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3 years ago

What is the mass of an object that has an acceleration of 2.63m/s2 when a unbalanced force of 112N is applied to it?

Physics

1 answer:

exis [7]3 years ago

4 0

Answer:

42.58kg

Explanation:

By Newton's second law, F = ma.

F is the force being applied, in this case 112N. a is the acceleration, in this case 2.63 m/s^2.

Thus, with some simple algebraic manipulation, we get the mass to equal:

m = F/a = 112N / 2.63 m/s^2 = 42.58kg

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An object accelerating at 16 m/s/s doubles its mass and triples its net force acting on it. What will the new acceleration be? (

nataly862011 [7]

Answer:

24 m/s²

Explanation:

The given parameters are;

The initial acceleration of the object, a = 16 m/s²

Let 'm' represent the initial mass of the object

The initial force acting on the object, F = m × a

∴ F = 16 × m = 16·m

When the mass is doubled, we have;

The new mass of the object, m₂ = 2 × m = 2·m

When the net force acting on the object triples, we have;

The new net force acting on the object, F₂ = 3 × F = 3 × 16·m = 48·m

From F = m × a, we have;

a = F/m

∴ The new acceleration of the object, a₂ = F₂/m₂

From which, by plugging in the values, we have;

a₂ = 48·m/(2·m) = 24

The new acceleration of the object, a₂ = 24 m/s².

6 0

3 years ago

If two equal charges are separated by a certain distance, the force of repulsion is F. Given F1, if each charge in F1 is doubled

Alisiya [41]

Answer: $F_2=16\times F_1$

Explanation:

Given

Force of repulsion between two charge particle is given by force F

Electrostatic force is given by

$F=\frac{kq_1q_2}{r^2}$

where $q_1$ and $q_2$ is the charges of particle

r=distance between charge particle

$=\frac{kq^2}{r^2}$ when charges are doubled and distance is reduced to half

i.e. q become 2 q and r becomes 0.5 r

$F_2=\frac{k(2q)^2}{(0.5r)^2}$

$F_2=\frac{kq^2}{r^2}\times 4\times 4$

$F_2=16\times F_1$

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3 years ago

What is the greatest distance an image can be located behind a convex spherical mirror?

Afina-wow [57]

Answer:

Maximum distance of image from mirror is equal to focal length of the mirror

Explanation:

As we know by the equation of mirror we have

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

here we know for convex mirror

object position is always negative as it will be placed behind the mirror always

while the focal length of the convex mirror is always taken positive

So here we have

$\frac{1}{d_i} + \frac{1}{-d_o} = \frac{1}{f}$

$\frac{1}{d_i} = \frac{1}{d_o} + \frac{1}{f}$

so here maximum value of image distance is equal to focal length of the mirror

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2 years ago

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3 years ago