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3 years ago

Using the image provided, which moon phase is next?

Physics

2 answers:

ella [17]3 years ago

5 0

Answer:

Waning Crescent

Sergio [31]3 years ago

3 0

Answer: waxing crescent

Explanation:

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Vesna [10]

Complete Question

A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward.

From this motion what is the distance covered

What is the magnitude and direction of the displacement

Answer:

$D = 72 \ m$

Magnitude

$d = - 20 \ m$

Direction

West

Explanation:

From the question we are told that

The first distance covered westward is $d_w_1 = 18 \ m /tex] The first distance covered eastward is [tex]d_e1 = 12 \ m /tex] The second distance covered westward is [tex]d_w_2 = 28 \ m /tex] The second distance covered eastward is [tex]d_e2 = 14 \ m /tex] Generally the distance covered is mathematically represented as [tex]D = d_w1 + d_w2 + d_e1 + d_e2$

=> $D = 18 + 28 + 12 + 14$

=> $D = 72 \ m$

For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative

The magnitude of the displacement is

$d = -d_w1 -d_w2 + d_e1 + d_e2$

=> $d = -18-28 + 12 + 14$

=> $d = - 20 \ m$

The direction is west

7 0

3 years ago

Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the ________.

Black_prince [1.1K]

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Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the c</span>ollision model.

-Hope this helps.

7 0

3 years ago

A 3858 N piano is to be pushed up a(n) 3.49 m frictionless plank that makes an angle of 31.6 ◦ with the horizontal. Calculate th

uranmaximum [27]

The work done will be equal to the potential energy of the piano at the final position

P.E=m.g.h

.consider the plank the hypotenuse of the right triangle formed with the ground
.let x be the angle with the ground=31.6°
.h be the side opposite to the angle x (h is the final height of the piano)
.let L be the length of the plank

sinx=opposite side / hypotenuse
= h/L

then h=L.sinx=3.49×sin31.6°=0.638m

weight w=m.g
m=w/g=3858/10=385.8kg

Consider Gravity g=10m/s2

then P.E.=m.g.h=385.8kg×10×0.638=2461.404J

then Work W=P.E.=2451.404J

8 0

3 years ago

Read 2 more answers

Would a bond between potassium and iodine most likely be covalent or ionic? Explain your answer

Ugo [173]

A covalent bond is between two nonmetals. An ionic bond is between a metal and a nonmetal. Potassium is a metal and iodine is a nonmetal, so their bond would most likely be ionic.

4 0

3 years ago

a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^

posledela

Answer:

Approximately $0.8\; \rm s$ (assuming that air resistance is negligible.)

Explanation:

Let $v_0$ denote the initial velocity of this ball. Let $\theta$ denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

Initial vertical velocity: $v_0(\text{vertical}) = v_0 \cdot \sin(\theta)$ .
Initial horizontal velocity: $v_0(\text{vertical}) = v_0 \cdot \cos(\theta)$ .

If air resistance on this ball is negligible, $v_0(\text{vertical})$ alone would be sufficient for finding the time of flight of this ball.

Calculate $v_0(\text{vertical})$ given that $v_0 = 8 \; \rm m \cdot s^{-1}$ and $\theta = 30^\circ$ :

$\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}$ .

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since $v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}$ , the vertical velocity of this ball right before landing would be $v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}$ .

Calculate the change to the vertical velocity of this ball:

$\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}$ .

In other words, the vertical velocity of this ball should have change by $8\; \rm m \cdot s^{-1}$ during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is $g = 10\; \rm m \cdot s^{-2}$ . In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by $10\; \rm m\cdot s^{-1}$ each second. What fraction of a second would it take to change the vertical velocity of this ball by $8\; \rm m \cdot s^{-1}$ ?

$\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}$ .

In other words, it would take $0.8\; \rm s$ to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be $0.8\; \rm s\!$ .

5 0

3 years ago