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2 years ago

What is the acceleration of a 10kg mass pushed by a 5N force?

Physics

1 answer:

GenaCL600 [577]2 years ago

5 0

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

From the question we have

$a = \frac{5}{10} = \frac{1}{2} \\ = 0.5$

We have the final answer as

Hope this helps you

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An upright object 2.80 cm tall is placed 16.0 cm away from the vertex of a concave mirror with a center of curvature of 24.0 cm.

horrorfan [7]

Answer:

f = 12 cm

Explanation:

<u>Center of Curvature</u>:

The center of that hollow sphere, whose part is the spherical mirror, is known as the ‘Center of Curvature’ of mirror.

<u>The Radius of Curvature</u>:

The radius of that hollow sphere, whose part is the spherical mirror, is known as the ‘Radius of Curvature’ of mirror. It is the distance from pole to the center of curvature.

<u>Focal Length</u>:

The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’.

The focal length of the spherical (concave) mirror is approximately equal to half of the radius of curvature:

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2 years ago

A long solenoid that has 1,130 turns uniformly distributed over a length of 0.430 m produces a magnetic field of magnitude 1.00

sweet-ann [11.9K]

Given Information:

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Explanation:

The current flowing in the solenoid winding can be found using

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Answer:

a) motion is PARABOLIC, b) positive particle is accelerated towards the negative plate, c) x = 6.19 10⁹ m

Explanation:

This exercise looks at the motion of a positively charged particle in an electric field.

a) Since the field is vertical the acceleration in this direction is

F = m a

the electric force is

F = q E

we substitute

q E = m a

a = qE / m

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a = 1.6 10⁻¹⁹ 2.02 10³ / 2.00 10⁻¹⁶ kg

a = 1,616 m / s²

on the x-axis there are no relationships because there are no forces.

Since the particle has velocities in both axes, its motion is PARABOLIC,

b) the positive particle is accelerated towards the negative plate,

The field is descending, for which the event is down

c) where hit the particle on the x-axis

they indicate that the particle leaves the center of the negative plate, for which we will fix our reference system at this point.

Let's find the components of the initial velocity.

sin θ = v_{oy} / v

cos θ = v₀ₓ / v

v_{oy} = v₀ sin θ

v₀ₓ = v₀ cos θ

v_{oy) = 1.02 10⁵ sin 37 = 0.6139 10⁵ m / s

v₀ₓ = 1.02 10⁵ cos 37 = 0.8146 10⁵ m / s

Let's find the time it takes to hit the negative plate

y = y₀I + v_{oy} t + ½ a and t2

in this case the positions are y = y₀ = 0 and the accelerations

a = - 1,616m/s2,

we substitute

0 = 0 + v_{oy} t - ½ a_y t²

v_{oy}= ½ a_y t

t = 2v_{oy} / a_y

let's calculate

t = 2 0.6139 10⁵ / 1.616

t = 7.597 10⁴s

in this time the particle travels a horizontal distance

x = v₀ₓ t

x = 0.8145 10⁵ 7.597 10⁴

x = 6.19 10⁹ m

the particle falls off the plate

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What is the acceleration of a 10kg mass pushed by a 5N force?​

What is the acceleration of a 10kg mass pushed by a 5N force?