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Scilla [17]
3 years ago
13

Match each type ol energy with its description

Physics
2 answers:
mel-nik [20]3 years ago
6 0
It is B I sloved it
miskamm [114]3 years ago
4 0

Answer:

B

Explanation: yes

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A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force
galina1969 [7]

Answer:

μ =tanθ

Explanation:=

The ratio of the force of static friction and the normal reaction is equal to tanθ. F=mgsinθ. R = mgcosθ.

μ=tanθ

6 0
3 years ago
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What are the SI units of thermal conductivity?​
kotykmax [81]

Answer:

The SI unit of thermal conductivity is watts per meter-kelvin (W/(m⋅K)).

Explanation:

hope this will help u

7 0
2 years ago
What is the magnitude of velocity for a 2000 kg car possessing 3000 kgm/s of momentum
Alex

For this case we have that by definition, the momentum is given by:

p = mv

Where,

  • <em>m: mass </em>
  • <em>v: speed </em>

Therefore, replacing values we have:

3000 = (2000) v

From here, we clear the value of the speed:

v = \frac {3000} {2000}\\v = 1.5 \frac {m} {s}

Answer:

The magnitude of velocity is:

v = 1.5 \frac {m} {s}

4 0
3 years ago
Read 2 more answers
I NEED HELP ASAP PLEASE!!!!!
Gre4nikov [31]

Answer:

answer is 2 option because more force is applied

6 0
2 years ago
9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the
mojhsa [17]

Answer:

Moment about SHOULDER  ∑ τ = 3.17 N / m,

Moment respect to ELBOW   Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

         ∑ τ = I α

   

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

          ∑ τ = I α

           I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

            I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

            I_mass = m L²

we substitute

           ∑ τ = (⅓ m L² + M L²) α

let's calculate

          ∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

           ∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

         Στ = I α

         I = I_ forearm + I_mass

         I = ⅓ m (L-0.03)² + M (L-0.03)²

         

let's calculate

        i = ⅓ 2.3 0.47² + 0.5 0.47²

        I = 0.2798 Kg m²

        Στ = 0.2798 10

        Στ= 2.80 N m

3 0
3 years ago
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