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Mariana [72]
3 years ago
7

A 12 kg hockey puck is traveling at a velocity of 15 m/s. What is the momentum?

Engineering
1 answer:
Ilya [14]3 years ago
6 0
180 is the correct answer
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10. Technician A says checking the automatic
Naya [18.7K]

Answer: a

Explanation:

4 0
3 years ago
A 3.7 g mass is released from rest at C which has a height of 1.1 m above the base of a loop-the-loop and a radius of 0.2 m . Th
Dmitrij [34]

Answer:

Normal force = 0.326N

Explanation:

Given that:

mass released from rest at C = 3.7 g = 3.7 × 10⁻³ kg

height of the mass = 1.1 m

radius = 0.2 m

acceleration due to gravity = 9.8 m/s²

We are to determine the normal force pressing on the track at A.

To to that;

Let consider the conservation of energy relation; which says:

mgh = mgr + 1/2 mv²

gh = gr + 1/2 v²

gh - gr = 1/2v²

g(h-r) = 1/2v²

v² = 2g(h-r)

However; the normal force will result to a centripetal force; as such, using the relation

N =mv²/r

replacing the value for v² = 2g(h-r) in the above relation; we have:

Normal force = 2mg(h-r)/r

Normal force = 2 × 3.7 × 10⁻³ × 9.8 ( 1.1 - 0.2 )/ 0.2

Normal force = 0.065268/0.2

Normal force = 0.32634 N

Normal force = 0.326N

6 0
3 years ago
For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5
bonufazy [111]

Answer:

The total force resisted by the flange bolts is  163.98 N

Explanation:

Solution

The first step is to find  the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ =  diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of  the pipe at the inlet section

A₂ = π /4 D₂²

D₂ =  diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265  * 5 = 1.963 * V₂

V₂ = 12.8 m/s

Note: Kindly find an attached copy of the part of the solution to the given question below

8 0
3 years ago
Sandra is holding a piece of tissue that has a negative charge and a feather that has a neutral charge. The two objects have sim
Murljashka [212]
The correct answer would be d
The objects will not move towards or away from each other
Hope this helps
4 0
2 years ago
The anchor shown is used to tie tower guy cables to the ground and is supported by a distributed force from the soil, which can
ahrayia [7]

Answer:

See attachment for complete answer step by step solving

Explanation:

Given that:

The anchor shown is used to tie tower guy cables to the ground and is supported by a distributed force from the soil, which can be approximated as shown. The anchor also is subjected to the loads shown, where P1 = 5 kNand P2 = 3 kN . The anchor is made from aluminum with E = 69 GPa . It is a cylindrical rod with diameter d1 = 4 cm with three segments of length L = 2 m . One end of the rod is linearly tapered to a diameter of d2 = 2 cm.

a. Calculate the intensity of the reaction load, Po

b. Use the reaction found in Part A, po = 13 kN/m , and calculate the total change in length for the anchor.

5 0
3 years ago
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