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Sergio [31]
2 years ago
14

What is the acceleration of an object that has a mass of 10kg and is pushed with a force of 50n

Physics
1 answer:
worty [1.4K]2 years ago
7 0

Answer:

5m/s/s

Explanation:

force = mass x acceleration

50 = 10a

a=5m/s/s

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How do you know that forces are balanced when static friction acts on an object?
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By looking at the acceleration of the object.
In fact, Netwon's second law states that the resultant of the forces acting on an object is equal to the product between the mass m of the object and its acceleration:
\sum F = ma

So, when static friction is acting on the object, if the object is still not moving we know that all the forces are balanced: in fact, since the object is stationary, its acceleration is zero, and so the resultant of the forces (left term in the formula) must be zero as well (i.e. the forces are balanced).
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Something that can not be used up or depleted​
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oxygen

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Much power does a 10 Ohm bulb have in a series circuit with a battery of 12V and nothing else in the the circuit?
Sedaia [141]

Answer:

it will be d) 14.4W

Explanation:

potential difference (v) = 12 volts

resistance (r) = 10 ohms

now, we know

=》

power =  \frac{v {}^{2} }{r}

=》

power \:  =  \frac{12 {}^{2} }{10}

=》

power =  \frac{144}{10}

=》

power = 14.4 \: watt

8 0
2 years ago
If an experiment used 945 j of heat, how many calories (cal) is that?
irina1246 [14]
There are approximately two hundred and twenty-five point seven five calories from using 945 j of heat

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5 0
3 years ago
A sphere of mass of 1.55 kg is accelerated upwards by a string to which the sphere is attached. Its speed increases from 2.81 m/
Mama L [17]

Answer:

The tension in the string is 16.24 N

Explanation:

Given;

mass of the sphere, m = 1.55 kg

initial velocity of the sphere, u = 2.81 m/s

final velocity of the sphere, v = 4.60 m/s

duration of change in the velocity, Δt = 2.64 s

The tension of the string is calculated as follows;

T =  ma  + mg\\\\T = m(a + g)\\\\where;\\\\a \ is \ upward \ acceleration \ of \ the \ sphere\\\\g \ is \ acceleration \ due \ to \ gravity =9.8 \ m/s^2\\\\a = \frac{\Delta V}{\Delta t} = \frac{v- u}{ t} = \frac{4.6 - 2.81 }{2.64} = 0.678 \ m/s^2

T = 1.55(0.678 + 9.8)

T = 1.55(10.478)

T = 16.24 N

Therefore, the tension in the string is 16.24 N

5 0
2 years ago
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