All categories

2 years ago

Which moon phase comes before the phase in this picture?

Physics

2 answers:

hodyreva [135]2 years ago

8 0

Waxing gibbous
Ushjdjsidjnd

mafiozo [28]2 years ago

3 0

Waxing gibbous moon phase

You might be interested in

Letícia leaves the grocery store and walks 150.0 m to the parking lot. Then, she turns 90° to the right and walks an additional

Alexxx [7]

Answer:

165.529454

Explanation:

According to the Pythagorean Theorem for calculating the lengths of a right angle triangle's sides, a^2 + b+2 = c^2, where c is the longest side (and the side opposing the right angle). So in your case it would be 150*150 + 70*70 = 27400. And √ 27400 is your answer.

7 0

2 years ago

12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it

vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

$v_{f} = v_{o}+a\cdot (t-t_{o})$ (1)

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$v_{f}$ - Final velocity, measured in meters per second.

$a$ - Acceleration, measured in meters per square second.

$t_{o}$ - Initial time, measured in seconds.

$t$ - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust ( $v_{o} = 0\,\frac{m}{s}$ , $a = 30\,\frac{m}{s^{2}}$ , $t_{o} = 0\,s$ , $0\,s\le t< 30\,s$ )

$v = 30\cdot t$ (2)

Free fall ( $v_{o} = 900\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s}$ , $t_{o} = 30\,s$ , $30\,s \le t \le 120\,s$ )

$v = 900-9.81\cdot (t-30)$ (3)

Now we created the graph speed-time, which can be seen below.

5 0

3 years ago

A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

$\sum F_x = 0$

$f-N_w = 0$

$N_w = f$

The forces in the horizontal direction would be,

$\sum F_y = 0$

$N_f -W =0$

$N_f = W$

The sum of Torques at equilibrium,

$\sum \tau = 0$

$Wdcos\theta - N_wLsin\theta = 0$

$WdCos\theta = fLSin\theta$

$f = \frac{Wd}{Ltan\theta}$

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

$f_{max} = \mu W=\frac{Wd}{Ltan\theta}$