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PtichkaEL [24]
3 years ago
13

Physics questions please help me

Physics
1 answer:
Ahat [919]3 years ago
5 0
Hi,Find  answers from Task 5

1.(X+4)+(X)+(X+4)+(X)=50cm

4x+8=50cm

4x=42

X=10.5cm

Length=10.5+4=14.5cm

Width=10.5cm

Area= length × width=(10.5/100) × (14.5/100) =0.0152m2

2. Volume of a sphere= 4/3 ×π×r³

4/3 ×π×r³=3.2×10^-6 m³

r³=3.2×10^-6 m³/1.33×π

r³=7.64134761e-7

r=0.00914m

Surface area of the blood drop= 4πr²

=4×3.142×0.00914×0.00914=0.00105m²

3.

Equation of an ideal gas     = PV =n RT  

Equation for pressure, = P= n RT/V

Equation for the volume of an ideal gas= V= n RT/P

If the volume of gas doubles ,V(new)=  2n RT/P

Equation for temperature of an ideal gas, T = PV/n R

If temperature of gas triples, T (new)= 3PV/n R

New Equation for Pressure, = n× R× (3PV/n R)/(2n RT/P)

Pressure factor increase= P(new)/P(old)  ={ n× R× (3PV/n R)/(2n RT/P)}/{ n RT/V}

=3PV²/2n RT
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Answer: A) Deceleration of the car is -6.6667 m/s² while it came to stop.

B) The total distance the car travels is 200 meter during the 10 s period.

Explanation:

Given Data

Initial velocity of the car ($$v_{i}$$) =   20.0 m/s

Final velocity of the car (v_{f}) = 0 m/s

Time (in motion) =7.00 s

Time (in rest) =3 s

To find - A) car's deceleration while it came to a stop

              B) the total distance the car travels in 10 s

A) The formula to find the deceleration is

Deceleration = (( final velocity - initial velocity ) ÷ Time)        (m/s²)

Deceleration = ((v_{f}) - ($$v_{i}$$)) ÷ time     (m/s²)

Deceleration =  ( 0.0 - 20 ) ÷ 3      (m/s²)

Deceleration =   (- 20) ÷ 3  (m/s²)

Deceleration   =  - 6.6667 m/s²

(NOTE : Deceleration is the opposite of acceleration so the final result must have the negative sign)

The car's deceleration is  - 6.6667 m/s² while it came to a stop

B) The formula to find the distance traveled by the car is  

Distance traveled by the car is equals to the product of the speed and time

Distance = Speed × Time  (meter)

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Relationship between voltage, resistance and current in a circuit
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9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the
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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

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The study 'characterizing vibration-assisted atomic force based nanomachining' aims to elucidate nanomachine properties for heterogeneous materials.

<h3>What is nanomachining?</h3>

The expression nanomachining makes reference to the study of nanometric machines (nanomachines) and related materials, which can be achieved by different approaches including sensor-based strategies related to acoustic auditive phenomena.

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