The period-luminosity relation is critical in finding distances with

Using software, a simulation of a javelin being thrown both on Earth and on the moon is created. In both cases, the velocity and

mario62 [17]

Horizontal distance covered by a projectile is X = Vix *T

where Vix is the initial horizontal component of velocity and T is time taken by the projectile

Vix = ViCos theta

In question they said that initial velocity and angle is same on earth and moon

so Vix would remains same

now let's see about time taken T

time taken to reach the highest point

Vfy = Viy +gt

at highest point vertical velocity become zero so Vfy =0

0 = Vi Sin theta + gt

t = Vi Sintheta /g

Total time taken to land will be twice of that

On earth

Te= 2t

Te = 2Sinθ/g

on moon g is one-sixth of g(earth)

Tm = 2Sinθ/(g/6)

Tm = 6(2Sinθ/g)

Tm = 6Te

so total time taken by the projectile on moon will be six times the time taken on earth

From first equation X = Vix*T

we can see that X will also be 6 times on moon than earth

so projectile will cover 6 times distance on moon than on earth

4 0

2 years ago

Read 2 more answers

When white light passes through a prism, a red light is least changed in direction? True or false

Goryan [66]

True because the picture below proves this....

* from which red color is least deviated and violet most.

* Hopefully this helps:) Mark me the brainliest:) !!

∞ 234483279c20∞

4 0

3 years ago

A small block with mass 0.0400 kg is moving in the xy-plane. The net force on the block is described by the potentialenergy func

OlgaM077 [116]

Answer:

A= 148.92 m/s²

Explanation:

Given that

U(x,y) = (6.00 )x² - (3.75 )y ³

m= 0.04 kg

Now force in the x-direction

Fx= - dU/dx

U(x,y) = (6.00 )x² - (3.75 )y ³

dU/dx= 12 x

When x=0.4 m

dU/dx= 12 x 0.4 = 4.8

So we can say that

Fx= - 4.8 N

From Newtons law

F= m a

- 4.8 = 0.04 x a

a = -120 m/s²

Acceleration in x direction ,a = -120 m/s²

In y -direction

F= - dU/dy

U(x,y) = (6.00 )x² - (3.75 )y ³

dU/dy = 0 - 3.75 x 3 y²

When y = 0.56 m

dU/dy = - 3.75 x 3 x 0.56 x 0.56

dU/dy = - 3.52

So we can say that force in y -direction

F= 3.52 N

F= m a'

3.52 = 0.04 x a'

a'=88.2 m/s²

acceleration in y direction is 88.2 m/s²

The resultant acceleration

$A=\sqrt{a^2+a'^2}$

$A=\sqrt{120^2+88.2^2}$

A= 148.92 m/s²

7 0

3 years ago

An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 34.5

Inessa05 [86]

Answer:

the acceleration of the airplane is 5.06 x 10⁻³ m/s²

Explanation:

Given;

initial velocity of the airplane. u = 34.5 m/s

distance traveled by the airplane, s = 46,100 m

final velocity of the airplane, v = 40.7 m/s

The acceleration of the airplane is calculated from the following kinematic equation;

v² = u² + 2as

$2as= v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(40.7)^2 -(34.5)^2}{2 \times 46,100} \\\\a = 5.06 \ \times \ 10^{-3} \ m/s^2$

Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²

5 0

2 years ago

GETTING TIMED PLS HELP! Do the runners in the picture above represent kinetic or potential energy? you need to explain why.

prohojiy [21]

Answer:

They represent kinetic energy

Explanation:

Kinetic Energy

A body can do work due to some of its attributes or states. For example, its mass can do work if used to provide energy, if the object is at a certain height respect to some reference level, it can do work when going downwards (potential energy), if the object moves at a certain speed, it can do work when transferring part of its speed to other objects. It's called kinetic energy and is given by

$\displaystyle K=\frac{mv^2}{2}$

Both runners are moving in a horizontal path, thus they have kinetic energy, given by the above equation. If they could jump below ground level, then they will also have potential energy

8 0

2 years ago