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xenn [34]
3 years ago
8

A certain comet has a parallax 1/40 as large as the moon. How far away is it (compared to the Moons distance)? Show Work

Physics
1 answer:
Ad libitum [116K]3 years ago
3 0
The distance and parallax are inversely related. We can find the distance using the following equation:

d= \frac{1}{p}

where d is distance and p is parallax.

We are given the parallax of the comet relative to the moon, and we are looking for the distance to the comet relative to the moon's distance, so wee can plug in the following value:

d= \frac{1}{ \frac{1}{40} }

The distance is 40 times as far away as the moon.
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What is the momentum of a vehicle that has a mass of 800 kg and is moving at a velocity of 5 m/s
yarga [219]

Answer:

4000 kg.m/s

Explanation:

p=mv

m=800 kg

v=5 m/s

p=(800)(5)= 4000 kg.m/s

3 0
3 years ago
5. What is the period of a vertical mass-spring system that has an amplitude of
s2008m [1.1K]

Answer:

T = 0.638 s

Explanation:

vmax = Aω

ω = 7.02/0.713 = 9.8457 rad/s

T = 2π/ω = 2π/9.8457 = 0.63816

8 0
3 years ago
Grace, Erin, and Tony are on a seesaw. Grace has a mass of 45kg and is seated 0.7m to the left of the fulcrum. Nicole has a mass
salantis [7]
Use Scoratic it works with any time of subject
5 0
3 years ago
The formula h = -16t squared + 64t squared gives the height, h, and time, t, of an object launched from the ground with a speed
padilas [110]

Answer:

<em>Details in the explanation</em>

Explanation:

<u>Vertical Launch</u>

When an object is thrown vertically in free air (no friction), it moves upwards at its maximum speed while the acceleration of gravity starts to brake it. At a given time and height, the object stops in mid-air and starts to fall back to the launching point until reaching it with the same speed it was launched.

We are given an expression for the height of an object in function of time t

h = -16t^2 + 64t

<em>Please note we have deleted the second 'squared' from the formula since it's incorrect and won't describe the motion of vertical launch.</em>

We now have to evaluate h for the following times, assuming h comes in feet

At t=1 sec

h = -16(1)^2 + 64(1)=64-16=48\ ft

The object is at a height of 48 feet

At t=2 sec

h = -16(2)^2 + 64(2)=128-64=64\ ft

The object is at a height of 64 feet. This is the maximum height the object will reach, as we'll see below

At t=3 sec

h = -16(3)^2 + 64(3)=192-144=48\ ft

The object is at a height of 48 feet. We can clearly see it's returning from the maximum height and is going down

At t=4 sec

h = -16(4)^2 + 64(4)=256-256=0\ ft

The object is at ground level and has returned to the launch point.

5 0
3 years ago
A truck travels to and from a stone quarry that is located 2.5 km to the east. What is the total distance traveled by the truck?
scoray [572]

Answer:

5 km

0

Explanation:

The question says the truck travelled to and fro from the stone quarry.

When going to the stone quarry, the distance was 2.5 km to the east.

On returning, the distance back is also 2.5km

Therefore total distance travelled is 2.5km + 2.5km

= 5km

The displacement 8s dependent on what is the initial and also the final position. The initial and the final position are one and the same thing. Because the truck starts and goes back to the same position, the displacement is zero.

5 0
3 years ago
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