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3 years ago

How does the light model help you change the location of the Shadow

2 answers:

5 0

Light models help because depending ion where it is, it changed the way the shadow appears. That's why youre shadow can appear small or large

timama [110]3 years ago

3 0

The light model helps you change the location of the shadow is because when light comes the shadow moves farther and farther that is what I learned in science today

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If the ball leaves the projectile launcher at a speed of 2.2 m/s at an angle of 30ᴼ, and the projectile launcher is on a table a

IgorC [24]

Answer: 1.12 m

Explanation:

This situation is related to parabolic motion, hence we can use the following equations:

$y=y_{o}+V_{o}sin \theta t-\frac{g}{2}t^{2}$ (1)

$x=V_{o} cos \theta t$ (2)

Where:

$y=0 m$ is the ball final height (when it hits the ground)

$y_{o}=1.1 m$ is the ball initial height

$V_{o}=2.2 m/s$ is the initial velocity

$\theta=30\°$ is the angle at which the ball was launched

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the horizontal distance the ball travels

Rewriting (1) with the given values:

$0 m=1.1 m+(2.2 m/s)(cos 30\°)t-\frac{9.8 m/s^{2}}{2}t^{2}$ (3)

Multiplying all the eqquation by -1 and rearranging:

$4.9 m/s^{2} t^{2}-1.1 m/s t-1.1 m=0$ (4)

So, since we have a quadratic equation here (in the form of $0=at^{2}+bt+c$ , we will use the quadratic formula to find $t$ :

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ (5)

Where $a=4.9$ , $b=-1.1$ , $c=-1.1$

Substituting the known values and choosing the positive result of the equation, we have:

$t=\frac{-(-1.1)\pm\sqrt{(-1.1)^{2}-4(4.9)(-1.1)}}{2(4.9)}$

$t=0.59 s$ (6)

Now, substituting (6) in (2):

$x=(2.2 m/s)(cos 30\°)(0.59 s)$ (7)

$x=1.12 m$ (8) This is the horizontal distance at which the ball hits the ground.

3 0

4 years ago

A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant a

RideAnS [48]

Answer:

-414.96 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.85^2}{2\times 0.75}\\\Rightarrow a=-9.88\ m/s^2$

$F=ma\\\Rightarrow F=42\times -9.88\\\Rightarrow F=-414.96\ N$

The force the ground exerts on the parachutist is -414.96 N

If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase

5 0

4 years ago

What must be applied to move a load?

Misha Larkins [42]

An applied force must be applied to move a load. This applied force must be large enough to overcome any opposing forces in order to move the load.

3 0

4 years ago

Read 2 more answers

A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the

marysya [2.9K]

Answer:

Explanation:

1 ) the volume of helium in the balloon when fully inflated

= 4/3 π R³

= 4/3 x 3.14 x 5.3³

= 623.3 m³

2 ) force of gravity on the entire system ,but with no people

= (mass of the empty balloon and basket + mass of helium )x g

=( 129 + 623.3 x.18 )x 9.8

= 2363.7 N

3 ) magnitude of the buoyant force on the entire system (but with no people )

( Volume of empty balloon + basket + volume of helium ) density of air x g

( 0.057 + 623.3 ) x 1.23 x 9.8

= 7513.94 N

4 ) magnitude of the force of gravity on each person

Mass of each person x g

= 71 x 9.8

= 695.8 N

6 0

3 years ago

Pelden pushes a box of mass 30 kg with a force of 60 N. What is the

Firlakuza [10]

Answer:

Mass = 30kg

Force = 60N

Acceleration = ?

F = m/a

a = m/f

a = 30kg/ 60N

0.5ms^2 Answer

7 0

3 years ago