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castortr0y [4]
3 years ago
7

Which condition must be met in order for an equation to be balanced?

Physics
1 answer:
vitfil [10]3 years ago
4 0

Answer:

There must be an equal amount of each element on both sides of the equation. Hope this helps and please marks as the brainliest.

Explanation:

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James Bond is trying to escape his enemy on a speedboat but
SVEN [57.7K]

Answer:

100 m/s

Explanation:

Mass the mass of Bond's boat is m₁. His enemy's boat is twice the mass of Bond's i.e. m₂ = 2 m₁

Initial speed of Bond's boat is 0 as it won't start and remains stationary in the water. The initial speed of enemy's boat is 50 m/s. After the collision, enemy boat is  completely stationary. Let v₁ is speed of bond's boat.

It is the concept of the conservation of momentum. It remains conserved. So,

m_1u_1+m_2u_2=m_1v_1+m_2v_2

Putting all the values, we get :

0+(2m_1)50=m_1v_1+(2m_2)(0)\\\\100m_1=m_1v_1\\\\v_1=100\ m/s

So, Bond's boat is moving with a speed of 100 m/s after the collision.

3 0
4 years ago
Describe the energy of a playground swing at its highest point
Korvikt [17]
At the hightest point it would be potential energy, because the energy is stored and then when it is rocking back and forth it has kinetic energy
8 0
4 years ago
A CD spins at a constant angular velocity of 5.0 revolutions per second clockwise.
Lera25 [3.4K]

The true statement about the CD is:

<h3><em>b. No net torque acts on it at all.</em></h3>

\texttt{ }

<h3>Further explanation</h3>

Centripetal Acceleration can be formulated as follows:

\large {\boxed {a = \frac{ v^2 } { R } }

<em>a = Centripetal Acceleration ( m/s² )</em>

<em>v = Tangential Speed of Particle ( m/s )</em>

<em>R = Radius of Circular Motion ( m )</em>

\texttt{ }

Centripetal Force can be formulated as follows:

\large {\boxed {F = m \frac{ v^2 } { R } }

<em>F = Centripetal Force ( m/s² )</em>

<em>m = mass of Particle ( kg )</em>

<em>v = Tangential Speed of Particle ( m/s )</em>

<em>R = Radius of Circular Motion ( m )</em>

Let us now tackle the problem !

\texttt{ }

<em>Complete Question:</em>

<em>A CD spins at a constant angular velocity of 5.0 revolutions per second clockwise. Which of the following statements about the CD is true?</em>

<em>a. A net torque acts on it clockwise to keep it moving</em>

<em>b. No net torque acts on it at all.</em>

<em>c. A net torque acts on it counterclockwise to keep it moving</em>

<u>Given:</u>

angular velocity = ω = 5.0 revolutions per second

<u>Asked:</u>

net torque = Στ = ?

<u>Solution:</u>

Constant angular velocity → angular acceleration = α = 0 rad/s²

\Sigma \tau = I \alpha

\Sigma \tau = I (0)

\Sigma \tau = 0 \texttt{ Nm}

\texttt{ }

<h3>Conclusion:</h3>

The true statement about the CD is:

<em>b. No net torque acts on it at all.</em>

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

\texttt{ }

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

#LearnWithBrainly

8 0
4 years ago
A car travels along a horizontal road which is an arc of radius 125m. the greatest speed at which the car can travel without sle
hammer [34]

The coefficient of friction between the Tyre and the ground is 0.11

<u>Explanation:</u>

Given:

Radius of the track (r)=125 m.

Speed with which the car travels (v) =42 km/hr

                                                             =11.67 m/s

To Find:

Coefficient of friction between the Tyre and the ground.

Formula to be used:

Centripetal\ force\ Fc= mv^2/r\\\\=m\times 11.67^2/125\\\\=1.08 m

We know that,Frictional force is equal to centripetal force

Frictional force=μmg

therefore 1.08 m=μmg

Cancelling "m" on both sides we get,

μ=1.08/g=1.08/9.8

=0.11

Thus the coefficient of friction between the Tyre and the ground is 0.11

7 0
3 years ago
You exert a force of 15 newtons while you move a rock 2 meters. How much work did you perform?
grandymaker [24]
The formula to find work is Work = Force * Distance. In this case, the force exerted on the rock is 15 Newtons and the distance the rock moved was 2 meters, so 15 * 2 = 30. The answer to your question is D) 30 n-m.
4 0
3 years ago
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