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aivan3 [116]
2 years ago
13

Please write ONE complete sentence describing the relationship between Kinetic Energy and Gravitational Potential Energy in a sy

stem
where energy is conserved:
Physics
1 answer:
Scilla [17]2 years ago
3 0

Answer:

Explanation:

The sum of kinetic and potential energy is the mechanical energy of the system. Assuming the energy of the system is conserved, which it is in your case, then the initial mechanical energy will equal the final mechanical energy. This relationship is very useful in mechanics.

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How much force is needed to accelerate a 1,100 kg car at a rate of 1.5 m/s2?
Anna [14]
Assuming there is no force of friction...

F = ma
F = (1300kg)(1.5m/s^2)
F = 1950N
Just multiply mass by acceleration.
1300 x 1.5 = 1950N.
7 0
2 years ago
When two adjacent lights blink on and off in quick succession, we perceive a single light moving back and forth between them. th
makkiz [27]
This is called the Phi Phenomenon.
This is an illusion of movement created when two or more adjacent lights blink on and off in quick succession; when two adjacent stationary lights blink on and off in quick succession; we perceive a single light moving back and forth between them. It is an optical illusion of perceiving a series of still images, when viewed in rapid succession, as continuous motion. 
5 0
3 years ago
If the focal length of a concave mirror is 18cm, find its radius of curvature.
Galina-37 [17]

Given :

The focal length of a concave mirror is 18 cm.

To Find :

The radius of curvature of the concave mirror.

Solution :

We know,

\text{Focal length}=\dfrac{\text{Radius of curvature}}{2}\\\\F=\dfrac{R}{2}\\\\R = 18\times 2\ cm\\\\R = 36 \ cm

Therefore, the radius of curvature of concave mirror is 36 cm.

Hence, this is the required solution.

8 0
3 years ago
Pls help me with this problem!!
Olenka [21]

Answer:

v = 19.6 m/s.

Explanation:

Given that,

The radius of the circle, r = 5 m

The time period of the ball, T = 1.6s

We need to find the ball's tangential velocity.

The formula for the tangential velocity is given by :

v=\dfrac{2\pi r}{T}

Putting all the values in the above formula

v=\dfrac{2\pi \times 5}{1.6}\\\\v=19.6\ m/s

So, the tangential velocity of the ball is 19.6 m/s. Hence, the correct option is (c).

7 0
3 years ago
A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe
Lina20 [59]

The final speed of the block after the collision with the obstacle is \boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}.

Further Explanation:

Given:

The mass of the block is 6.0\,{\text{kg}}.

The initial momentum of the block is 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}.

The impulse imparted by the obstacle is 10\,{\text{N}} \cdot {\text{s}}.

Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

{p_f} - p{ & _i} = I               …… (1)                                    

Substitute 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} for {p_i} and - 10\,{\text{N}} \cdot {\text{s}} for I  in equation (1).

 \begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}

The final momentum of the block can be expressed as:

{p_f} = m{v_f}                   …… (2)                                  

Substitute 20\text{kg}\;\text{m/s} for {p_f} and 6.0\,{\text{kg}} for m in equation (2).

 \begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}

Thus, the final speed of the block after the collision with the obstacle is \boxed{3.33\;\text{m/s}}.

Learn More:

  1. Choose the 200 kg refrigerator. Set the applied force to 400 n (to the right) brainly.com/question/4033012
  2. With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward brainly.com/question/9719731
  3. Which of the following is an example of a nonpoint source of freshwater pollution brainly.com/question/1482712

Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords:  Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

5 0
3 years ago
Read 2 more answers
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