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Murrr4er [49]
3 years ago
11

Which has more inertia - a 2,750 gram object or a 2,500 gram object?

Physics
2 answers:
noname [10]3 years ago
5 0

Question:

Which has more inertia - a 2,750 gram object or a 2,500 gram object?

Answer:

2,750 has more interia..

k0ka [10]3 years ago
3 0

Answer:

a 2,750 gram object has more inertia.

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As you move away from a positive charge distribution, the electric field:
GalinKa [24]

Answer:

The electric field always decreases.

Explanation:

The electric field due to a point charge is given by :

E=\dfrac{kq}{r^2}

Where

k = electric constant

q = charge

r = distance from the charge

It is clear from the above equation that as the distance from the charge particle increases the electric field decreases. As you move away from a positive charge distribution, the electric field always decreases. Hence, the correct option is (c) "Always decreases".

3 0
3 years ago
The atomic mass of gold is 0.197 kg/mole. how many moles are in 0.566 kg of gold.
iogann1982 [59]

Explanation:

0.566kg *(1mol/0.197 kg)= 2.87 mol gold

note how the units cancel out, if the units do not cancel out (kg/kg=1) then u did something wrong

7 0
3 years ago
Read 2 more answers
A mass is attached to a spring with an unknown spring constant. The spring gains 10 J of elastic potential energy if stretched b
Firlakuza [10]

From the given information in the question, the correct option is Option 1: 14 cm.

A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:

PE = \frac{1}{2} k x^{2}

Where PE is the energy, k is the spring constant and x is extension.

i. Given that: PE = 10 J and x = 10 cm, then;

PE = \frac{1}{2} k x^{2}

10 = \frac{1}{2} k 10^{2}

20 = 100k

k = 0.2 J/cm

ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.

PE = \frac{1}{2} k x^{2}

20 = \frac{1}{2} (0.2) x^{2}

40 = 0.2 x^{2}

x^{2} = 200

x = \sqrt{200}

  = 14.1421

x = 14.14 cm

So that;

x is approximately 14.00 cm.

Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.

For more information, check at: brainly.com/question/1352053.

8 0
3 years ago
Stan does 178 J of work lifting himself 0.5 m. What is Stan’s mass? The acceleration of gravity is 9.8 m/s 2 . Answer in units o
Rom4ik [11]
Work= (force)(distance)
178= m(9.81)x0.5
178=m(4.905)
178/4.905=m

His mass is 36.3 kg
4 0
3 years ago
Read 2 more answers
The force F1, 10.0N acts at 10.0cm. What is the magnitude of the torque due to F1 about an axis through Point A perpendicular to
Nookie1986 [14]
Thank you for posting your question here at brainly. I hope the answer will help you. Below is the solution. Feel free to ask more question. 

<span>torque = rF
= 0.1(10)
=1 Nm</span>
4 0
3 years ago
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