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love history [14]
3 years ago
9

Which of the following objects is in dynamic equilibrium? A - a man standing in one place without moving B- a bicycle accelerati

ng forward at 1.25 m/s^2 C- a car driving in a circle at a constant speed of 20 m/s D- a motorcycle with a constant velocity
Physics
2 answers:
Bas_tet [7]3 years ago
5 0
The answer is <span>D- a motorcycle with a constant velocity.

Velocity is much more stable in terms of speed and distance that can be associated with the dynamics of equilibrium (balance) compared to speed alone. It brings the constant relationship between the two elements. 

Speed is only focused on time of travel and not with direction. While acceleration does not show equilibrium but only a change in speed.
</span>
Anettt [7]3 years ago
3 0

Answer:

Option (D)

Explanation:

A dynamic equilibrium is the state in which the net force on the body is zero but the body is moving.

For option (a), A man is standing without moving, it means the man is in static equilibrium.

For option (b), a bicycle accelerating forward with acceleration, it means it is not in dynamic equilibrium.

For option (c), a car driving in circle with a constant speed, it means the direction of speed is changing, so the acceleration is there and the car is not in dynamic equilibrium.

For option (d), a motorcycle with a constant velocity, it means the net force is zero and the motorcycle is in dynamic equilibrium.

You might be interested in
Three charges are located at a different position in a plane: q1= 10μC at →r1=(5,6)cm q2=−27μC at →r2=(−6,10)cm and q3=−12μC at
sasho [114]

Answer:

 E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

 E_{total} = 2,467 10⁶ N / A       θ = -21.8      

Explanation:

For this exercise we will use that the electric field is a vector quantity, so the total field is

        E_total = E₁₃ + E₂₃

bold font vectors .  We can work with the components of the electric field in each axis

X- axis

       E_ total x = E₁₃ₓ + E_{23x}

y-axis  

      E_{total y} = E_{13y} + E_{23y}

the expression for the electric field is

       E = k q / r²

where r is the distance between the charge and the positive test charge

       

in this exercise

Let's find the field created by charge 1

q₁ = 10 μC = 10 10⁻⁶ C

x₁ = 5 cm = 0.05 m

x₃ = 21 cm = 0.21 m

         E_{13x} = 9 10⁹ 10 10⁻⁶ / (0.21 -0.05)²

         E_{13x} = 3.516 10⁶ N / C

y₁ = 6 cm = 0.06 cm

y₃ = -12 cm = -0.12 m

        E_{13y} = 9 10⁹ 10 10⁻⁶ / (-0.12 - 0.06)²

        E_{13y} = 2,777 10⁶ N / C

let's find the field produced by charge 2

q₂ = -27 μC = - 27 10⁻⁶ C

x₂ = -6 cm = -0.06 m

x₃ = 0.21 m

        E_{23x} = 9 10⁹ 27 10⁻⁶ / (0.21 + 0.06)²

        E_{23x} = 1.23 10⁶ N / A

y₂ = 10 cm = 0.10 m

y₃ = -0.12 m

        E_{23y} = 9 10⁹ 27 10⁻⁶ / (-0.12 - 0.10)²

        E_{23y} = 1.86 10⁶ N / C

Taking the components we can calculate the total electric field, we must use that charge of the same sign repel and attract the opposite sign, remember that the test charge is always considered positive.

       E_{total x} = E_{13x} - E_{23x}

       E_{total x} = (3.516 - 1.23) 10⁶

       E_{total x} = 2.29 10⁶ N / A

       

       E_{total y} = -E_{13y} + E_{23y}

       E_{total y} = (-2.777 +1.86) 10⁶ N / A

       E_{total y} = -0.917 10⁶ N / A

we can give the result in two ways

         E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

or in the form of modulus and angle, let's use the Pythagorean theorem to find the modulus

                E_{total} = √ (E_{total x}^2 + E_{total y}^ 2)

                 E_{total} = √ (2.29² + 0.917²) 10⁶

                E_{total} = 2,467 10⁶ N / A

let's use trigonometry for the angle

                tan θ = E_total and / E_totalx

                θ = tan⁻¹ E_{total y} / E_{total x}

                θ = tan⁻¹ (-0.917 / 2.29)

                θ = -21.8

The negative sign indicates that the angle is measured with respect to the x-axis in a clockwise direction.

7 0
3 years ago
1 point
White raven [17]

Answer:

Waves transfer energy, not motion

8 0
2 years ago
Read 2 more answers
The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo
ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

So

(V_{f})^{2}=(V_{o})^{2}+2aS\\  So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}

As we find acceleration.Now we need to find time

So

V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s

Now for distance

So

Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m

7 0
3 years ago
redacteaza un text in care sa iti exprimi parerea in legaturacurtamentul golegilor din textul oracolul
Ivahew [28]

Answer:

Seriously I have no idea. I need help with my homework.

Explanation:

I really need help with my homework. Sorry

4 0
3 years ago
Tốc độ v của một vật được cho bởi phương trình v = At
ki77a [65]

If <em>v(t)</em> is speed measured in meters per second (m/s), and <em>t</em> is time measured in seconds (s), then the constants <em>A</em> and <em>B</em> in

<em>v(t)</em> = <em>At</em> ³ - <em>Bt</em>

must have units of m/s⁴ and m/s², respectively; otherwise, the equation is dimensionally inconsistent.

[m/s] = <em>A</em> [s]³ - <em>B</em> [s]

[m/s] = [m/s⁴] [s]³ - [m/s²] [s]

[m/s] = [m/s] - [m/s]

[m/s] = [m/s]

5 0
2 years ago
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