All categories

2 years ago

About how many days in the moon’s cycle is it waxing?.

Physics

2 answers:

Monica [59]2 years ago

8 0

Answer: 29.5 days

The waxing moon phase comprises three separate moon phases: the Waxing Crescent, First Quarter, and the Waxing Gibbous.

mylen [45]2 years ago

6 0

Answer:

its called a lunar cycle, and it lasts about 29 1/2 days.

You might be interested in

A wildebeest runs with an average speed of 4.0\,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction, start text, m, en

devlian [24]

Answer:

60m

Explanation:

4 0

3 years ago

A satellite with a mass of 2,000 ㎏ is inserted into an orbit that is twice the Earth' s radius. W hat is the force of gravity on

dolphi86 [110]

Answer:

option (b) 4900 N

Explanation:

m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R

F = G Me x m / (R + h)^2

F = G Me x m / 2R^2

F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2

F = 4900 N

4 0

3 years ago

A 200 N trash can is pulled across the sidewalk by a person at constant speed by a force of 75 N. What is the coefficient of fri

Pavel [41]

Answer:

μ = 0.375

Explanation:

F = Applied force on the trash can = 75 N

W = weight of the trash can = 200 N

f = frictional force acting on trash can

Since the trash can moves at constant speed, force equation for the motion of can is given as

F - f = 0

75 - f = 0

f = 75 N

μ = Coefficient of friction

frictional force is given as

f = μ W

75 = μ (200)

μ = 0.375

5 0

3 years ago

A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic

Ilia_Sergeevich [38]

Answer:

$\mu_k=0.27$

Explanation:

According to the free body diagram, in this case, we have:

$\sum F_x:-F_k=ma\\\sum F_y:N=mg$

Recall that the force of friction is given by:

$F_k=\mu_k N$

Replacing and solving for the coefficient of kinetic friction:

$-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}$

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

$a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}$

Finally, we calculate $\mu_k$ :

$\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27$

4 0

3 years ago

Find the potential energy associated with a 61-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level. Take the

Natali [406]

Answer:

The potential energy of the hiker is $1.13\times 10^6\ J$ .

Explanation:

Given that,

Mass of the hiker, m = 61 kg

Height above sea level, h = 1900 m

We need to find the potential energy associated with a 61-kg hiker atop New Hampshire's Mount Washington. The potential energy is given by :

$E=mgh$

g is the acceleration due to gravity

$E=61\ kg\times 9.8\ m/s^2\times 1900\ m\\\\E=1.13\times 10^6\ J$

So, the potential energy of the hiker is $1.13\times 10^6\ J$ . Hence, this is the required solution.

8 0

3 years ago