All categories

3 years ago

If we rotate on our axis does that count as a day

1 answer:

s2008m [1.1K]3 years ago

5 0

Yes if we rotate on our axis one full time it is considered a day. but if we go around the entire sun one full time it is considered a year. hope this helps

You might be interested in

PLEASE ANSWER THANKS

Veseljchak [2.6K]

The answers are
1.no
2. they will attract each other
3.b
4.neither direction

8 0

3 years ago

Read 2 more answers

A cannon fires a cannonball 500.0 m downrange when set at a 45.0o angle. At what velocity does the cannonball leave the cannon?

kobusy [5.1K]

Answer:

v = 70 m/s

Explanation:

Range on the cannon ball is given as

$d = 500.0 m$

here the angle of the projection of the ball is given as 45 degree

now we know that if the velocity of the ball is "v" then its two components will be given as

$v_x = vcos45$

$v_y = vsin45$

so here time of flight of the motion is given as

$T = \frac{2v_y}{g}$

$T = \frac{2vsin45}{g}$

also the range is given as

$R = v_x T$

$R = (vcos45)(\frac{2vsin45}{g})$

now plug in all data in this equation

$500.0 = \frac{v^2(2sin45cos45)}{g}$

$v = 70 m/s$

4 0

3 years ago

When an object is lifted against the force of gravity it has gravitational potential energy. The energy depends on the object's

USPshnik [31]

The answer is actually True, I just took the test and it was correct.

5 0

4 years ago

Read 2 more answers

A fairgrounds ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follo

cestrela7 [59]

Answer:

13.37 rev/min

Explanation:

acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration ( $a_c$ ) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².

r = 9 m

Centripetal acceleration ( $a_c$ ) is given by:

$a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s$

The velocity (v) is given by:

v = ωr; where ω is the angular velocity

Hence:

ω = v/r = 12.6 / 9

ω = 1.4 rad/s

ω = 2πN

N = ω/2π = 1.4 / 2π

N = 0.2228 rev/s

N = 13.37 rev/min

4 0

3 years ago

A 150-newton force, applied to a wooden crate at an angle of 30° above the horizontal, causesthe crate to travel at constant vel

omeli [17]

Answer:

Explanation:

A component of 150 N in vertical direction will reduce the magnitude of reaction force.

reaction force exerted by the floor

= mg - 150 sin 30

where m is mass of the crate .

the magnitude of the horizontal component of the 150-newton force

150 cos30

= 130 N

This force tries to pull the crate in forward direction with acceleration but it has no acceleration . It is so because frictional force of equal magnitude acts on it in opposite direction which makes the net force acting on it equal to zero.

Hence frictional force is equal to 150 cos 30.

= 130 N .

3 0

3 years ago