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3 years ago

Explain giving an example for;

Physics

1 answer:

Bingel [31]3 years ago

6 0

$\boxed{\large{\bold{\red{ANSWER~:) }}}}$

The motion of a ball falling through the atmosphere, or a model rocket being launched up into the atmosphere are both excellent examples of Newton’s 1st law of motion.

Riding a bicycle is a good example of Newton’s 2nd law of motion.In this example, the bicycle is the mass. The leg muscles pushing on the pedals of the bicycle is the force.

You hit a wall with a certain amount of force, and the wall returns that same amount of force. This is an example of Newton’s 3rd law of motion

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A truck is moving at 2.0 m/s accelerates at a rate of +2.00 m/s2. It does this over a distance of 400.0m. Find the final velocit

otez555 [7]

If v is the truck's final velocity, then

v² - (2.0 m/s)² = 2 * (2.00 m/s²) * (400.0 m)

v² = 1604 m²/s²

v = 40.05 m/s ≈ 40 m/s

3 0

3 years ago

What’s the answer for this problem?

pickupchik [31]

The answer is always true a

4 0

3 years ago

A revolutionary war cannon, with a mass of 2090 kg, fires a 16.7 kg ball horizontally. The cannonball has a speed of 113 m/s aft

OLEGan [10]

Answer:

0.90291 m/s

0.45055 m/s

Explanation:

$m_1$ = Mass of canon = 2090 kg

$m_2$ = Mass of ball = 16.7 kg

$v_1$ = Velocity of canon

$v_2$ = Velocity of ball = 113 m/s

In this system the momentum is conserved

$m_1v_1=m_2v_2\\\Rightarrow v_1=\dfrac{16.7\times 113}{2090}\\\Rightarrow v_1=0.90291\ m/s$

The velocity of the cannon is 0.90291 m/s

Applying energy conservation

$\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2=\dfrac{1}{2}m_2v^2\\\Rightarrow m_1v_1^2+m_2v_2^2=m_2v^2\\\Rightarrow v_2=\sqrt{\dfrac{m_1v_1^2+m_2v_2^2}{m_2}}\\\Rightarrow v_2=\sqrt{\dfrac{2090\times 0.90291^2+16.7\times 113^2}{16.7}}\\\Rightarrow v_2=113.45055\ m/s$

The ball would travel 113.45055-113 = 0.45055 m/s faster

6 0

4 years ago

Describe what happens in a crash according to newton's first law

Nostrana [21]

According to Newton's first law, an object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force. It is the natural tendency of objects to keep on doing what they're doing. All objects resist changes in their state of motion. In the absence of an unbalanced force, an object in motion will maintain its state of motion. This is often called the law of inertia.

The law of inertia is most commonly experienced when riding in cars and trucks. In fact, the tendency of moving objects to continue in motion is a common cause of a variety of transportation injuries - of both small and large magnitudes. Consider for instance the unfortunate collision of a car with a wall. Upon contact with the wall, an unbalanced force acts upon the car to abruptly decelerate it to rest. Any passengers in the car will also be decelerated to rest if they are strapped to the car by seat belts. Being strapped tightly to the car, the passengers share the same state of motion as the car. As the car accelerates, the passengers accelerate with it; as the car decelerates, the passengers decelerate with it; and as the car maintains a constant speed, the passengers maintain a constant speed as well.

But what would happen if the passengers were not wearing the seat belt? What motion would the passengers undergo if they failed to use their seat belts and the car were brought to a sudden and abrupt halt by a collision with a wall? Were this scenario to occur, the passengers would no longer share the same state of motion as the car. The use of the seat belt assures that the forces necessary for accelerated and decelerated motion exist. Yet, if the seat belt is not used, the passengers are more likely to maintain its state of motion. The animation below depicts this scenario.

If the car were to abruptly stop and the seat belts were not being worn, then the passengers in motion would continue in motion. Assuming a negligible amount of friction between the passengers and the seats, the passengers would likely be propelled from the car and be hurled into the air. Once they leave the car, the passengers becomes projectiles and continue in projectile-like motion.

7 0

3 years ago

The temperature of 10 kg of a substance rises by 55oC when heated. Calculate the

guajiro [1.7K]

Given :

The temperature of 10 kg of a substance rises by 55oC when heated.

To Find :

The temperature rise when 22 kg of the substance is heated with the same quantity of heat.

Solution :

We know, change in temperature when heat is given to object is given by :

$\Delta H = m S\Delta T$

It is given that same amount of heat is given in both the cases also the substance is same.

So,

$M_1 S \Delta T_1 = M_2 S \Delta T_2\\\\10\times 55 = 22 \times \Delta T_2\\\\\Delta T_2 = 25^o\ C$

Hence, this is the required solution.

7 0

3 years ago