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aksik [14]
2 years ago
8

Sachi wants to throw a water balloon to knock over a target and win a prize. The target will only fall over if it is hit with a

force of 0.035 N. The water balloon has a mass of 11.4 grams. How fast will Sachi have to make the water balloon accelerate to hit the target with enough force and win the prize?
Physics
2 answers:
Verizon [17]2 years ago
8 0
3.1 the only reason i know this is cause i got it wrong 
Mashutka [201]2 years ago
8 0
Remember that acceleration (a) equals force (F) over mass (m) or a=F/m. In this equation, both the force, 0.035N, and the mass, 11.4 grams, are provided. However, the mass is provided in the wrong unit, as it should be calculated in kilograms. To set up this equation, you will have to convert 11.4 grams into .0114 kilograms then divide .035 N over 0.114 kg to get your answer of 3.07 m/s^2 rounded to 3.1 m/s^2
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Answer:

It gives our light which we need for probably everything.

Explanation:

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A negatively charged object is located in a region of space where the electric field is uniform and points due north. the object
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\Delta U = e d

- The second largest increase is when the charge is moving east. In this case, actually, the variation of potential energy is zero. This is because the charge is moving perpendicular to the field, and so it is moving along points with same potential. Therefore, in this case the variation of potential energy is zero:
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2 years ago
A constant braking force of 10 newtons applied for 5 seconds is used to stop a 2.5-kilogram cart traveling at 20 meters per seco
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Alchen [17]

Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat (U_{g}), in joules, is equal to the sum of the final translational kinetic energy (K), in joules, and work losses due to air resistance (W_{l}), in joules:

U_{g} = K +W_{l} (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l} (2)

Where:

m - Mass of the cat, in kilograms.

g - Gravitational acceleration, in meters per square second.

h - Initial height of the cat, in meters.

v - Final speed of the cat, in meters per second.

If we know that m = 2.70\,kg, g = 9.807\,\frac{m}{s^{2}}, h = 5.20\,m and W_{l} = 120\,J, then the final speed of the cat is:

v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }

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The speed of the cat when it hits the ground is approximately 7.586 meters per second.

4 0
2 years ago
A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with incompressib
tatiyna
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