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2 years ago

13

What is the word for a group of stars that’s not a constellation?.

1 answer:

Sever21 [200]2 years ago

7 0

Asterism, a pattern of stars that is not a constellation. An asterism can be part of a constellation, such as the Big Dipper, which is in the constellation Ursa Major, and can even span across constellations, such as the Summer Triangle, which is formed by the three bright stars Deneb, Altair, and Vega.

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A projectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above the horizontal. It strikes

Alchen [17]

<h2>Answer: x=125m, y=48.308m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=54.5m/s$ is the projectile's initial speed

$\theta=35\°$ is the angle

$t=2.80s$ is the time since the projectile is launched until it strikes the target

$x$ is the final horizontal position of the projectile (the value we want to find)

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the projectile (we are told it was launched at ground level)

$y$ is the final height of the projectile (the value we want to find)

$g=9.8m/s^{2}$ is the acceleration due gravity

Having this clear, let's begin with x (1):

$x=(54.5m/s)cos(35\°)(2.8s)$ (3)

$x=125m$ (4) This is the horizontal final position of the projectile

For y (2):

$y=0+(54.5m/s)sin(35\°)(2.8s)-\frac{(9.8m/s^{2})(2.8s)^{2}}{2}$ (5)

$y=48.308m$ (6) This is the vertical final position of the projectile

4 0

3 years ago

As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.33 m l

deff fn [24]

Answer: 62 μT

Explanation:

Given

Length of rod, l = 1.33 m

Velocity of rod, v = 3.19 m/s

Induced emf, e = 0.263*10^-3 V

Using Faraday's law, the induced emf of a rod can be gotten by the formula

e = blv where,

e = induced emf of the rod

b = magnetic field of the rod

l = length of the rod

v = velocity of the rod. On substituting, we have

0.263*10^-3 = b * 1.33 * 3.19

0.263*10^-3 = b * 4.2427

b = 0.263*10^-3 / 4.2427

b = 0.0000620 T

b = 62 μT

Thus, the strength of the magnetic field is 62 μT

8 0

3 years ago

Read 2 more answers

A nature photographer is using a camera that has a lens with a focal length of 3.06 cm. The photographer is taking pictures of a

sleet_krkn [62]

Answer:

film is at distance of 3.07 cm from lens

Explanation:

Given data

focal length = 3.06 cm

distance = 10.4 m = 1040 cm

to find out

How far must the lens

solution

we apply here lens formula that is

1/f = 1/p + 1/q

here f = 3.06 and p = 1040 so we find q

1/f = 1/p + 1/q

1/3.06 = 1/1040 + 1/q

1/ q = 0.3258

q = 3.0690 cm

so film is at distance of 3.07 cm from lens

6 0

3 years ago

Acceleration usually has the symbol a. It is a vector. What is the correct way

Ghella [55]

Explanation:

a straight line under the letter

3 0

3 years ago

Who's really good at physical science?

34kurt

Definitely not me!!!! (unless it's magnetism or circuits)

3 0

3 years ago