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3 years ago

1. The following can be inferred from Newton’s second law of motion except:

Physics

1 answer:

lora16 [44]3 years ago

7 0

Answer: 1.d) The acceleration of an object is always less than the acceleration due to gravity, g (9.81m/s^-2)

2.a)acceleration decreases

Explanation:

Newton's second law:

Newton's second law states that the acceleration of an object is defined by two variables which is the total force acting on the object and the mass of that object. The acceleration is directly proportional to the net force that is applied on an object and inversely proportional to the mass of that object.

When the force applied on an object is increased so does the acceleration of an object however if the mass increase the acceleration decreases.

This can be felt when you look at the truck which usually carry heavy loads they seem to drive slow due to the load hence their acceleration is decreased by the mass that these truck carry .

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A ray of laser light travels through air and enters an unknown material. The laser enters the material at an angle of 36 degrees

V125BC [204]

Answer:1.27

Explanation:

Given

incident angle $i=36^{\circ}$

refracted angle $r=27.5^{\circ}$

Suppose $n_2$ is the refractive index of material then using Snell's law we can write

$n_1\sin i=n_2\sin r$

where $n_1$ =refractive index of air

$1\times \sin (36)=n_2\times \sin (27.5)$

$n_2=\dfrac{0.5877}{0.4617}$

$n_2=1.27$

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3 years ago

Help please an observer at the North Pole sees the moon phase shown below. What Moon phase will be observed approximately one we

ziro4ka [17]

Answer:

it should be a new moon

Explanation:

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3 years ago

A ball is ejected to the right with an unknown horizontal velocity from the top of a pillar that is 50 meters in height. At the

dimulka [17.4K]

Answer:

15.67 m/s

Explanation:

The ball has a projectile motion, with a horizontal uniform motion with constant speed and a vertical accelerated motion with constant acceleration g=9.8 m/s^2 downward.

Let's consider the vertical motion only first: the vertical distance covered by the ball, which is S=50 m, is given by

$S=\frac{1}{2}gt^2$

where t is the time of the fall. Substituting S=50 m and re-arranging the equation, we can find t:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(50 m)}{9.8 m/s^2}}=3.19 s$

Now we now that the ball must cover a distance of 50 meters horizontally during this time, in order to fall inside the carriage; therefore, the velocity of the carriage should be:

$v=\frac{d}{t}=\frac{50 m}{3.19 s}=15.67 m/s$