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3 years ago

What is a similanty between the first quarter moon and the third quarter moon?

Physics

2 answers:

Nezavi [6.7K]3 years ago

4 0

Answer:

The moon is located perpendicular to Earth and the sun.

Explanation:

You rule out all of the other answers and this is what you get. If you dont believe me check it for yourself.

Hope this helped! Please mark brainliest!

Korvikt [17]3 years ago

3 0

They are both moons

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Help me please!!!!!..............

qaws [65]

Answer:

<em>The car took 3 hours to travel 180 miles</em>

Explanation:

<u>Constant Speed Motion</u>

An object travels at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

$\displaystyle v=\frac{d}{t}$

Where

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we can solve for t:

$\displaystyle t=\frac{d}{v}$

The car travels at v=60 mi/h for a distance of d=180 miles. The time taken is:

$\displaystyle t=\frac{180\ mi}{60\ mi/h}$

t = 3 hours

The car took 3 hours to travel 180 miles

6 0

3 years ago

A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is

pychu [463]

Answer:

The final velocity of the runner at the end of the given time is 2.7 m/s.

Explanation:

Given;

initial velocity of the runner, u = 1.1 m/s

constant acceleration, a = 0.8 m/s²

time of motion, t = 2.0 s

The velocity of the runner at the end of the given time is calculate as;

$v = u + at$

where;

v is the final velocity of the runner at the end of the given time;

v = 1.1 + (0.8)(2)

v = 2.7 m/s

Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.

7 0

3 years ago

A penny is kicked horizontally off the roof of a 10-story building (33.3 m high) and lands 52 m away on the ground. A) what is t

yanalaym [24]

A) The penny was kicked horizontally off the building. By this very statement, the penny had 0 initial vertical velocity.

B) Apply the following kinematics equation to the penny's vertical motion:

D = Vt + 0.5At²

D = vertical distance traveled, t = time, V = initial vertical velocity, A = vertical acceleration

Given values:

D = 33.3m, V = 0m/s, A = 9.81m/s²

Plug in and solve for t:

33.3 = 4.905t²

t = 2.61s

C) The penny fell for 2.61 seconds, therefore it moved horizontally for 2.61 seconds.

v = x/t

v = horizontal velocity, x = horizontal distance traveled, t = time

Given values:

x = 52m, t = 2.61s

Plug in and solve for v:

v = 52/2.61

v = 19.9m/s

D) Let's calculate the penny's vertical speed right before it hits the ground:

v = at

v = final vertical speed, a = vertical acceleration, t = time

Given values:

a = 9.81m/s², t = 2.61s

Plug in and solve for v:

v = 9.81(2.61)

v = 25.6m/s

Use the Pythagorean theorem to find the final speed:

V = √(Vx²+Vy²)

V = final speed, Vx = final horizontal speed, Vy = final vertical speed

Given values:

Vx = 19.9m/s, Vy = 25.6m/s

Plug in and solve for V:

V = √(19.9²+25.6²)

V = 32.4m/s

5 0

3 years ago

Plz help ASAP I'll mark as brainliest

gogolik [260]

Hi there!

Hooke's law states that:

F = -kx

k = Spring constant (N/m)

x = DISPLACEMENT from equilibrium (m)

Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.

The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)

The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.

We can begin by looking at the given graph.

When the spring force = 4N, the total length of the spring is 35 cm.

Now, the EQUILIBRIUM length is 30 cm, so the total elongation is:

35 - 30 = 5 cm.

5.1.

If the spring elongates by 10 cm, the total length of the spring is:

30 + 10 = 40 cm

According to the graph, a length of 40 cm corresponds to a force of 8N.

5.2.

We can solve for the weight of the ball using the following:

W (weight) = m (mass) · acceleration due to gravity (10N/kg)

Using a summation of forces:

∑F = T - W

The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:

0 = T - W

T = W = 8 N

5.3.

0 = T - Mg

T = Mg

Use the prior value of T and gravity to solve:

8 = 10M

m = 0.8 kg

8 0

3 years ago

A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of 2.00 m/s . The coefficient o

kobusy [5.1K]

2.04 meters distance is traveled by the sled before stopping.

Mass of the sled = m

The initial speed of the sled = 2 m/s

Coefficient of kinetic friction between sled and ice = 0.100

Let the distance the sled moves before it stops be d.

Gravity = 9.8 m/ s²

Let the initial kinetic energy sled be

$= K _{i}$

$K_{i} = \frac{1}{2} mv ^{2}$

The work done by the frictional force is,

$Work \: done \: by \: frictional \: force =W_{f}$

$W _{f} = μ_{k}mgd$

Work done by frictional force= Initial kinetic energy of the sled

$W_{f} = K_{i}$

$μ_{k}mgd= \frac{1}{2} mv ^{2}$

So, the distance traveled by the sled before stopping is

$d= \frac{1mv ^{2} }{2 \:μ_{k}mg}$

$d= \frac{1v ^{2} }{2 \:μ_{k}g}$

$d= \frac{2^{2} }{2 \times \:0.100 \times 9.8}$

$d= 2.04 \: m$

Therefore, the distance traveled by the sled before stopping is 2.04 meters.

To know more about work done, refer to the below link:

brainly.com/question/13662169

#SPJ4

5 0

2 years ago