All categories

3 years ago

How long is a year on Earth and what journey does the Earth make in this space of time?

2 answers:

8 0

A year on Earth is 365 and 1/4 days (rounded). It corresponds to
one revolution of the Earth in our orbital path around the sun.

hichkok12 [17]3 years ago

4 0

A year is 365 days and the earth moves around the sun in an ORBIT

You might be interested in

if a ball with an original velocity of 0 is dropped from a tall structure and takes 7 Seconds to hit the ground what velocity do

krok68 [10]

$a_y=\dfrac{v_y-v_{0y}}t\implies-9.81\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{v_y-0}{7\,\mathrm s}\implies v_y=-68.7\,\dfrac{\mathrm m}{\mathrm s}$

5 0

3 years ago

A 7kg object is dropped on Earth,. Assuming it has not yet hit the ground, what is the velocity of the object after 2 seconds of

kirza4 [7]

The answer would be 9.8 meters/ sec^2.

7 0

3 years ago

Read 2 more answers

3 In a television tube, an electron starting from rest experiences a force of 4.0 × 10−15 N over a distance of 50 cm. The final

MAXImum [283]

Answer:

The final speed of the electron = 2.095×10⁸ m/s

Explanation:

From newton's fundamental equation of dynamics,

F = ma ........................Equation 1

Where F = force, m = mass of the electron, a = acceleration of the electron.

making a the subject of the equation,

a = F/m.................... Equation 2

Given: F = 4.0×10⁻¹⁵ N,

Constant: m = 9.109×10⁻³¹ kg.

Substituting into equation 2

a = 4.0×10⁻¹⁵/9.109×10⁻³¹

a = 4.39×10¹⁶ m/s².

Using newton's equation of motion,

v² = u²+2as .......................... Equation 3

Where v = final velocity of the electron, u = initial velocity of the electron, a = acceleration of the electron, s = distance covered by the electron.

Given: u = 0 m/s(at rest), s = 50 cm = 0.5 m, a = 4.39×10¹⁶ m/s²

Substituting into equation 3

v² = 0² + 2(0.5)(4.39×10¹⁶)

v = √(4.39×10¹⁶)

v = 2.095×10⁸ m/s

Thus the final speed of the electron = 2.095×10⁸ m/s

7 0

3 years ago

!!please help !!!!! I’ll give brainliest

Oxana [17]

The answer is 45 degrees

7 0

3 years ago

Read 2 more answers

On level ground a shell is fired with an initial velocity of 49.0 m/s at 70.0 โ above the horizontal and feels no appreciable ai

xenn [34]

Answer:

Horizontal component = 16.8 m/s

Vertical component = 46.0 m/s

Explanation:

If we denote the initial velocity by <em>v</em> and the angle above the horizontal by <em>θ</em>,

the horizontal component of this initial velocity is given by

$v_x = v\cos \theta$

$v_x = (49.0\text{ m/s})\cos\,(70.0^\circ) = 16.8\text{ m/s}$

The vertical component is given by

$v_y = v\sin\theta$

$v_x = (49.0\text{ m/s})\sin\,(70.0^\circ) = 46.0\text{ m/s}$

3 0

3 years ago