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2 years ago

What is the purpose of a free body diagram

2 answers:

4 0

<span>A free body diagram is a representation of how the forces that are acting on a point or particle interact. You place your point at the origin and then draw your forces.

</span>

Inga [223]2 years ago

4 0

Answer:

The purpose of a free body diagram is to make a visual representation of all the forces acting on an object.

In a free-body diagram, the object is represented at the centre; every force acting on the object is represented as an arrow, starting from the centre of the object, and with a length proportional to the magnitude of the force and direction equal to the direction of the corresponding force.

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A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average

Kaylis [27]

Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

3 0

2 years ago

A discus thrower turns with angular acceleration of 50 rad/s2, moving the discus in a circle of radius 0.80m. Find the radial an

anyanavicka [17]

Answer:

The value of tangential acceleration $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

The value of radial acceleration $\alpha_{r} = 80 \frac{m}{s^{2} }$

Explanation:

Angular acceleration = 50 $\frac{rad}{s^{2} }$

Radius of the disk = 0.8 m

Angular velocity = 10 $\frac{rad}{s}$

We know that tangential acceleration is given by the formula $\alpha_{t} =$ $r \alpha$

Where r = radius of the disk

$\alpha$ = angular acceleration

⇒ $\alpha_{t} =$ 0.8 × 50

⇒ $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

This is the value of tangential acceleration.

Radial acceleration is given by

$\alpha_{r} = \frac{V^{2} }{r}$

Where V = velocity of the disk = r $\omega$

⇒ V = 0.8 × 10

⇒ V = 8 $\frac{m}{s}$

Radial acceleration

$\alpha_{r} = \frac{8^{2} }{0.8}$

$\alpha_{r} = 80 \frac{m}{s^{2} }$

This is the value of radial acceleration.

7 0

2 years ago

One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi

Masteriza [31]

Answer:

a) $S_1=-9.65}\ J/K$

b) $S_2=71.18\ J/K$

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

The entropy change of reservoir 745 K

$S_1=-\dfrac{7190}{745}\ J/K$

Negative sign because heat is leaving.

$S_1=-9.65}\ J/K$

The entropy change of reservoir 101 K

$S_2=\dfrac{7190}{101}\ J/K$

$S_2=71.18\ J/K$

The entropy change of the rod will be zero.

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

3 0

2 years ago

An electron is released a short distance above earth's surface. a second electron directly below it exerts an electrostatic forc

slamgirl [31]

The mass of an electron is 9.109 x 10⁻³¹ kg

The weight of the electron is (mass) x (g) = 8.926 x 10⁻³⁰ Newton

The charge on an electron is -1.602 x 10⁻¹⁹ Coulomb

The repelling force between the two electrons is (K · q₁ · q₂ / r²) =

(8.98755 x 10⁹ N-m²/C²) x (1.602 x 10⁻¹⁹ C)² / D²

In order for the bottom one to just exactly hold the top one up at a distance 'D', the repelling force has to be exactly equal to the weight of the upper electron.

8.926 x 10⁻³⁰ N = (8.98755 x 10⁹ N-m²/C²)·(1.602 x 10⁻¹⁹ C)² / D²

We have to solve THAT ugly mess for ' D '.

Clean up the units first:

Cancel the C² on the right side, then divide each side by Newton:

8.926 x 10⁻³⁰ = (8.98755 x 10⁹ m²) x (1.602 x 10⁻¹⁹)² / D²

Now, let's multiply both sides by (D² x 10²⁹) :

D² x 8.926 x 10⁻¹ = (8.98755 m²) x (1.602)²

Divide each side by (0.8926):

D² = (8.98755 x 1.602²) / (0.8926) meter²

D² = 25.84 m²

Take the square root of each side:

<em>D = 5.08 meters</em>

I am shocked, impressed, and amazed !

Are you shocked, impressed, or amazed ?

3 0

3 years ago

Question 1 of 7What is the change in internal energy if 90 J of thermal energy is added to asystem, and the system does 30 J of

Dmitriy789 [7]

Given:

The thermal energy added to the system is Q = 90 J

The work done by the system on the surroundings is W = 30 J

To find the change in internal energy.

Explanation:

According to the first law of thermodynamics, the change in internal energy can be calculated by the formula

$\Delta U=\text{ Q-W}$

On substituting the values, the change in internal energy will be

$\begin{gathered} \Delta U\text{ =90-30} \\ =60\text{ J} \end{gathered}$

Final Answer: The chage in internal energy is 60 J (option D)

5 0

4 months ago