Which one is not a derived unit? 1. Hertz 2. mol 3. Watt 4. Newton

A new planet is discovered beyond Pluto at a mean distance to the sun of 4004 million miles. Using Kepler's third law, determine

AVprozaik [17]

Answer:

103239.89 days

Explanation:

Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

a³ / T² = 7.496 × 10⁻⁶ (a.u.³/days²)

where,

a is the distance of the semi-major axis in a.u

T is the orbit time in days

Converting the mean distance of the new planet to astronomical unit (a.u.)

1 a.u = 9.296 × 10⁷ miles

$\frac{4004 * 10^{6}}{9.296 * 10^{7}} = 43.07\ a.u.$

Substituting the values into Kepler's third law equation;

$\frac{(43.07)^{3}}{T^{2}} = 7.496 * 10^{-6}$

$T^{2} = \frac{(43.07)^{3}}{7.496 * 10^{-6}}$ (days)²

$T^{2} = \sqrt{\frac{(43.07)^{3}}{7.496 * 10^{-6}}}$

T = 103239.89 days

An estimate time T for the new planet to travel around the sun in an orbit is 103239.89 days

7 0

3 years ago

A block of mass 0.08 kg is pushed against a spring with spring constant k=31 N/m. The spring is compressed 0.15 meters from its

ELEN [110]

Answer:

1.11 meters

Explanation:

As the spring is compressed, elastic potential energy is built up in the spring. The total elastic potential energy can be found using the following formula

Ep = 1/2 x k x s²

where k = 31 N/m (spring constant)

s = 0.15 m (compression)

Ep = 3.4875 J

When the block of mass is released, the elastic potential energy (Ep) is converted to kinetic energy (Ek). From this we can find the initial velocity of the mass of block after release

Ek = 1/2 x m x u²

where Ek = Ep = 3.4875J

m = 0.08 kg (mass of block)

u = unknown (initial velocity)

u = 2.9526 m/s

Now that we know the initial velocity we need to find the deceleration of the mass of block due to friction. We will first find the force of friction from the following formula

F = ∪ x m x g

where F = unknown (frictional force)

∪ = 0.4 (coefficient of friction)

m = 0.08 kg (mass of block)

g = 9.81 m/s² (acceleration due to gravity)

F = 0.31392 N

From this force we calculate the deceleration based on the following formula

F = m x a

where F = 0.31392 (frictional force)

m = 0.08 kg (mass of block)

a = unknown (acceleration)

a = -3.924 m/s² -

*the negative sign is due to this value being deceleration

Now to find the total distance traveled we use the equation for motion

v² = u² + 2as

where v = 0 (final velocity)

u = 2.9526 m/s (initial velocity

a = -3.924 m/s² (deceleration due to friction)

s = unknown (distance traveled)

s = 1.11 meters

3 0

3 years ago

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 86.0 m/s2

Harman [31]

When the fuel of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be going up until all the forces of gravity would dominate and change the direction of the rocket. We need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction.

Given:
a = 86 m/s^2
t = 1.7 s

Solution:

d = vi (t) + 0.5 (a) (t^2)
d = (0) (1.7) + 0.5 (86) (1.7)^2
d = 124.27 m

vf = vi + at
vf = 0 + 86 (1.7)
vf = 146.2 m/s (velocity when the fuel is consumed completely)

Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at
0 = 146.2 + (-9.8) (t)
t = 14.92 s

Then, the second distance
d= vi (t) + 0.5 (a) (t^2)
d = 146.2 (14.92) + 0.5 (-9.8) (14.92^2)
d = 1090.53 m

Then, we determine the maximum altitude:
d1 + d2 = 124.27 m + 1090.53 m = 1214.8 m

5 0

3 years ago

A 120.0 kg crate is placed on a 15.00°

Citrus2011 [14]

F = 2820.1 N

Explanation:

Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as

Fnet = ma = 0 (a = 0 no sliding)

= F - mgsin15°

= 0

or

F = mgsin15°

= (120 kg)(9.8 m/s^2)sin15°

= 2820.1 N

7 0

2 years ago

a boy is riding a bicycle at a velocity of 5.0 m/s. he applies the brakes and uniformly decelerates to a stop at a rate of 2.5 m

JulijaS [17]

The working equation would be Vf (final velocity) = Vi (initial velocity) + a (acceleration) t (time). The given data are the initial velocity (5.0 m/s), acceleration (-2.5 m/s^2, negative since it is said to decelerate) and the final velocity (0 m/s, since it will put to a stop). The time would be 2 seconds.

3 0

3 years ago

Which one is not a derived unit?1. Hertz2. mol3. Watt4. Newton​

Which one is not a derived unit?1. Hertz2. mol3. Watt4. Newton