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Bingel [31]
3 years ago
8

A student uses a spring loaded launcher to launch a marble vertically in the air. The mass of the marble is

Physics
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer:

Part a)

When spring compressed by 2 cm

H = 1.47 m

Part b)

When spring is compressed by 4 cm

H = 5.94 m

Explanation:

Part a)

As we know that the spring is compressed and released

so here spring potential energy is converted into gravitational potential energy at its maximum height

So we will have

\frac{1}{2}kx^2 = mg(H + x)

0.5(220)(0.02)^2 = 0.003(9.81)(H + 0.02)

so we have

H = 1.47 m

Part b)

Similarly when spring is compressed by 4 cm

then we have

\frac{1}{2}kx^2 = mg(H + x)

0.5(220)(0.04)^2 = 0.003(9.81)(H + 0.04)

so we have

H = 5.94 m

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d1i1m1o1n [39]

Answer:

The mass and velocity for kinetic energy. Potential Energy: How high an object is and the mass in kilograms or it is the weight in and how high an object is. There are two formulas to calculate potential energy, but the one with grams is used more often.

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5 0
3 years ago
What is the force on a 1000kg elevator that is falling freely at 9.8 m/sec^2
Advocard [28]
From Newton's second law, we know F = ma, where a is the acceleration and m is the mass in kg.

F = 1000kg * 9.8m/s = 9800N

F = 9800 N

Hope this helps!
3 0
2 years ago
A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0
3 years ago
How is work related to potential energy and kinetic energy?
NemiM [27]
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Work is the force on the object as it changes a distance. Interestingly, as work is done on an object, potential energy can be stored in that object. For example, if you carry a load up the stairs. Now that load will have potential energy that can be transformed into kinetic energy and so on
3 0
3 years ago
If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front o
andrezito [222]
The answer is

18 / x = 12 / 4 
12x = 72
x = 6mm
4 0
2 years ago
Read 2 more answers
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