286389=9.8066500286389\ N

1 \lb=2.2046226218488\ kg

It is known that

1\ lbs=\dfrac{9.8066500286389}{2.2046226218488}=4.4482216282508\ N

So,

1900\ lbs=1900\times 4.4482216282508\\\Rightarrow 1900\ lbs=8451.6210936765\ N

The force in Newtons is 8451.6210936765 Newtons

4 0
3 years ago
They realize there is a thin film of oil on the surface of the puddle. If the index of refraction of the oil is 1.81, and they o
Sphinxa [80]

Answer:

The right solution is "165.8 nm".

Explanation:

Given:

Index of refraction,

n = 1.81

Wavelength,

λ = 600 nm

We know that,

⇒ t=\frac{\lambda}{2\times n}

By putting the values, we get

      =\frac{600}{2\times 1.81}

      =165.8 \ nm

3 0
3 years ago
The greater the difference in electronegativity between two covalently bonded<br><br> atoms
katrin [286]

Answer:

The greater the difference in electronegativity between two covalently bonded atoms, the greater the bond's percentage of ionic character.

Explanation:

Bond polarity (i.e the separation of electric charge along a bond) and ionic character (amount of electron sharing) increase with an increasing difference in electronegativity.

Therefore, we can say that, the greater the difference in electronegativity between two covalently bonded atoms, the greater the bond's percentage of ionic character.

7 0
3 years ago
If an electron is released at PP , what is the magnitude of the net force that these rods exert on it?
pishuonlain [190]

The magnitude of the net force that the rods exert after an electron is released at point P is 2.885 × 10⁻¹⁵ N.

Given values:

Length of non-conducting rod, l = 1.20 m

Charge on positive rod, +Q = +2.50 μC = +2.50 × 10⁻⁶ C

Charge on negative rod, -Q = -2.50 μC = -2.50 × 10⁻⁶ C

Distance from point P of each rod, x = 60 cm = 0.60 m

Calculation of Net electric force exerted on point P:

Consider an electron released at point P, then the net electric force exerted will be given as:

F = e. E_net       - ( 1 )

Step 1:

The net electric field value is given as:

E_net  = E₁ cos Φ + E₂ cos Φ      

           = 2E₁ cos Φ                  -( 2 )

where, E₁ & E₂ are electric fields due to positive and negative rod                

            respectively.

            Φ is phase angle

Step 2:

The electric field due to positive rod is given as:

E₁ = k (λ/r)             - ( 3 )

where, k is Coulomb's force constant

            λ is linear charge density

            r is distance between point P and half of the rod.

Now, the linear charge density is given as:

λ = Charge/length = Q/x

The value of r is given as:

r = √x²-a²

where, x is length of rod

           a is half length of rod

Applying values in above equation, we get:

r = √x²-(x/2)²

r = √(1.20 m)²-(1.20/2)²

  = √1.08

  = 1.04 m

Substituting all the determined values in equation 3 we get:

E₁ = k (λ/r)

   = k [(Q/x)/r]

   = k [ Q/xr ]

   = (9×10⁹ Nm²/C²) [ |+2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

   = 1.803×10⁴ N/C

Step 3:

Similarly, the electric field due to negative rod is given as:    

E₂ = k [ Q/xr ]

    = (9×10⁹ Nm²/C²) [ |-2.50×10⁻⁶ C|/(1.20 m)(1.04 m)]

    = 1.803×10⁴ N/C

Step 4:

Consider equation 2:

E_net  = 2E₁ cos Φ

From the figure we get the phase angle as:

Φ = tan⁻¹ (0.60 m/0.60 m)

   = tan⁻¹ ( 1 )

   = π/4  

Now, the electric field produced due to each rod is equal and mutually perpendicular. Thus, the net electric field after applying values can be calculated as:

E_net = 2(1.803×10⁴ N/C) cos π/4

          = 2(1.803×10⁴ N/C) (0.5)

          = 18030 N/C

Step 5:

Consider equation 1 :

F = e. E_net

where, e is charge on an electron

Applying values in above equation we get:

F = (1.6 × 10⁻¹⁹ C)(18030 N/C)

  = 2.885 × 10⁻¹⁵ N

Therefore, the magnitude of the net force that the rods exert after an electron is released at point P is  2.885 × 10⁻¹⁵ N.

Learn more about electric force here:

<u>brainly.com/question/1634182</u>

#SPJ4      

8 0
2 years ago
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