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3 years ago

Which planet spins on it's side Uranus or neptune?

Physics

2 answers:

BartSMP [9]3 years ago

8 0

The answer is Uranus, as it has a tilted axis.

maxonik [38]3 years ago

6 0

I believe its Uranus because if you see some pictures of it theres something spinning around it.
Hope this helps! :)

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A playground is on the flat roof of a city school, 6.2m above the street below. The vertical wall of the building is h=7.30m hig

olya-2409 [2.1K]

Answer:

a. 18.13m/s

b. 0.84m

c. 2.4m

Explanation:

a. to find the speed at which the ball was lunched, we use the horizontal component.Since the point distance from the base of the ball is 24m and it takes 2.20 secs to reach the wall,we can say that

t=distance /speed

$speed v_{x}=vcos\alpha=24/2.2=10.9m/s\\V=10.9/cos53^{0}\\V=18.13m/s$

Hence the speed at which ball was lunched is 18.13m/s

b. from the equation

$y=v_{y}t-\frac{1}{2}gt^{2}\\ v_{y}=18.13sin53=14.48m/s\\\\y=(14.48*2.2)-\frac{1}{2}9.8*2.2^{2}\\y=8.14m\\$

the vertical distance at which the ball clears the wall is

y=8.14-7.3=0.84m

c. the time it takes the ball to reach the 6.2m vertically

$6.2=14.47t-4.8t^{2}\\\\14.47t-4.8t^{2}-6.2=0\\\using\\t=-b±\frac{\sqrt{b^{2}-4ac} }{2a}\\ where \\a=-4.8, b=14.47t, c=6.2\\hence t=2.4secs$

the horizontal distance covered at this speed is

$y+24=(18.13cos53)2.4\\y=26.186-24\\y=2.12m$

4 0

3 years ago

What piece of sports equipment is aerodynamic

Mashcka [7]

Aerobie. Frisbee. Discus. Javelin. I suppose an American football to some extent.
Pull! Clay pigeons. Arrows. Wingsuit. Kites. Hang gliders. Sails. sailboat keels/dagger boards. Water skis. Ski jumping skis. Boomerang. 
I'm excluding spheres and parachutes as bluff bodies even though aerodynamics often plays a big part in their motion.

6 0

3 years ago

Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t

n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

$y=\frac{\lambda (m+\frac{1}{2})D}{d}$

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light used

In this problem we have:

$\lambda=683 nm = 683\cdot 10^{-9} m$ is the wavelength of the light

$d=1.1 mm = 0.0011 m$ is the width of the slit

m = 13 is the order of the minimum

$y=8.57 cm = 0.0857 m$ is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

$D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m$

6 0

3 years ago

Propose a hypothesis for how the position of the ball will affect the amount of its gravitational pull energy

Ray Of Light [21]

throwing a ball up initially has a lot of kinetic energy because it is moving upwards ( kinetic energy is energy which a body possesses by virtue of being in motion.) this all then get converted to gravitational potential energy, and for a moment it is stationary before it begins to fall again. by the time it has returned again, all the gravitational potential energy has turned back into kinetic.

4 0

3 years ago

A 0.50 kg object is attached to a spring with force constant 157 N/m so that the object is allowed to move on a horizontal frict

Genrish500 [490]

Explanation:

We have,

Mass of an object is 0.5 kg

Force constant of the spring is 157 N/m

The object is released from rest when the spring is compressed 0.19 m.

(A) The force acting on the object is given by :

F = kx

$F=157\times 0.19\\\\F=29.83\ N$

(B) The force is simply given by :

F = ma

a is acceleration at that instant

$a=\dfrac{F}{m}\\\\a=\dfrac{29.83}{0.5}\\\\a=59.66\ m/s^2$

6 0

3 years ago