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3 years ago

"5 N, up" is an example of a ___. OA) force OB) mass OC) weight OD) magnitude

Physics

1 answer:

Sunny_sXe [5.5K]3 years ago

5 0

Answer:

A) Force

Explanation:

It is an example of force since force is a vector quantity so it has magnitude and direction. In this case the magnitude is equal to 5 [N] and the direction is upward.

The weight can not be, as it always acts downward.

Mass is not a force, its unit is given usually in kilogram [kg]

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A radio wave sent from the surface of the earth reflects from the surface of the moon and returns to the earth. The elapsed time

frez [133]

Answer:

7.92 × 10^8 M

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4 years ago

Starting with a new moon, name the phases the moon will pass through before the next new moon.

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3 years ago

The critical angle for a substance is measured at 53.7 degrees. light enters from air at 45.0 degrees. at what angle it will con

faust18 [17]

When light travels from a medium with greater refractive index $n_1$ to a medium with smaller refractive index $n_2$ , there exists an angle (called critical angle) above which the light is totally reflected, and the value of this angle is given by
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
In this problem, we know that the critical angle is $\theta_c = 53.7^{\circ}$ , so we can find the ratio between the refractive indices of the two mediums:
$\frac{n_2}{n_1} = \sin \theta_c = \sin 53.7^{\circ} =0.81$
and since the second medium is air (n=1.00), the refractive index of the first medium is
$n_1= \frac{n_2}{0.81}= \frac{1.00}{0.81}=1.23$

In the second part of the problem, we have light entering from air ( $n_i = 1.00$ ) at angle of incidence of $\theta_i = 45.0 ^{\circ}$ , into the second medium with $n_r = 1.23$ . By using Snell's law, we can find the angle of refraction of the light inside the medium:
$n_i \sin \theta_i = n_r \sin \theta_r$
$\sin \theta_r = \frac{n_i}{n_r} \sin \theta_i = \frac{1.00}{1.23} \sin 45^{\circ}=0.574$
$\theta_r = \arcsin(0.574)=35.1^{\circ}$

3 0

3 years ago

The highest barrier that a projectile can clear is 19.7 m, when the projectile is launched at an angle of 15.0o above the horizo

svlad2 [7]

Answer:

76 m /s

Explanation:

maximum height, H = 19.7 m

Angle of projection, θ = 15 degree

Let u be the projectile launch speed

Use the formula of maximum height

$H = u^{2}Sin\theta ^{2}/2g$

$19.7 = u^{2}Sin^{2}15/19.6$

u = 75.9199 = 76 m /s

6 0

3 years ago

A 10 kg runaway grocery cart runs into a spring with spring constant 250 n/m and compresses it by 60 cm. what was the speed of t

andrew-mc [135]

The initial kinetic energy of the cart is
$K= \frac{1}{2} mv^2$ (1)
where m is the mass of the cart and v its initial velocity.

Then, the cart hits the spring compressing it. The maximum compression occurs when the cart stops, and at that point the kinetic energy of the cart is zero, so all its initial kinetic energy has been converted into elastic potential energy of the spring:
$U= \frac{1}{2}kx^2$
where k is the spring constant and x is the spring compression.

For energy conservation, K=U. We can calculate U first: the compression of the spring is x=60 cm=0.60 m, while the spring constant is k=250 N/m, so
$U= \frac{1}{2}kx^2= \frac{1}{2}(250 N/m)(0.60 m)^2=45 J$

So, the initial kinetic energy of the cart is also 45 J, and from (1) we can find the value of the initial velocity:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 45 J}{10 kg} } =3 m/s$

7 0

3 years ago