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Dmitrij [34]
3 years ago
13

Determine the diameter of the largest circular hole that can be punched into a sheet of polystyrene 6 mm thick, knowing that the

force exerted by the punch is 45 kN and that a 55-MPa average shearing stress is required to cause the material to fail.
Physics
1 answer:
Yakvenalex [24]3 years ago
4 0

Answer:

43.41 mm

Explanation:

Given:

thickness of sheet, t = 6 mm

Force exerted by punch, F = 45 KN

Average shearing stress, T = 55 MPa

From average shearing stress T = Force F / Area A

Hence area = force/stress =45000/ 55 =818.18 mm^2

From area = pi*diameter*thickness

diameter = area/(pi* thickness)

= 818.18/(3.142*6)

= 43.41 mm

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Which of the following is a terrestrial habitat?<br>A) Pond B) Garden C) Lake D) River​
kenny6666 [7]

Answer:

garden

Explanation: All the other habitats are aquatic

8 0
3 years ago
How much tension must a rope withstand if it is used to accelerate a 6526 kg car vertically upwards at 8.9 m/s^2?
ExtremeBDS [4]

Answer:

The value is T =  122036.2 \  N

Explanation:

From the question we are told that

     The mass of the car is  m  = 6526 \  kg

      The acceleration  is  a=  8.9 \  m/s^2

Generally the net force applied on the rope is mathematically represented as

          F_{net}   =  T -  W

Here W is the weight of the car which is evaluated as

         W =  m * g

=>      W =  6526  * 9.8

=>       W =  63954.8 \  N

Generally the net force can also be mathematically represented as

       F =  m * a

So

        m * a  =  T  -  63954.8

=>     6526  *  8.9 =  T  -  63954.8

=>      T =  122036.2 \  N

7 0
3 years ago
What is the net force on a car moving in a straight line with a constant velocity
guapka [62]

Answer:

For example, when a car travels at a constant speed, the driving force from the engine is balanced by resistive forces such as air resistance and friction in the car's moving parts. The resultant force on the car is zero.

Explanation:

hope this helps

3 0
2 years ago
Read 2 more answers
Needddd helppppppp!!!
yulyashka [42]

Answer:

2/9 times as strong.

Explanation:

From the question given above, the following assumptions were made:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

Next, we shall obtain an expression to determine the new force. This can be obtained as follow:

F = GMm / r²

Cross multiply

Fr² = GMm

Divide both side by Mn

G = Fr² / Mm

Since G is constant, then we have

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Finally, we shall determine the new force as follow:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Fr² / Mm = F₂ × (3r)² / 2M × m

Fr² / Mm = F₂ × 9r² / 2Mm

Cross multiply

Fr² × 2Mm = F₂ × 9r² × Mm

Divide both side by 9r² × Mm

F₂ = Fr² × 2Mm / 9r² × Mm

F₂ = F × 2 / 9

F₂ = 2/9 F

Thus, the new force is 2/9 times the original force i.e 2/9 times as strong.

4 0
3 years ago
A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro
Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia =  1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0
3 years ago
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